Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: kagliostro on April 11, 2014, 11:26:07 am
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As I've told in one other tread I'm trying to develop, with a friend, the plan for a BIG PSE Amp
he has listen to a 40W SE amp and want one
During the planning (as cheap output tubes) I proposed him to use a pair of GU50 tubes
he assumed on the web some info about those tubes and was enthusiast about it
so he ordered some GU50 and some of the special sockets in Russia (and received it the past week)
but the appetite comes with eating
so he decided that he want to have a pair of the russian tubes and at the same time a pair of
KT88 (to be swapped on the octal sockets with 6L6 and EL34) on the amp, staying at the stated 40W max of output
We have think at various solutions and a blend of the tubes (something like in the Egnater Rebel) seems to meet his liking
The trouble is how to arrange the OT
a pair of dedicated OT (one for each type of tube)
a single OT with a single primary with an intake for the tube that requires a less primary impedance
a single OT with 2 primary windings, one for each type of tube
(this last, gives the way to supply each tube with a different level of B+ voltage)
I've a faint remembrance of a discussion on this matter here on the forum (as far as I can remember)
Please can someone give help on solving this problem
Here attached a pair of OT options and an extract of the EL152 Telefunken tube which is the civil version of the german military LS50 (the GU50 is the russian version of the LS50 - russian got plans at the end of II WW)
Datasheet of KT88 - 6L6 - EL34 are easily available, so I think is not necessary to post it
Many Thanks
K
p.s.: The McIntosh old amp used a double primary, but those were PP amps and the second primary was connected to the cathodes
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Could you change to a different ohm speaker when changing to the other output tubes, to reflect the different load back to the primary? Those GU50 seem to have a lot of fans...
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We are trying to use such a configuration as to avoid the change of impedance connection of the speaker
K
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I think the inherent design problem is the mismatch of the plate impedances of the different types of power tubes. E.g., 6V6's and El84's can be mixed together because they have the same plate impedance. If tubes with different plate impedances are mixed, current draw will be fouled-up. This would be like the bad practice of mixing speakers with different impedances. Could be even worse if a tube failed.
I suspect that two mismatched primaries operating simultaneously will duplicate the issue: current draw from the secondary side will force incorrect current draw in the mismatched tubes on the primary side. I believe the McIntosh amp used 4 of the same power tubes.
Two separate power amps will definitely work.
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As I've told in one other tread I'm trying to develop, with a friend, the plan for a BIG PSE Amp. he has listen to a 40W SE amp and want one
What will your supply voltage be, and what speaker impedance? There is a straightforward process to designing an output stage if you have decided on those. The process may even indicate a poor choice of supply voltage, but we'll figure that out when we get there.
so he decided that he want to have a pair of the russian tubes and at the same time a pair of KT88 (to be swapped on the octal sockets with 6L6 and EL34) on the amp, staying at the stated 40W max of output
The GU-50 data sheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/018/g/GU50.pdf) (I can't recommend Frank's (http://www.mif.pg.gda.pl/homepages/frank/) enough for finding data sheets; use the Search function).
The GU-50 data sheet says it's a 40w tube. You can't get 40w from a pair of them without going to push-pull. Being SE Class A, it is probably realistic to hope for a power output per tube of 30% of plate dissipation, so 24w total from a pair. Same for the KT88's. It seems like you'd need both pairs fully driven to get 40w of output in parallel single-ended.
Note that various data sheets for the KT88 only show 50-66w for push-pull ultralinear operation with cathode bias. Again, this seems to point to 40w from a pair of single-ended KT88's being unobtainable.
I know the theoretical maximum efficiency of class A is 50%, but that requires perfectly linear output tubes which can pull their plate to 0v, and such tubes don't exist.
The trouble is how to arrange the OT
a pair of dedicated OT (one for each type of tube)
a single OT with a single primary with an intake for the tube that requires a less primary impedance
a single OT with 2 primary windings, one for each type of tube
(this last, gives the way to supply each tube with a different level of B+ voltage)
Dedicated output transformers gives the most flexibility.
Any shared output transformer might as well have a single primary as far as core saturation is concerned, because all tubes' current will be taking up the core's flux capability.
You'd really like to have a single supply voltage for both tube types to simplify your amp. You'd really like the plate voltage to not be extreme, so you can have a G2 voltage equal to plate voltage.
You'd really like to have a Blend Control arrangement that is heavily center-weighted so that for most settings you have both pairs contributing to power output. That makes it easier to hit your goal of 40w output, while still offering the option of using exclusively either pair if there is some perceived tonal benefit.
Look at Figure 9 of this page on pots (http://sound.westhost.com/pots.htm) for a scheme for a center-weighted balance pot; this gives you a way to blend each pair of output tubes, while mostly having both contribute to output power. NOTE: the specific values for the components may be dictated by the maximum grid resistance listed on the tube data sheet.
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Could you change to a different ohm speaker when changing to the other output tubes, to reflect the different load back to the primary?
I think the inherent design problem is the mismatch of the plate impedances of the different types of power tubes. E.g., 6V6's and El84's can be mixed together because they have the same plate impedance.
Respectfully, this is a wrong way to perceive the required primary impedance for the OT.The output transformer doesn't match the tube's plate impedance (which for all power pentodes and beam power tubes is much higher than the OT primary impedance). The OT reflects the speaker impedance back to the primary to provide the desired primary impedance.
The reason 6V6's and EL84's use the same primary impedance in a given amp circuit is because they have the same plate dissipation ratings, similar maximum voltage ratings and can deliver similar plate current. As a result, they can control the same amount of power in a circuit. That power is really dictated by plate current and primary impedance (and the resulting plate voltage swing). As long as the 2 tube types can deliver the same plate current with the set plate and screen voltages, the same one primary impedance can be used for both tube types.
The primary impedance needed is determined by supply voltage and desired output power (which is why I asked for those earlier). You use the equation for power to determine the needed primary impedance as √(Voltage/Power), and voltage has to be the supply voltage minus whatever minimum plate voltage has to remain across the tube.
Ultimately, this will be an iterative process attacking the problem from a couple directions, as there are but so many primary impedances stocked by transformer manufacturers in SE transformers that can handle the currents we'll be dealing with. And only so many power supply voltages reasonable with existing transformers, which is again why I ask what the proposed supply voltage will be. It gives a starting point to start figuring everything else.
EDIT: The easiest way to understand the primary impedance needed is to pretend it is a resistor and the audio power output is the heat dissipation of the resistor under conditions of the tube's maximum a.c. output in the circuit.
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Hotblue: As long as the 2 tube types can deliver the same plate current with the set plate and screen voltages, the same one primary impedance can be used for both tube types.
I'm not sure this addresses my issue. I'm not talking about the plate impedances matching the OT primary, but rather (my perceived) need for the plate impedances to match one another. More specifically the issue is: can the GU50 tubes be used simultaneously with the 6L6/KT88/EL34 tubes?
The answer seems to be Yes, if and only if there is a specific plate voltage & a specific screen voltage supplied to all such tubes, which voltages produce the same plate impedances in all the disparate tube types being used simultaneously. The tube charts will tell; but I'm skeptical.
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I'm not sure this addresses my issue. I'm not talking about the plate impedances matching the OT primary, but rather (my perceived) need for the plate impedances to match one another.
Keep in mind the tube plate current flows only in one direction (out the plate, if you're thinking in terms of electrons). So it is best to think of the tube in British terms: it is merely a valve to control plate current drawn though the primary.
You're right that as long as the supply voltage isn't killing the little tube, you can put any set of tubes together on the same OT. It is only inadvisable if the OT primary impedance is so low that when signal is applied to the output tube grids, that one of the tubes tries to draw too much current. Otherwise, you could put a KT88 and a 6AU6 paralleled on the same OT... the 6AU6 simply won't contribute any meaningful output power.
The answer seems to be Yes, if and only if there is a specific plate voltage & a specific screen voltage supplied to all such tubes, which voltages produce the same plate impedances in all the disparate tube types being used simultaneously.
The only impedance that matter is the selected primary impedance. Pretend it is a resistor. With an applied voltage, the tube will flow plate current. The only thing that matters is the tube used can flow enough current into that load impedance to result in the power output you want.
EDIT: The thing to realize is the plate load impedance is selected in the design process; later when we worry about having the correct speaker impedance to match the OT tap, it's because we want to reflect the same impedance that was chosen in the design stage. Not that the tube necessarily has a particular impedance that we're trying to match.
I know this is standing our usual concepts on their head, as we start out with amps talking about matching OT impedances. But that's because the information given is targeting a user, not a designer. It's also why all this seems harder than it is.
I think Kagliostro's big problem will be the screen voltage limit of the GU-50.
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> mismatch of the plate impedances of the different types of power tubes
We don't "match" loads to pentodes.
Pentode is really a two-resistance device. It comes up from zero at about 200 Ohms, then knees-over to a much higher resistance like 50K.
Split-the-difference is "optimum" but the optimum is very broad. The real goal is to stay well away from either 200 or 50K.
> the different types
The plate circuit is not a horrible problem. What will be a big deal is the grid, getting proper *balance* of bias and drive for these very different tubes. This goes *roughly* as the Mu between G1 and G2. For 6L6 and KT88 it is about 10. GU50 is more like 20.
Forgetting the real problem of using such different tubes at the *same* time.... the OT is not a problem. Find a single combination of plate Voltage and Current that works for *all* tubes you will use.
I built an amp that could run 7027, 6550, or 6L6(GC). Each tube was run at 400V 50mA. The 400V and 20 Watts is well within the ratings of any of these. (*Well* within ratings because reliability was more important than sheer power.) 6L6(metal) was technically a few % over ratings but one did work for an hour.
> GU50 KT88 6L6 EL34
6L6GC is the lowest (claimed) voltage, 500V.
6L6GC and EL34 are both safe at 30 Watts actual dissipation.
So you may run any of these tubes very close to 500V and 60mA. 8.3K load. Expect, at best, 40% of Pdiss as Power Out, about 12 Watts per tube.
Yes, you could run the KT88 or GU50 near 575V 70mA in the same 8.3K load and get near 16 Watts per tube. Is the added 1.25dB more output worth the trouble of a dual-voltage supply? (Remember that you already need multiple G1 biases and G2 supply voltages.)
Yes, you can run 6L6GC at 575V, but drop the current to 52mA to stay within a 30W plate dissipation. Now you need an 11K load per tube. Simple 40W SE OTs are hard/impossible to buy and not something most custom winders are familiar with. Trying to do multiple impedances wastes valuable winding space and leaves more room for mistakes.
Your 40 Watt goal needs at least 3 tubes working.
I recently consulted on a "50W" SE amp which would not make close to 50 Watts without severe spiking in the waveform. They can be VERY fussy things. I would NOT add the complications of multiple different tube types.
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I did a little playing around with loadlines for the GU-50. I plotted a 40w dissipation curve (in red) on the plate curves of the GU-50 with 250v on G2 (the rated maximum voltage).
My first attempt (in blue) was to idle at max dissipation (400v plate, 100mA idle for 40w), draw a line with the left endpoint at 2x the idle current and 0v on the grid, and the right endpoint at about double the bias voltage. This gives a loadline for class A, but the resulting impedance is 3443Ω. The curves are for a single tube, so when you double the tubes you would halve the load impedance, which is 1722Ω.
Power output of the tube with this load is [(Voltagemax - Voltagemin)*(Currentmax - Currentmin)]/8. The resulting power is 13.6w per tube, or 27.2w for the pair (34% efficiency, so pretty close to the estimate I gave before).
I then looked at Hammond SE OT's (http://www.hammondmfg.com/1627.htm), but there is nothing close to ~1.8kΩ that they offer. I then noted that the 1640SEA is rated 30w, 1250Ω and 200mA maximum current. That is a bit of a problem, but I decided to make a 2nd try with the 1250Ω OT.
I drew a 2nd loadline in green for a single tube, so the slope corresponds to 2500Ω (I messed up the gradient a bit while working, but no matter). In order to stay below the dissipation curve with this line, I had to shift it left to a lower plate voltage. I settled on 300v as looking like a viable point, with -19v bias and 132mA plate current at idle.
The new loadline's endpoints give a power output of [(541v-59v)*(231mA-28mA)]/8 = 12.2w per tube or 24.4w per pair into 1.25kΩ. The problem is the idle current of the pair is 264mA, which well exceeds the rated d.c. limit of the transformer. At this point, you could contact Hammond and ask how/if the idle current draw will impact transformer performance.
Going the other way, I tried the 1627SEA (30w, 2.5kΩ, 160mA d.c.) with promising results. Plotting a 5kΩ line on the single-tube curves, at idling at 500v 80mA per tube, the idle bias looks like -27v to me. The peak plate current is 172mA and 25v plate, while the endpoint at -54v looks like it will be pretty close to plate current cutoff which should happen at 5000Ω * 0.172A = 860v. [(860v-25v)*(172mA)]/8 = ~18w per tube or 36w per pair (45% efficiency because the GU-50 can pull its plate so low).
At this point, K could try to give 527v plate and 250v screen, or simply go for 500v plate 250v screen and accept a bit less output power from these tubes. -27v / 0.164A = ~165Ω for a shared cathode bias resistor, or round up to the nearest standard value (there is a little extra current for screen draw added).
He could also use a 2nd 1627SEA for the KT88's, and the same 500v plate 250v G2. Power output and idle bias will probably be different. I'll need to plot another loadline to estimate that.
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Looking at plate curves for the KT88, the 5kΩ load per tube chokes the KT88's capability, even at 400v plate 200v G2 (which would be -16v bias for 80mA idle). The KT88 can pull it's plate to ~20v with this condition, and will make similar power as the GU-50.
But now you need 2 different power supplies. And as PRR mentioned, the drive voltages for full power will be different.
Maybe talk him into the GU-50 amp by itself. Comes close-enough to 40w that he won't know the difference. And a single tube type will simplify matters a lot for you.
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Sometimes I wonder if these projects are intended for actual usage or merely some design exercises of the "just because I can" variety. As HPB have kindly demostrated, even getting a pair of GU-50 to work (at below the require power output) is already a chore, so one has to question why PSE, why 40W from PSE and why two different types of tubes? Is it "just because I can"? :think1:
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Many Many Thanks to all, I higly appreciate your effort
First I want to say that I found news from multiple sources, that the GU50 are used with 300v at G2
the russian datasheet seems (according to what you read on the net) conservative with a level of 250v G2
the reason is the use was in military apparatus where the maximum reliability must be
Some people refers that they use the GU50 tube with the EL152 specs and there is no redplate
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We referred to the EL152 datasheet where a 18W in SE is reached with
300v on plate
250v on G2
130mA quiescent anode current
3.5mA polarization grid current
-24v cathode polarization voltage
2k load
17v required AC for full modulation
20mA grid current at full modulation
18W output (SE one tube)
10% distortion
46% anode efficiency
Unfortunately I've that page only in german and italian, no english
(http://i.imgur.com/fO3JQKk.jpg)
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The TA is a recovered Industrial Transformer (see attached image)
(http://i.imgur.com/31MBm5k.jpg)
with the following specs
430VA
Primary
0-230v-400v
Secondary
0-12v @ 4.16A (50VA)
0-18v-24v @ 2.8A
0-55v-75v @ 2A
0-95v @ 0.736A (70VA)
0-115v @ 0.695A (80VA)
This is a possible configuration suggested by PRR
(http://el34world.com/Forum/index.php?action=dlattach;topic=16497.0;attach=40770)
and this is one other possible configuration by me
(http://i.imgur.com/YytBfXN.jpg)
this a possible variant
(http://i.imgur.com/76zqR07.jpg)
About the OT we can have it build with custom specs, so, if a particular value isn't in the commercial offer, we can get it custom wrapped
Where we encountered the perplexity of the artisan was the request for a double primary
Grazie ancora
Franco
p.s.: I was writing this post and someone posted something I haven't still read, now I go to read
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Well, I showed a way with 500v plate 250v G2 and a readily-available 1.25kΩ OT from Hammond. I got the same 18w as shown by your data sheet.
First I want to say that I found news from multiple sources, that the GU50 are used with 300v at G2. the russian datasheet seems (according to what you read on the net) conservative with a level of 250v G2. the reason is the use was in military apparatus where the maximum reliability must be.
A 1967 6V6GTA data sheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/135/6/6V6GTA.pdf) shows maximum plate voltage is 350v and maximum screen voltage is 315v. But we both know Fender amps used them at ~420v or more on both plate and screen. Users don't report short ube life with old American tubes.
But when they use Russian tubes that carry the same specs as the 6V6 on the data sheet, there were reported failures. With the GU-50, maybe you can push, maybe you can't. It might be wise to not push especially if you're not gaining anything.
Some people refers that they use the GU50 tube with the EL152 specs and there is no redplate
We referred to the EL152 datasheet where a 18W in SE is reached with
300v on plate
250v on G2
130mA quiescent anode current
...
2k load
...
We (but mainly I) don't know for certain that all EL152 characteristics are the same as GU-50 characteristics (note that the bias voltage you cited from the EL152 is -24v, while the GU-50 curves showed a bias of -19v for the almost the exact same operating point). But the real issue is the 2kΩ load... That's for a single tube, and with 2 in parallel you need a 1kΩ primary impedance which can allow 260mA of d.c.
It is worthwhile to note this was essentially the same as the 2nd condition I tried (in my first post on loadlines), and the lowest readily available OT primary impedance was 1.25kΩ. The higher load impedance resulted in less output power. But that OT's maximum d.c. current limit was being exceeded by 32% which may compromise power output even more.
Either of your configurations for the power transformer will work (though you will not need the highest voltage if you use the 300v plate 250v G2 condition). Available current will be almost absurdly high for what will be drawn from the supply, even if both the KT88 and GU-50 share the same power supply. You'll still need 2 different filament transformers in addition to your industrial transformer to light up the different tubes.
Average current drawn from the supply should be essentially equal to the idle current of all tubes (132mA * 4 = 528mA) because the peak current is double the idle current.
Where we encountered the perplexity of the artisan was the request for a double primary
My condition for 300v plate 250v G2 and 18w output per tube results in 36w for a pair in SE, or 72w for all 4 tubes. A single 72w SE OT seems absurd, and will be enormous (you'll need a serious hunk of metal for the chassis just to support it). 528mA of unbalanced d.c. also seems impossibly high, and will result in an enormous (and expensive) transformer. A double-primary is a simple matter compared to having a core that can handle 72w with over 500mA of unbalanced d.c.
Two smaller transformers seems much more reasonable. I'd be interested to know the price difference of the custom transformer, even if he made two smaller transformers. There is no technical reason that both tube types need to be operating into the same transformer core.
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What'd you know, the configuration is perhaps do-able, I used 2k (or 1k for two tubes) for the load, but the Hammond 1.25k should be close enough. The bias for the KT88 needs to be lower just a tad, which could be done with the bias adjustment pot, or pre-configured via a relay/switch. Here are the load lines for the two tube types, I did not bother with the smaller EL34 or 6L6, they probably will require more adjustments.
[chart removed to avoid confusion]
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Is it OK to have such hi values of negative screen grid current? My understanding is that near -0- negative screen grid current is OK. But at the lowest operative KT88 plate voltage -- 250V -- screen current is already -22mA and rising to -28mA at 400 plate volts.
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Is it OK to have such hi values of negative screen grid current? My understanding is that near -0- negative screen grid current is OK. But at the lowest operative KT88 plate voltage -- 250V -- screen current is already -22mA and rising to -28mA at 400 plate volts.
Where are you getting those figures from? Which datasheet?
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Is it OK to have such hi values of negative screen grid current? My understanding is that near -0- negative screen grid current is OK. But at the lowest operative KT88 plate voltage -- 250V -- screen current is already -22mA and rising to -28mA at 400 plate volts.
This is an excellent question!
The screen is a wide open coil of wire, and when plate voltage is high cathode current largely gets sucked right past the screen and on to the plate. When you look at a KT88 data sheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/086/k/KT88.pdf) and see Ig2(o) = 1.7mA (as you do on Typical Operation on Page 2 of the linked sheet), you'e seeing that most of the cathode current is going to the plate (plate current is 64mA at idle for this condition). The "o" in the screen current symbol means "at idle" or "zero signal."
For the same condition, Ig2(max signal) = 9mA. But that's an average current.
When plate current swings from its idle value to its peak positive value, it creates a voltage drop across the plate load impedance which causes the plate voltage to swing low. We generally care about the point on the loadline when the control grid voltage is 0v, because that is also the point of peak plate current and low plate voltage.
Inside the tube, cathode current drawn has a weaker attraction to the plate, because it's now at 90v or 50v or 20v (whatever voltage corresponds to G1 = 0v and peak plate current). If the screen is pegged to an unchanging voltage (maybe 300v or 400v), some additional portion of cathode current can get diverted to the screen by virtue of its high attraction as it makes its way towards the plate. So screen current rises. Still, the bulk of the current goes to the plate.
Now look at the lower graph on Page 7 of the linked data sheet. The graph indicates conditions for a screen voltage of 300v. See the dashed line? That indicates screen current when G1 momentarily reaches 0v (or peak positive plate current). Where the loadline touches the 0v gridline, You would note the plate voltage. Then you would move straight down staying at the same plate voltage value, and read the indicated screen current.
You see that the curve of screen current only has a significant rise at very low plate voltage, which would only happen when the control grid is driven to 0v and you have a high load impedance (which generally cuts below the knee of the 0v gridline) which results in a low plate current at that moment.
The screen current indicated is only valid for the instant the plate voltage is at that value. So even the peaks at low plate voltage will only be brief blips of high current. The average current will not have risen so much.
It seems to me there should be a separate scale for the G2 current curve that is not shown on the KT88 data sheet. Such a separate scale is given on other tubes' data sheets.
Bottom line, screen current rise can and does happen in every amp you've ever played. That's why they have screen resistors (to limit screen dissipation). But the rise, if serious, only is significant when you're cranking maximum power output and hitting that 0v gridline. And depending on the choice of load impedance (and resulting plate voltage at G1 = 0v), the screen current may not rise much.
The possibility of screen current rise is also why screen resistance is generally kept to a low value (so that screen voltage stays largely constant).
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Increased screen current at low plate voltage (but steady screen voltage) is more explicitly shown in the GU-50 data sheet (http://www.mif.pg.gda.pl/homepages/frank/sheets/018/g/GU50.pdf), page 2, middle graph.
NOTE: I've answered what I think you were asking, rather than the specific words you typed. Otherwise, Jazbo is right: where'd ya get those numbers?
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Ancora molte grazie A TUTTI per l'aiuto
Guys you put a lot of meat on the grill, that is really a BIG Barbecue to be digested :icon_biggrin:
@ HotBluePlates
I've just seen you sent one two other post, now I go to read it
It might be wise to not push especially if you're not gaining anything.
I agree with you
But when they use Russian tubes that carry the same specs as the 6V6 on the data sheet, there were reported failures. With the GU-50, maybe you can push, maybe you can't.
Yes you are right, but often happens that datasheet are "domesticated" for commercial reasons and also one thing to consider is that a russian civil commercial tube isn't a rugerizzed military tube
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It is worthwhile to note this was essentially the same as .............
:worthy1: :worthy1: :worthy1:
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Either of your configurations for the power transformer will work (though you will not need the highest voltage if you use the 300v plate 250v G2 condition). Available current will be almost absurdly high for what will be drawn from the supply, even if both the KT88 and GU-50 share the same power supply
Here I want to be sure I've understand correctly (also if I understand english, my level of understanding is far from perfect)
You are saying that the transformer is enough big to supply the 2 x GU50 + 2 x KT88 tubes, correct ?
The voltage we must use are B1 (266v) and B2 (299v) on my first PS schematic, B1 for G2 and B2 for plates ?
http://i.imgur.com/YytBfXN.jpg (http://i.imgur.com/YytBfXN.jpg)
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You'll still need 2 different filament transformers in addition to your industrial transformer to light up the different tubes.
Err, why ? I don't follow you ???
the 0-12v winding has 4.16A (50VA)
GU50 filament requirement is 12.6v 0.8A each = 1.6A
KT88 filament requirement is 6.3v 1.6A, in series 12.6v 1.6A = 1.6A
1.6A + 1.6A = 3.2A
4.16A - 3.2A = 0.960A for preamp tubes
is that incorrect ??
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My condition for 300v plate 250v G2 and 18w output per tube results in 36w for a pair in SE, or 72w for all 4 tubes. A single 72w SE OT seems absurd, and will be enormous (you'll need a serious hunk of metal for the chassis just to support it). 528mA of unbalanced d.c. also seems impossibly high, and will result in an enormous (and expensive) transformer.
The intention of our blend pot is to have always a 40W (36W) max output power, this power can came from GU50 50% + KT88 50% or GU50 100% + KT88 0% or any other combination such as GU50 10% + KT88 90%
we never push the 4 tubes at full roar
(http://el34world.com/Forum/index.php?action=dlattach;topic=16943.0;attach=42631)
Two smaller transformers seems much more reasonable. I'd be interested to know the price difference of the custom transformer, even if he made two smaller transformers. There is no technical reason that both tube types need to be operating into the same transformer core.
I can agree with you, and about the speaker have you think to parallel both output on a single speaker or to use separated speakers ?
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There is one thing that isn't so clear to me, forget for a moment the dual OT option
300v plate 250v G2 260mA for 2 x GU50 on a 1k load
and about KT88 ?
you mean is possible to use same 300v plate 250v G2 260mA for 2 x KT88 on which load ?
1k in a separate primary winding (two separated primary windings)
or 500R in a single primary winding where the 4 tubes are all connected ??
or a 1k single winding with CT (500-CT-500) (but build as an usual SE transformer) where to connect the B+ and plates of the tubes at the extremities (GU one side - KT other side) ?
Be patient, I'm losing control of this thing :w2: :sad2:
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@ Jazbo8
Many thanks for the plot, I'm always in trouble behind one, but I hope that I'll understand it looking with attention
Grazie ancora
Franco
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jjasilli - could you also clarify what you meant by negative screen current?
Franco - the charts suggest that it is possible to use the same OP transformer (1k-1.25k) for a pair of GU50 or KT88 in PSE as you planned, I did not looked at the EL34 or 6L6, but if you really need it, I can run some more simulations.
In any case, the characteristic charts I posted should be treated as ball park figures, since the "diode line", i.e., the Ra line at low plate voltage is seldom below 280 Ohms, so the minimum plate voltage (@ the maximum plate current) at Eg=0V should be higher than those figures shown on the charts, so the screen current would be lower as a result.
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Hi people,
I'm Kagliostro's friend. The GUILTY of this crazy project.
First of all: forgive my ignorance on english language and i hope on your benevolence for wrong concept or mistaken. I shall try to do possible.
Now, my compliments for yours cognisance and knowledge! This tread is very detailed and informative reply.
Let me show you something about the T.A. and the reasons of my choices for this project.
What I wanted to do was to take advantage of a transformer retrieved for create the high anodic tension of plates and much current as possible. Hence the start of this insane creation: an amplifier S.E. with 2 GU50 + 2 KT88 with the relative, appropriate, out transformer. The difficult is to choice the best equilibrium between impedence, current and tension for a better result of watt and sound.
This is what i've imagined thus far whit the help of Kagliostro:
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But when they use Russian tubes that carry the same specs as the 6V6 on the data sheet, there were reported failures. With the GU-50, maybe you can push, maybe you can't.
Yes you are right, but often happens that datasheet are "domesticated" for commercial reasons and also one thing to consider is that a russian civil commercial tube isn't a rugerizzed military tube
You are assuming the military tube somehow has higher ratings than what is shown on the data sheet.
On the whole, that was not the case for American military tubes. Some even carried lower maximum ratings than the prototype tube. But then the military version was subjected to certain tests to validate the tube. The new, lower, ratings may also give a false perception the tube is superior to the original. Really, the manufacturer simply promised less and delivered the same as always.
It is your choice. My choice as a designer would be to adhere tightly to the manufacturer's ratings. That may simply mean the tubes in my design will live a very long time compared to some other designs that push the ratings.
Also, you don't gain anything by pushing. The load impedance determines the power output, and we're just verifying the tube can adequately drive the load. So why strain the tube? What do you gain with the new condition? If you can't give a solid answer, it's not worth pushing the ratings (in my opinion).
Either of your configurations for the power transformer will work (though you will not need the highest voltage if you use the 300v plate 250v G2 condition). Available current will be almost absurdly high for what will be drawn from the supply, even if both the KT88 and GU-50 share the same power supply
Here I want to be sure I've understand correctly (also if I understand english, my level of understanding is far from perfect). You are saying that the transformer is enough big to supply the 2 x GU50 + 2 x KT88 tubes, correct ? The voltage we must use are B1 (266v) and B2 (299v) on my first PS schematic, B1 for G2 and B2 for plates ?
Yes. 132mA * 4 = 528mA.
Half the time, the tubes are hitting a positive peak of double the idle current. The other half the time, they are near-zero mA. The average current is very close to the idle current, or 528mA total. Your drawing for B2=299v and B1=266v uses the 115v, 75v and 24v windings. The lowest-rated winding is the 115v winding at 0.695A, well in excess of the needed average current for all 4 output tubes.
You'll still need 2 different filament transformers in addition to your industrial transformer to light up the different tubes.
Err, why ? I don't follow you ??? the 0-12v winding has 4.16A (50VA).
GU50 filament requirement is 12.6v 0.8A each = 1.6A
KT88 filament requirement is 6.3v 1.6A, in series 12.6v 1.6A = 1.6A
1.6A + 1.6A = 3.2A
4.16A - 3.2A = 0.960A for preamp tubes
is that incorrect ??
No, you are correct. I overlooked your 12v winding, and that you could use both tube types with it. The industrial transformer is all you'll need.
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My condition for 300v plate 250v G2 and 18w output per tube results in 36w for a pair in SE, or 72w for all 4 tubes. A single 72w SE OT seems absurd, and will be enormous (you'll need a serious hunk of metal for the chassis just to support it). 528mA of unbalanced d.c. also seems impossibly high, and will result in an enormous (and expensive) transformer.
The intention of our blend pot is to have always a 40W (36W) max output power, this power can came from GU50 50% + KT88 50% or GU50 100% + KT88 0% or any other combination such as GU50 10% + KT88 90%we never push the 4 tubes at full roar
Your friend is silly enough to want 2x GU-50 and 2x KT-88 in parallel single-ended. Sooner or later, he will be silly enough to want all output tubes roaring at the same time.
There is still 528mA of idle current through a single transformer if all 4 tubes use the same OT. 158w of waste heat in the output tubes just sitting (300v * 528mA). The transformer will have to have a HUGE core just to overcome the unbalanced d.c. The Hammond 1642SEA (http://www.hammondmfg.com/1627.htm) which can have 300mA of unbalanced d.c. (yes, at 75w output) is 28 pounds.
You might save a little on transformer size by not needing all 80w at the same time, but I don't think you'll get around the unbalanced d.c.Or, the single transformer will not save any size or weight over using 2 smaller transformers.
There is one thing that isn't so clear to me, forget for a moment the dual OT option300v plate 250v G2 260mA for 2 x GU50 on a 1k load. and about KT88 ? you mean is possible to use same 300v plate 250v G2 260mA for 2 x KT88 on which load ?
1kΩ for a pair of single-ended GU-50's. 1kΩ for a pair of single-ended KT-88's. The tubes will need different bias voltages, and will require a little different input signal voltage for the same output. But with the same supply voltage and same load, both tube types make essentially the same power output (the KT88's characteristics aren't different-enough from the GU-50's to need a different load or operating point).
IF you run all 4 tubes on the same winding, you can use a 500Ω winding for the 4 tubes. BUT all four tubes have to be driven at the same time to see 2kΩ per tube. Otherwise, 2x driven tubes on a 500Ω winding means each driven tube sees a 1kΩ impedance, or half what it needs to make the output power you want. They may also redplate.
You can use 2 OTs, each with a 1kΩ primary and only 2x tubes in parallel or you can use 1 OT with 2x 1kΩ primaries with only 2x tubes per primary. This gives the each pair of tubes the load they need regardless of how many are driven. If you pull out a tube (say 1 GU-50), you'd need to double the secondary load to reflect 2kΩ back on the primary for the remaining tube.
I'd tend to go for 2x separate output transformers. I would also use separate speakers for each output transformer. There's just no sense in trying to tie the pair of output stages together. Or for the complications which don't benefit the amp's operation.Also, your friend may as well have the ability to drive both pairs to 36w at the same time, and get 72w output.
As I said, he's already going to have 158w of heat coming off the output tube plates, and another 38w of heat from the output tube heaters. Either side could be turned to zero output. I figured he might as well be able to turn both output sections up full if desired.I hope it's cold in Italy during the summer... :l2: Your friend won't need a winter coat.
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it's not worth pushing the ratings
May be, previously, I was not very clear about, I agree with you
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So the best way will be to have a pair of OT and a pair of speakers
and I'm convinced about
so, is to be decided if to use a single GU and a single KT on 2 x 20W 2k primary OT
or 2 x GU and 2 x KT on 2 x 40W 1K primary OT
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You put a flea on my ear
The lowest-rated winding is the 115v winding at 0.695A
Yes, but those are 0.695A AC not DC ......... May this be a problem ??
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Half the time, the tubes are hitting a positive peak of double the idle current. The other half the time, they are near-zero mA.
are you speaking about an SE circuit or a PP circuit ??
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Your friend won't need a winter coat
that's why the test rooms are not heated in winter :l2: :l2:
(no, I'm only kidding :icon_biggrin: )
Thanks
Franco
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Jazbo8 asked me to clarify what I mean by negative screen current. Both HotBlue & Jazbo8 asked me where I got my numbers. So here goes!
Lowest KT88 plate voltage is from this tube chart: http://www.mif.pg.gda.pl/homepages/frank/sheets/084/k/KT88_GEC.pdf (http://www.mif.pg.gda.pl/homepages/frank/sheets/084/k/KT88_GEC.pdf)
Negative screen current info can be found on this website: http://www.ok1rr.com/tubes/burle/TP-122_Screen-Grid-Current_Loading_and_Bleeder_Considerations.pdf (http://www.ok1rr.com/tubes/burle/TP-122_Screen-Grid-Current_Loading_and_Bleeder_Considerations.pdf)
I took the vertical negative numbers on the right hand side of Jazbo8's charts to be screen current draw. But maybe those negative numbers are voltages.
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I took the vertical negative numbers on the right hand side of Jazbo8's charts to be screen current draw. But maybe those negative numbers are voltages.
Right, those numbers are G1 (control grid) voltage.
Jazbo8 asked me to clarify what I mean by negative screen current. ...
Negative screen current info can be found on this website: http://www.ok1rr.com/tubes/burle/TP-122_Screen-Grid-Current_Loading_and_Bleeder_Considerations.pdf (http://www.ok1rr.com/tubes/burle/TP-122_Screen-Grid-Current_Loading_and_Bleeder_Considerations.pdf)
Look at Figure 1 on page 3 of your linked document. The x-axis is marked "DC Plate Voltage" and the curves drawn start at point M, dip below the x-axis, then rise up again at low plate voltage (to continue essentially off the graph. The portion when those curves are below the x-axis is "negative" screen current. That's because the y-axis represents current values, so the current at low plate voltage is high positive screen current.
Each of the curves are labeled "IC2 for EC1" meaning "Screen Current for Control Grid Voltage". The 3 curves then show how screen current changes for a G1 voltage of -5v, -10v and -20v.
So a careful reading of the concerns raised (about the peak in screen current at low plate voltage) shows the paper is advocating using a screen resistor. It also notes that case 4 (where there is a high negative screen current [opposite normal screen current]) causes screen voltage to rise (not drop, as in every other time someone looks at screen operation with high screen current) and is a case that happens only under extreme fault conditions.
The negative current, where the screen grid becomes an electron emitter the way the cathode is an electron emitter, only happens if the screen grid is contaminated (faulty from the factory), if the tube has been heavily abused (run so hot the cathode coating evaporated to some degree and deposited on the screen), etc. Negative screen grid current, just like negative control grid current, under normal conditions is exceedingly small and only becomes significant when you're basically melting the tube anyway.
I had originally overlooked the "negative" in your question, then on a 2nd reading assumed you misspoke.
Jazbo8 asked me to clarify what I mean by negative screen current. Both HotBlue & Jazbo8 asked me where I got my numbers. So here goes!
Lowest KT88 plate voltage is from this tube chart: http://www.mif.pg.gda.pl/homepages/frank/sheets/084/k/KT88_GEC.pdf (http://www.mif.pg.gda.pl/homepages/frank/sheets/084/k/KT88_GEC.pdf)
Tell us what page and condition you're looking at. There's a lot of stuff in that sheet, and I'm not sure to which info you're directing us.
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Right, those numbers are G1 (control grid) voltage. :BangHead:
Tell us what page and condition you're looking at. Page 1, Par's 3 & 4 However the negative current issue was a red herring, because Jazbo's -#'s were Voltages, NOT current .
Also, it seems that if Kagliostro is willing to sacrifice 1/2 of his desired power output: then running the minimum plate voltage of 250 will bring the loadline closer to the knee of the curves. This might improve the prospect of power tube compression/overdrive.
(Also, the new Forum format is wreaking havoc with my font sizes & spontaneous highlighting of text. Maybe cutting & pasting through MS Notepad will work - it did)
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Tell us what page and condition you're looking at. Page 1, Par's 3 & 4 ...
This is for the KT88 screen current? That's what I'm not seeing. Page 1 of the KT88 is maximum ratings data and seems unrelated to what you were pointing to.
Also, it seems that if Kagliostro is willing to sacrifice 1/2 of his desired power output: then running the minimum plate voltage of 250 will bring the loadline closer to the knee of the curves. This might improve the prospect of power tube compression/overdrive.
I am not comprehending where the concept of "minimum plate voltage" is coming from. The minimum plate voltage is 0v I suppose.
If you look back at the first set of curves I posted, I plotted a load for 300v on the plate. The load impedance for 2x tubes is 1.25kΩ, and was the lowest impedance offered in a readily-available transfomer. If the loadline gets more-vertical (if you drop plate voltage, a more-vertical loadline would be a good tradeoff as it gives a larger peak current), that corresponds to a lower load impedance. And as I said, there just isn't a lower value available off-the-shelf.
Also, such steep loadlines don't make sense in Class A (which he must be using as the design criteria is parallel single-ended operation). Those steep loadlines are much more at home with class AB operation, where plate voltage is raised, idle current dropped, peak plate currents are much greater and the tube shuts off part of the time to keep dissipation in check.
Additionally, once the other side of the push-pull output stage cuts off, the side which remains on sees a load equal to the plate-to-plate impedance divided by 4. Which is how those steep loadlines even exist when we just don't see them in available impedances.
I think this seals the deal: I'm gonna write a thread on how a class A output stage works, starting with idealized devices. There are so many books available that I didn't want to spend the time repeating what others have already written. But I think Receiving and addressing individual questions posted to the thread will give a lot of folks an "A-Ha!" moment that isn't happening with existing book chapters.
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> news from multiple sources, that the GU50 are used with 300v at G2
OK, but at "only" 250V the GU50 is able to provide 240mA peak, which implies 120mA at idle. 333V on plate gives 40 Watts Pdiss.
Do you really want so much current?
I see little reason to go to higher G2 voltage.
Yes, there "may" be a happy compromise near 300V plate 290V on G2 (simple power). Idle at 133mA(!). The higher G2 voltage should cover the 266mA peak current. About 2.2K load.
If you go to a higher plate voltage you must run less than 120mA idle current, don't need even 240mA peak current, and can run a *lower* G2 voltage. This is less stress on G2, less drive needed at G1. Of course a G2 voltage much different from plate voltage is a power supply complication.
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Audio is assumed symmetric.
Although much audio is asymmetric, we do not know which side is higher, so we must provide equal swing either way.
The up-swing and down-swing should be the same.
They never are, but we like to be close.
10% (1.1:1) or so is plenty close enough for many purposes.
This graph shows nearly 2:1 (187:96) asymmetry. That's gross distortion, and will sound quite different on different sources, or possibly for the same e-guitar strumming up or strumming down.
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There is still a flea on my ear
The lowest-rated winding is the 115v winding at 0.695A
Yes, but those are 0.695A AC not DC ......... May this be a problem ??
if the DC current required is 132mA * 4 = 528mA
695mA AC are enough to satisfy the request ??
Thanks
Franco
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Audio is assumed symmetric.
Although much audio is asymmetric, we do not know which side is higher, so we must provide equal swing either way.
The up-swing and down-swing should be the same.
They never are, but we like to be close.
10% (1.1:1) or so is plenty close enough for many purposes.
This graph shows nearly 2:1 (187:96) asymmetry. That's gross distortion, and will sound quite different on different sources, or possibly for the same e-guitar strumming up or strumming down.
You are right, with Eg2=300V, when I drew the load lines, I had to cram them under the max Pda which resulted in a pretty high asymmetry, I agree the better way is to lower the Eg2 and increase the plate voltage, but I am not sure the existing PT can support it without adding more parts and complications.
[chart removed to avoid confusion]
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The lowest-rated winding is the 115v winding at 0.695A
Yes, but those are 0.695A AC not DC ......... May this be a problem ??
if the DC current required is 132mA * 4 = 528mA
695mA AC are enough to satisfy the request ??
The VA rating divided by a.c. voltage of the winding gives RMS current. 80VA / 115v = ~0.696A RMS.
"RMS" (root-mean-square, the root of the mean of the squares) is a method of specifying alternating current which provides the same heating of a resistor as a direct current. Therefore, the RMS current output will be equal to the d.c. that can be drawn from the winding.
An audio amp doesn't draw d.c. from the power supply except at idle. So you have to figure the average current drawn from the supply when that draw consists of peaks of current, and also dips which can be down to 0mA (and for class AB, could be down at 0mA for some time).
That's an involved process to handle on paper for a class AB stage where peak current is much more than double-idle, and when 0mA can last for a significant time.
The limit of class A is a peak that is double-idle current and just-touches 0mA for an instant (or never quite goes to 0mA). So for an ideal class A output stage, the average current at full power is equal to idle current (2x idle + zero current / 2 = idle current). Reality is that since there will be some distortion, and the stage is never quite "ideal", the average current will be somewhat higher than the idle current.
You have 695mA RMS available, and the output tubes will only draw 528mA (or a hair more). You have plenty.
Cross-check against how Hammond rates their power transformers (http://www.hammondmfg.com/300series.htm). The 363CX is rated for 119VA and has a 5v 3A winding, and also a 360v winding at 288mA. 360v * 0.288A = 103.68 VA plus 15VA for the 5v winding, or 118.68 VA total (same-as 119 VA). You will see that the "288mA d.c." could also be called 288mA RMS. This confirms your transformer's winding is sufficient for your output tubes.
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Many Thanks HotBluePlates
really a good explanation and I can say that I've understand all
Franco
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Sorry to throw a monkey wrech into the mix, but I just noticed that the PT voltages shown on PPR's and your wiring examples were the peak un-loaded voltages, so the actual DC voltage on the plate would be different depending on the exact load, which could be quite a bit lower if you are talking about 530mA.
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I don't know exactly the voltages present in an unloaded condition with the windings arranged for 266v and 299v
but with all the windings in series (excluding only the 12v winding) the theoretical AC voltage is
24v+75v+95v+115v = 309v AC
309v AC x 1.4 = 432.6v DC
the measured unloaded DC voltage with 24v+75v+95v+115v windings in series is 471v a bit more than the theoretical
K
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I hope it's cold in Italy during the summer... :l2: Your friend won't need a winter coat.[/font]
You not consider a obvious thing: if i wanted to cook a beefsteak while playing some accord on my guitar ??? :l2: :l2: :l2: ...or if i wanted a cup of coffee ??? It could, in fact, be made a very innovative mod ! :laugh: :laugh:
...Ok. i'm kidding.
Thank you to everyone for the solutions and clarifications!
So, the best way for the coexistence of 2 GU50 and 2 KT88 is two T.O. twins, but with primarys slightly different for the impedence and with 250 mA at the minum, it's that correct ?. Ok, let's do this! I think that it can be possibile and more simple for the T.O.'s constructor.
If we accept that one primary for the KT88 be 2K and the primary for the GU50 be 1k, with 250 mA for everyone, we can agree that it could be a reasonable way for a S.E. with 80 w and yours relatives stages ?
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Hotblue: This is for the KT88 screen current? That's what I'm not seeing. Page 1 of the KT88 is maximum ratings data and seems unrelated to what you were pointing to.
This is per the section of the tube chart for typical operation. I don't mean to imply that it's an "absolute minimum". However I consider the tube charts state this info for good reason, based upon significant research & development by the manufacturer. Running the tube out of spec, and doing do over time, may complicate a DIY project with additional research & development aspects. But, if it feels good, do it!
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If we accept that one primary for the KT88 be 2K and the primary for the GU50 be 1k, with 250 mA for everyone, we can agree that it could be a reasonable way for a S.E. with 80 w and yours relatives stages ?
Using two OPTs, the design becomes much simpler, but please verify the loaded voltage of your power supply, the B+ voltage from PSU II is a lot lower than those indicated above.
(https://dl.dropboxusercontent.com/u/1326040/PS%20Sim.gif)
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If we accept that one primary for the KT88 be 2K and the primary for the GU50 be 1k, with 250 mA for everyone, we can agree that it could be a reasonable way for a S.E. with 80 w and yours relatives stages ?
Using two OPTs, the design becomes much simpler, but please verify the loaded voltage of your power supply, the B+ voltage from PSU II is a lot lower than those indicated above.
My recommended operating condition would not use the 95v winding at all. We'd be looking for roughly 300v plate and 250v screen of all output tubes. That loads the windings closer to their rated current. If anything, I almost expect the output voltage to be high, but we won't know for certain without loading the windings to find out.
I'd also make life easy by using 2x OT's with 1kΩ primaries, each connected a pair of output tubes. That load would cut below the knee on the KT88's, but rising screen current could be controlled with screen resistors.
[/size][size=78%]Hotblue: This is for the KT88 screen current? That's what I'm not seeing. Page 1 of the KT88 is maximum ratings data and seems unrelated to what you were pointing to.
This is per the section of the tube chart for typical operation. I don't mean to imply that it's an "absolute minimum". However I consider the tube charts state this info for good reason, based upon significant research & development by the manufacturer. Running the tube out of spec, and doing do over time, may complicate a DIY project with additional research & development aspects. But, if it feels good, do it!
I'm lost as to what operating conditions you're referring to. You mentioned:
But at the lowest operative KT88 plate voltage -- 250V -- screen current is already -22mA and rising to -28mA at 400 plate volts.
Then you linked a data sheet, but I cannot find the numbers you posted for screen current anywhere on the sheet. That's why I keep asking you to point me to exactly what you're concerned about (i.e., "2nd Page, 1st operating condition, Ig2 = ...") so we're on the same page.
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On further study.... a 2:1 asymmetry blows-up the equation, but on the graph it's 'only' about 17% THD. Which I suppose is acceptable in guitar-flavor. There's still an issue with the unknown asymmetry of the signal interacting with the amplifier asymmetry in unexpected ways.
And it *won't* actually stay at 400V 100mA. The bias will re-adjust when the amp is pushed to FULL output. The voltage can't change much, so the current will go to the average of the extremes, here (285-2)/2 or 140mA (from 100mA). The OT will thump as its average current changes, and its bass response will be less.
This is actually a common cheat. The 6550 (same-as KT88) datasheet boasts a "20 Watt" SE condition which shows huge THD and current-shift. But the dynamics of that with a plucked input may be hard to get "sweet".
> voltage from PSU II
Total guessing now.... if this PT is the lump shown in another thread, it is HUGE, so PSU's default 20 Ohms DCR may be way off.
I think this plan will need a LOT of extended breadboarding to find happy voltages and sounds. I would budget a LOT of time in the lab. I would be prepared to toss large chunks of iron and find/wind other lumps of iron.
Or build a nice 5 Watt Champ and mike it into a 600W transistor amp and a Full Stack speaker.
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My recommended operating condition would not use the 95v winding at all. We'd be looking for roughly 300v plate and 250v screen of all output tubes. That loads the windings closer to their rated current. If anything, I almost expect the output voltage to be high, but we won't know for certain without loading the windings to find out.
I seemed to recall that the 40W output wasn't possible when the plate voltage is lowered, but will run the sims to see.
On further study.... a 2:1 asymmetry blows-up the equation, but on the graph it's 'only' about 17% THD. Which I suppose is acceptable in guitar-flavor. There's still an issue with the unknown asymmetry of the signal interacting with the amplifier asymmetry in unexpected ways.
I think this plan will need a LOT of extended breadboarding to find happy voltages and sounds. I would budget a LOT of time in the lab. I would be prepared to toss large chunks of iron and find/wind other lumps of iron.
Only 17% THD! That's just too low for guitar amps... :laugh:
I was looking at a table showing the regulation for various PT VA rating, and 18-20% is about what can be expected based on the transformer in question, and if you plug that into PSU II, 28 Ohms is the resultant winding resistance, in any case, as you said, the best way is actually do a load test BEFORE getting too far into the design process - but I think B+ of ~300V should be doable.
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This tread is very interesting - Thanks friends
My recommended operating condition would not use the 95v winding at all. We'd be looking for roughly 300v plate and 250v screen of all output tubes.
Those are the condition that I think is better to follow
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On further study.... a 2:1 asymmetry blows-up the equation, but on the graph it's 'only' about 17% THD
Here I need a bit of clarification, for 2:1 asymmetry do you mean the fact that a GU50 has an MU of 20 and KT88 am MU of 10 ?
knowing that the two tubes haven't the same MU my idea, as you can see in the schematic I posted previously and here, was to use between the MV and G1 a trimmer as to try to "balance" or better "mitigate" this difference
do you think this will have no effect on the whole equation ?
(http://i.imgur.com/mXv7xDA.jpg)
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One other thing, my friend's idea is to use cathode bias, and I usually prefer that way
but will fixed bias be a way to solve problems in this arrangement ?
Ringrazio ancora molto tutti quanti per l'aiuto
Franco
p.s.: I apologize, on my previous #1 and #19 post my schematic reported a wrong OT connection, now is corrected
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The "2:1 asymmetry" refers to the ratio of the positive vs the negative swing of the output voltage across the load - the more asymmetry, the higher the distortion. In any case, please take a look at the following sims, notice that the asymmetries are lower now, based on HBP's recommended operating conditions, please also note the different bias points and the loads used for each tube types. The output power is also lower, so the 40W target is no longer met as suggested eariler in the thread.
(https://dl.dropboxusercontent.com/u/1326040/KT88_GU50_SE_250V.gif)
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Jazbo's latest plots look decent to me; PRR may have some comment regarding them based on his experience with these circuits. I'd highly recommend following his suggestion about breadboarding.
Remember when you look at the loads in Jazbo's latest plots (RL=2kΩ for KT88, RL= 2.5kΩ for GU-50) that those are for a single tube. Halve them when you attach 2 tubes to a single primary winding (bias voltage and idle current per tube stay the same).
Jazbo: What program are you using for the plots? It looks much better than what I'm using (screen captures of data sheet curves), and I'd love to play with it.
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Jazbo: What program are you using for the plots? It looks much better than what I'm using (screen captures of data sheet curves), and I'd love to play with it.
I use Ayumi's pctube program (http://jazzbo8.wordpress.com/2013/02/12/vacuum-tube-amplifier-design-in-the-cyber-age/) for my charts - but it's a bit difficult to use unless you want to build your own SPICE models. Otherwise, you should probably use LTSpice, which with the correct tube models, allows you to plot the characteristic curves easily, e.g., vary the screen voltage and grid voltages at will.
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Many tanks of all of you, guys !
Ok. I'm agree with you for two Output Transformer. I will do.
I think that lower the plate's voltage and obtain ( with others values ) a load line close symmetry as close as possible, will be best way for lower THD of second harmonic and lower the consumption of mA.
...But i don't understand why lower so much the plate voltage.
I will try to explain that better.
For example, i would show you an hypothetical stage: KT88 example with VG2 at 300 V.jpg
We have less usage of current and a load line close symmetry as close as possible ( ...with more fantasy! ). For two tube in paralel we can estimate an impedence of 1,9 K ( 2K ).
W% = ( 340 - 24 ) X 100 mA : 42 = 75%
At about the wattage shall be:
Wout = [( 340 - 25 ) X 0.707 ] X [( 195-100 ) X 0.707 ] = about 15 W. It can be reasonable for current, tension and wattage or did I miss a something of yours advice ?
For the average screen current : ( 0.25 X 0.068 ) + ( 0.5 X 0.008 ) = about 21 mA. Than can estimate an average screen power of 21 mA X 300 V = 6,3 W. 6,3 : 8 = 78 %
I'm not sure for the THD, but with a "non tested" formula i've calculated a THD of second harmonic of 4 %. But i thing that's impossible!
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...But i don't understand why lower so much the plate voltage.
For example, i would show you an hypothetical stage: KT88 example with VG2 at 300 V.jpg [and 340v on the plate].
My suggestion was for 300v plate and 250v screen. You're proposing 340v plate and 300v screen, with essentially the same load impedance I recommended (~2.5kΩ per tube, or 1.25kΩ of the primary when used with 2x output tubes in parallel).
How will you bias your tubes? I assumed cathode bias, as it is the safest method, and works best with class A operation. You need about 24v of bias for the KT88's (for your loadline) and about 19v of bias for the GU-50's (for my loadline).
I assumed you're using the power transformer Kagliostro posted (http://el34world.com/Forum/index.php?topic=16943.msg168375#msg168375). You can have either 399v and 266v, 432v and 300v, or 428v and 295v and 238v as the easy supply voltage options. For the GU-50, I picked the last option for 295v plate and 238v screen. In reality, I drew the line for 300v plate and for cathode bias I need 300v + 19v (bias) = 319v. I also notice the transformer is big and heavy, the windings are not fully loaded, and I believe the output voltage will be higher than calculated. Therefore, I believe you will end up with nearly 319v on the GU-50 plate when the amp is done.
MY suggested condition gives you 18w per tube, compared to 15w per tube from your condition. Either power level is fine and would probably only be noticed as the difference in exact loudness when distortion sets in. But your condition needs 340v plate + 24v bias = 364v supply for running the KT88 in cathode bias. So the needed supply voltages are 364v and 300v, which is not a readily-available option with your power transformer.
There is no reason your operating condition cannot work, and well. Sooner or later, you will have to think about designing the system (meaning the whole amp, including power supply), and consider what you have to work with. That was the real reason for the choice for the operating condition of the GU-50, and everything is easier for you if you select the same condition for the KT88 (though you don't have to). You should probably try to at least design the KT88's to use the same supply voltages the GU-50 will use.
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Sooner or later, you will have to think about designing the system (meaning the whole amp, including power supply), and consider what you have to work with.
+1, you need to perform a load test on the PT to see what you have to work with, especially since you are wiring all the secondaries together to get the HV. Once you are sure of the B+ it can deliver, then the rest of the design can follow.
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You forum readers are aware of my suggestion of self-split class A push-pull.
This probably belongs in a separate thread. Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.
Self-split push-pull operation is kind of a second-class circuit: it was only ever practically used in the old days in cheap student amps or table radios, when the desired price-point of the finished product didn't allow the extravagance of paying for an actual phase inverter (think about that; they couldn't justify the cost of a single extra triode and some resistors). IT requires double or more of the normal peak input voltage to the output tubes, and never gives the full output power that the output tube and transformer would normally be capable of. Modern use of the self-split stage is mostly in the realm of homebrew as a curiosity, or when full power output is not really a goal.
Has any one heard of any application where an ultra linear o/t is used in a self split application. A tap at 20% provides more power than a 40% tap.
Ultralinear doesn't apply to this amp project, as it requires G2 voltage be essentially the same as plate voltage. Maximum G2 voltage for the GU-50 is limited to a value well below the supply voltage available from the power transformer Dom and Franco want to use. Limiting the plate voltage to that value (~250v) would also reduce power output, and fail to fully utilize the transformer they have. So they're looking in a different direction.
Self-split ultralinear would be problematic at best, as self-split pentodes/beam power tubes rely on a largish screen resistance for the non-driven tube to help derive the "drive signal" for that tube. You pretty much need/want the tubes to be operating as pentodes (not midway between pentode/triode as with ultralinear) so that screen voltage variation of the non-driven tube will modulate its plate current. Last, if the designer was willing to pay for ultralinear taps and sought the higher output power, he'd also spring for a phase inverter.
Consider using a anode resistor to reduce voltages to the plates on the power tubes. I have seen a few older amps that have anode resistors.
Reducing supply voltage doesn't seem to be necessary with the power transformer to be used.
Additionally, wouldn't this allow you to artificially raise the reflected primary impedance?
Yes, but probably not to any meaningful degree.
Any anode resistor will simply turn some portion of output power to heat in the chassis. Even 100-200Ω seems impractically large when average current for a pair of tubes will be ~260mA, which implies as much as 13.5w of possible audio output lost as heat. 200Ω would only change the suggested 1kΩ primary for a pair of tubes by 20%. And if selecting the ideal load was to deliver more power, or hit a specific design goal for power, why waste it in anode resistors? There seems to be nothing to gain with their use.
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Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.
First part of the sentense is correct, but the second part isn't - in fact, it's rather odd to use a SE design to achieve high output power - all the fidgetting we are doing with the OPTs could be easily avoided with a PP design - not to mention at a lower cost and weight.... yadda, yadda, yadda.
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I guess that's true: a push-pull design would simplify the output transformer requirements.
36w+36w from parallel SE is not my idea of fun for my back, but it seems to be what they want to do...
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36w+36w from parallel SE is not my idea of fun for my back, but it seems to be what they want to do...
Yes, at the moment is so, previously was 40W (or better, 36W) in PSE
K
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Just make a Ferrari-themed cart to wheel the amp around on, and all will be well! :l2:
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Did someone say Ferrari, va-va-voom!
(http://www.reussenzehn.de/images/inhalt/tube-history/v8_parade.jpg)
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(http://www.circletracksupply.com/images/P/crown-industries-brake-lines-350.jpg)
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Just make a Ferrari-themed cart to wheel the amp around on, and all will be well!
Oh, may be my friend didn't want to move around the amp, it is just a practice amp for bedroom playing
:l2: :l2: :l2:
(http://www.reussenzehn.de/images/inhalt/tube-history/v8_parade.jpg)
may be I can use those layout for the 8 x 6ak5 amp we discussed in one other tread
but as you can see B+ has some difficulties on reaching the tube that is more far from PS :wav: :blob10:
:laugh: :laugh: :laugh:
K
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If your friend wants an amp with 2X KT88's to play in the bedroom, I assume he lives alone! :icon_biggrin:
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Not so far from GU50 Tank Sockets :icon_biggrin:
K
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Of the previous amp I haven't a schematic
but I've of this one
K
p.s.: Ahem, the G2 voltage seems a bit too much exaggerated (or so looks to me)
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There is no reason your operating condition cannot work, and well. Sooner or later, you will have to think about designing the system (meaning the whole amp, including power supply), and consider what you have to work with.
Your consideration about "designing the system" will be taken into account very seriously from now on. ...I didn't know that your condition give a power about 18 W with only 300 V on plate! I' have not do graphical's calculation for this condition, but if you tell me this I presume that it's true. Then your condition could it be very interesting.
Now, you suggest ( in a other post about the P.T. ) not use the 95v winding at all. I think understand the reason of this choice, but there are also the particular preamp stages which is required more voltage than 300 V. For an example you can visualize some points of SLO preamp stages: the first 4 stages ( 364 V and 353 V respectively ).
My project is similar to this for voltage dedicated to stages ( i can post something if is required ), so i think that this winding is necessary. After this B+ dedicated to stages, i must take the voltage for the KT88 and GU 50. That's another reason for "why" KT88's tension was so high: first tension for tubes amp, second tension for preamptubes. Now i would make the inverse.
...Sorry for the late or my answers, but now i have a very very hard week !!! ( i work 13 / 14 hours all days :sad2: :sad2: :sad2: )
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p.s.: Ahem, the G2 voltage seems a bit too much exaggerated (or so looks to me)
Yes, but he's using ~4kΩ screen resistors. He will lose some of the full power output, but with ~700v plate, 350v screen was probably an easily-derived voltage.
...I didn't know that your condition give a power about 18 W with only 300 V on plate! I' have not do graphical's calculation for this condition, but if you tell me this I presume that it's true. Then your condition could it be very interesting.
Look at the first graph I posted with the blue and green lines. The idle point of the green line (2374Ω per tube) is on the vertical 300v line @ 132mA. The bias is -19v, so with a symmetrical input signal, you use the points on the green line 19v on either side (at G1=0v and -38v). I labeled the endpoints (59v, 231mA) and (541v, 28mA). The difference of the voltages times the difference of the currents divided by 8 gives output power for a single tube.
[(541v-59v)*(231mA-28mA)]/8 = 12.23w, or 24.46w for the pair.
Sorry, I got confused with my own loadlines... I plotted a condition for 500v plate and 250v screen with a 5kΩ load for a single tube (or a 2.5kΩ primary for 2x tubes) which gave ~18w. You could use that load with the highest voltage you can manage with the power transformer you have (keep the screen ~250v) and you'll get a little less than 18w per tube.
Now, you suggest ( in a other post about the P.T. ) not use the 95v winding at all. I think understand the reason of this choice, but there are also the particular preamp stages which is required more voltage than 300 V. For an example you can visualize some points of SLO preamp stages: the first 4 stages ( 364 V and 353 V respectively ).
You may not need the high voltage.
The highest "preamp" voltage needed would usually be for the phase inverter, or in your case the last preamp stage. You select a voltage for that stage based on how large an input signal is to drive the output tubes to full power.
The drive signal needed from the phase inverter is equal to the bias voltage of the output tubes. If you take my condition for the GU-50 with a ~1kΩ load for the pair, 300v plate 250v screen and -19v bias, you only need a 19v peak output signal from the preamp. Very easy to drive!
100w amps with 6L6's might have a bias voltage of -40v to -50v. They also have push-pull output stages, so you need a peak-to-peak output signal from the phase inverter of double the bias voltage of each tube. So the SLO would need to develop an 80-100v peak output from the phase inverter. Doing that requires a large supply voltage.
Design the system. So the trick is to design your output tube stage with an eye towards the supply voltages (and available output transformers?) and your required power output. Then you work backwards through the amp designing each stage to meet the drive needs of the stage you just designed, until you get to the input jack.
All this does not mean you cannot use a high supply voltage for the preamp... it is just to explain why you may not need a high preamp supply voltage.
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Yes, but he's using ~4kΩ screen resistors. He will lose some of the full power output, but with ~700v plate, 350v screen was probably an easily-derived voltage.
For sure that is a reasonable motivation
K
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Our Italian friends want parallel-SE in order to make use of a salvaged industrial transformer with tons of current on tap. So no push-pull.
Yes. I know that if I wanted to have a Push pull, many problems and "inanity's ideas " for this project would never have had. But i love the sound of Single Ended and I wanted one amp that could be at the same level of push pull ( ...ok, 40 watt for a S.E. isn't 100 W on push pull. But, for me, this is a heavy strike on the stomach ! Now, we can try 72 W !) or next to this, but WITH YOURS DISTINCTIVE SOUND'S.
Just make a Ferrari-themed cart to wheel the amp around on, and all will be well!
Oh, may be my friend didn't want to move around the amp, it is just a practice amp for bedroom playing
:l2: :l2: :l2:
You are very funny guys ! :l2:
...But "de gustibus", sometimes, can bring some kind of madness to a difficult mind like mine!
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...If you take my condition for the GU-50 with a ~1kΩ load for the pair, 300v plate 250v screen and -19v bias, you only need a 19v peak output signal from the preamp. Very easy to drive!
100w amps with 6L6's might have a bias voltage of -40v to -50v. They also have push-pull output stages, so you need a peak-to-peak output signal from the phase inverter of double the bias voltage of each tube. So the SLO would need to develop an 80-100v peak output from the phase inverter. Doing that requires a large supply voltage.
...All this does not mean you cannot use a high supply voltage for the preamp... it is just to explain why you may not need a high preamp supply voltage.
You have right! I never tinked about this because my aim was to replicate 3 type of famous preamps ( with some different voice and mods respect the originals ) and listen to them on one amp in S.E..
Naively, I copyed this legendary circuits as they was, but not taking account of this difference!
Now, i suppose, i must re-designer my preamp stages's.
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Hi everyone.
I've tried to do something after studying and analysed your proposal. With maximum respect for your advice, I've deliberately choosed high anodic tension because for 4 type of tube in parallel the current was too much, i thought. But I have also tried to conform all that i learned on this tread ( symmetry of the load impedence, different type of impedence for tube, best compromise between power and sound, ect. ect. )
After so many unsuccessful and successful attempts at mediation to this possibile coexisting of tubes, I think that should be possible doing this solution.
My limit whith the language does not permit me to expain so good my intention, than i hope the schematic can do this.
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Stages that I think can do reasonable way.
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Whith option on 4 ohm in the 2° secondary of T.U. :
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A little gift to all:
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Example for KT88
DATA:
VA= 400 V; IA = 95 mA; VAp = 780 V; IAp = 200 mA; Vpp = 18. VAmax = 765V; VAmin = 30V; IAmax = 192 mA; IAmin = 5 mA.
Vg2 = 250 V; Z = 3,9K ; Vg1 = - 18 V; Ig2Q = about 7 mA; Ig2max = about 95 mA;
WA max = 42 W ; WG2 max = 8 W ;
[( VAmax – VAmin ) X ( IAmax – IAmin ) ] : 8
Wout /8 = [( 765 – 30) X ( 0.192 – 0.005 )]: 8 = 17 W
W% = ( VA x IA ) : Wmax = ( 400 x 0,095 ) : 42 = 90,4%
IG2 AV = Average screen current: ( 0,25 X IG2max ) + ( 0,5 X IG2min )
IG2 AV = ( 0,25 X 0,095 ) + ( 0,5 X 0.007 ) = 27,2 mA
WG2 % = IG2AV X VG2 // WG2 % = (0.00272 X 250) : 8 = 84%
IK = IA + IG2 = 95 + 7 = 102 mA
RK = VG1 : IK = 18 : 0,102 = 176,47 Ohm = 180r
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Please take a closer look at the datasheets, the estimated screen current at low anode voltage is going to be a lot higher, e.g., the GU50 datasheet used for Ig2 has the condition of Va of 800V, when the anode voltage swings low under maximum drive, Ig2 will not stay at 12mA as you have shown. Likewise, for EL34, you somehow ignored the peak in Ig2 when you use max Ig2 of 28mA. Therefore in both cases, the maximum/average Pdg2 would be higher than your estimates suggest.
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Damn! :BangHead: :BangHead: :BangHead:
You have right.
For the GU50 i have found a graphic similar to the 6l6gc's graphic (...whith curves of screen grid at different values), and i can approssimate maximum peak, average current and the power of G2.
But i can't find a similar datasheet for EL34. How i can stimate this values of G2?
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Damn! :BangHead: :BangHead: :BangHead:
You have right.
For the GU50 i have found a graphic similar to the 6l6gc's graphic (...whith curves of screen grid at different values), and i can approssimate maximum peak, average current and the power of G2.
But i can't find a similar datasheet for EL34. How i can stimate this values of G2?
Good question, I have not seen one either. But I thought you were going to use the KT88s.
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Please take a closer look at the datasheets, the estimated screen current at low anode voltage is going to be a lot higher ...Therefore in both cases, the maximum/average Pdg2 would be higher than your estimates suggest.
Dom, which of the many posted operating conditions have you chosen?
Average screen current probably has little relevance. You need to know idle screen current (very small) to accurately calculate a bias resistor (but that precision will be swamped when you have to round up to a standard value).
You also need to know the worst-case maximum peak screen current. Because you want to know if the screen will melt when you hit full power output. If that peak screen current * idle screen voltage is less than the screen dissipation rating, you need do nothing. If it exceeds the screen dissipation rating, figure how much the screen voltage would have to drop to just touch the dissipation rating; calculate the implied screen resistor value.
If the calculated screen resistor seems large (3-4kΩ), you may want to investigate more and determine an accurate average screen current. It will be less than (IG2 max + IG2 idle)/2, because most of the time the screen barely rises above the idle value and is only very briefly at its peak value.
That accurate method involves finding the screen current for 5 or 9 or 11 equal portions of the input sine signal. Meaning figuring screen current when G1 is at Sin 0 degrees, then G1 = Sin 10 degrees * peak input signal, then G1 = Sin 20 degrees * peak input signal, etc.
Yes, it's a pain in the @$$. But it's what you have to do if the simple approach doesn't work and you can't use a loadline that cut G1=0v above the knee of the curves.
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Dom - I use the SPICE model to generate the Ig2 curves to give you an idea how high it can go once the load line crosses "below the knee", the grid lines are from 0V to -30V, and Eg2=250V in the chart below.
(https://dl.dropboxusercontent.com/u/1326040/EL34_P_G2.gif)
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Tank you very much for the model, Jazbo8!
...This help me to imagine that the peak screen grid current will be very high, but i can't figure a sort of approximation for this high value. ... Anyway I think I should safe:
The limiting value for screen grid is 8W. If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V ) the max current possible for the screen grid would be 8W:35v = 228 mA, i think is hardly likely that my operating condition could to take the screen current so high. If i suppose the worst-case whith 110 mA of max peak current for the grid and 40 V of min. votage, i have 40 X 0,11 = 4,4 W. Is it correct ?
...But I thought you were going to use the KT88s.
Yes, certainly. I want use KT88 with the operating condition that i've posted. When i want try another tube like 6l6 or EL34, i change the bias ( bias rotary switch on the schematic ) and secondary of T.U. to 4 ohm at the same time. ...Maybe i must post a jpg.
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Dom, which of the many posted operating conditions have you chosen?
All of them. Now ( ...with your important advices) with some correction that i propouse in the next attachment.
.....You need to know idle screen current (very small) to accurately calculate a bias resistor
Yes. You have right.
In the schematic "Finale Aggiornato.jpg", i have calculated 10 mA for one screen grid by way of the implied screen resistor value: 260V - 250V = 10v. 10V : 0.01mA = 1K.
EL34 needs about 5 mA ( quiescent ), 6L6gc needs about 3 mA, KT88 needs about 7 mA and GU50 needs about 5 mA.
That accurate method involves finding the screen current for 5 or 9 or 11 equal portions of the input sine signal. Meaning figuring screen current when G1 is at Sin 0 degrees, then G1 = Sin 10 degrees * peak input signal, then G1 = Sin 20 degrees * peak input signal, etc.
Yes, it's a pain in the @$$. But it's what you have to do if the simple approach doesn't work and you can't use a loadline that cut G1=0v above the knee of the curves.
...It's too complex for me, but thanks for the procedure.
I hope that will be correct new grapich for GU50.
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If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V ) the max current possible for the screen grid would be 8W:35v = 228 mA, i think is hardly likely that my operating condition could to take the screen current so high. If i suppose the worst-case whith 110 mA of max peak current for the grid and 40 V of min. votage, i have 40 X 0,11 = 4,4 W. Is it correct ?
I don't think it would be that high... So assuming you are using the same OPT as the KT88, I can try to calculate the Pdg2 for you, but just to make sure, have you decided on the OPT that you are going to use or wind yourself? If so, what's the primary impdeance?
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Yes i did. The OPT can be made from a great man that works in Italy.
In view of the points described above, 2 twins OPT will be:
Zprim. = 2K ( for 2 tube in parallel with a relative impedence fo 4K);
max power = 40W;
Max current = 250 mA ( 200 mA is required, but i think that it's better );
1° sec. = 8 ohm;
2° sec. = 4 Ohm ( for el34 and 6l6g. this secondary reflect primary impedence to 4K with a load of 8 ohm)
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Thanks for your patience!
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Dom - I use the SPICE model to generate the Ig2 curves ...
What is the scale for the screen current? It won't be the same as the plate current numbers on the y-axis.
What are the multiple screen current lines? I presume these are for differing values of G1 voltage, but only the G1=0v line is important (well, maybe include the G1=idle bias, to figure idle screen current)
Dom, which of the many posted operating conditions have you chosen?
All of them.
Well, you have to analyze one at a time, and the explanation is a lot of work. PICK ONE, and there will be much less confusion.
For example, look at the EL34 curves Jazbo posted. You need to have a loadline plotted, because screen current will change with the changing plate voltage (which itself is due to plate current variation creating a voltage drop across the selected load impedance).
If the load impedance crosses the G1=0v line at Ip=100mA, that is about 20v or so on the plate. The screen voltage is supplied by a steady d.c. power supply point, and doesn't drop from the G2=250v Jazbo says the curves are for. The Screen Current must have a different scale than the plate current scale marked, because the uppermost G2 curve (assumed to be for G1=0v) also appears to cross 100mA at the same time the plate is passing 100mA. 250v * 100mA = 25w, and the screen would have vaporized before passing that current.
You also need to pick a condition because I'm lost with all the proposed operating conditions, and I assume we have to pick one which makes sense with the power transformer you have available (otherwise, why bother with it when you could pick an ideal power transformer or have one made?). There are too many overlapping steps and considerations to work without any boundaries.
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> The limiting value for screen grid is 8W. If I suppose that in a minimum voltage value ( maybe 25V or 35V or 45V )
The _Plate_ may fall to 25V-50V.
The Screen (G2) stays near 250V (or whatever the Screen supply is).
8 Watts at 240 Volts is 33mA average screen current.
Plotting is difficult and confusing. It looks to me (for EL34) like Screen current is under 25mA for 3/4 of the full-power cycle, peaking to 100mA for 1/4 of the cycle. Assume 15mA for 3/4 of the time and 63mA for 1/4 of the time. It is about 26mA average over the full cycle.
So the rough approximation says "safe" but the counting was very rough. Close enough to go ahead and breadboard; close enough to meter the Screen current on the first full-power test. (The men who designed the EL34 were surely thinking about Screen dissipation, and gave you "enough", but not "too much".)
GU50 seems to be less Screen current, but again there probably is not much safety-factor in the Screen dissipation.
Some series resistance is wise. 100 Ohms won't do much. 1,000 Ohms may reduce maximum power output (but it will be hard to overheat the Screen).
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What is the scale for the screen current? It won't be the same as the plate current numbers on the y-axis.
What are the multiple screen current lines? I presume these are for differing values of G1 voltage, but only the G1=0v line is important (well, maybe include the G1=idle bias, to figure idle screen current)
Ig2 has the same scale as Ip so a lot of screen current can flow when operating below the knee, however, due to the SPICE model's inaccuracy and the fact that modern tubes' diode line are seldom below 280R, so Ep is unlikely to swing as low as the curves suggest. And as correctly pointed out by you and PRR, the SPICE model assumes zero sag on the plate and screen supplies (nor the voltage drop by Rg2), so in the real circuit, the dissipation tends to be somewhat lower, perhaps this exercise can be viewed as the worst case scenario.
Below is how I go about estimating the screen grid dissipation:
1) draw a 280R diode line and note the intercept with the load line, in this case it falls at 40V;
2) pick out the maximum Ig2 from the Eg1=0V curve at Ep=40V, in this case, it's 61.6mA;
3) from RDH4, Pdg2 can be estimated by, (Ig2 idle/2 + Ig2 max/4) x Eg2, so (0.0072/2 + 0.0616/4) x 250 = 4.75W.
(https://dl.dropboxusercontent.com/u/1326040/EL34_SE_4k.gif)
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First of all: I’m very sorry for confusion I have caused and for my stupid mistake about calculation of Wg2!
It’s evident that if the average power of screen grid is calculated regarding a fixed tension ( 250V in this case ), the peak current ( max ) is calculated with THE same tension. For example 23 mA = 250V * 0.023 = 5,75W. This is not the same thing like plate dissipation. Ok.
Then… with a peak of 63 mA, I suppose that for ¼ of full power cycle the screen grid will be at 15,75 W ( 0.063 * 250V = 15.75W ), right ? Even if it will be for a little instant, can this current fuse the screen grid ?
If that were so, only 32 mA should be the “max” peak current ( screen grid current )reasonable for a tube like EL34. …This seems more strange for me, and I’ve somes dubt! This means that all operating conditions exceed this threshold can damage/fuse the screen grid ?
HotBluePlates says “If it exceeds the screen dissipation rating, figure how much the screen voltage would have to drop to just touch the dissipation rating; calculate the implied screen resistor value.”, that’s good. But I presume that it’s good for a “little” drop voltage, because if I decrease VG2 from 250 to 200 ( for example ) I change ALL the operating condition ( VG1 curves will be much squashed and the plate current will much rise at the same load impedance).
In other words, I would minor output and more supply current for the desired result.
For the average screen current and its relative power, this should not be a big problem : ( 0.25 * 0.063) + (0.5 * 0.005) * 250V = 4.55W.
From here I’m very confused by this question: What is the point for calculating the average screen current if the peak current ( max ) can damage screen grid with a few mA ?
Please, let me understand!
@ HotBluePlate
You also need to pick a condition because I'm lost with all the proposed operating conditions, and I assume we have to pick one which makes sense with the power transformer you have available (otherwise, why bother with it when you could pick an ideal power transformer or have one made?). There are too many overlapping steps and considerations to work without any boundaries.
My apologies for “bother” and bring confusion.
My intentions were not “bother”, but show you last possible solution which I came to that, ( by your help as to all others ) after reading your consideration and suggestion, for evaluating its various aspects and possibilities and If it was okay.
I came to the choice of 2 separates OPT after reflect and read all your opinions about the issue.
I have changed the anodic tension for various tubes ( and I redesigned my power transformer ) for a specific reason: the operating condition that you suggest can work perfectly ( 300V - 130 mA ), but with 130 mA for a single tube. If 4 tubes works at the same time, we have about 130 mA * 4 tubes, as you have written earlier. I have accepted and verified ( on my limited knowledge ) your operating condition, but I have chose this only for saving current and for a little more watts.
For other suggested operating condition that can work with 500V and 5k of load impedance, I think that I cannot do with my power transformer under full load.
If I bring high tension and lower current a little, maybe I can reach a reasonable compromise between power output and current saving.
After, I tried many and many operating conditions ( for the 4 type of tubes that I want use ) with the aim to coexist KT88, GU50, EL34 and 6L6 with the same impedance, voltage and current employed. This because I would work an amplifier with 2 KT88 and 2 GU50 or 2 EL34 and 2 6L6gc in the same time, with the opportunity to hear how sounds only a pair of KT88 or GU50 ( same thing for EL34 or 6L6gc with the 4 ohm secondary )
.
My hope was that can be work without fuse the screen grid with operating condition posted, but now I’ve some doubts!
I just wanted to show my proposal to know if it was okay, because I am not an expert on that like you.
Moreover, it seemed to me correct and respectful to all of you show these various proposals and the continuation of the project.
I’m sorry for mistakes, but I'm trying.
@ PRR
Thank you very much for explanations. I begin to understand something.
Some series resistance is wise. 100 Ohms won't do much. 1,000 Ohms may reduce maximum power output (but it will be hard to overheat the Screen).
I don’t know exactly what you mean. Do you mean about some resistance on my schematic ?
@ Jazbo8
Many thanks for graphic simulation!
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It’s evident that if the average power of screen grid is calculated regarding a fixed tension ( 250V in this case ), the peak current ( max ) is calculated with THE same tension.
Then… with a peak of 63 mA, I suppose that for ¼ of full power cycle the screen grid will be at 15,75 W ( 0.063 * 250V = 15.75W ), right ? Even if it will be for a little instant, can this current fuse the screen grid ?
No, it's not quite right... At quarter power, the screen grid current is completely different, it is certainly nowhere near 63mA, so the dissipation is not 15.75W! See if you can figure out the actual dissipation, hint, take a look at the signal swing needed for 1/4 power output.
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> Even if it will be for a little instant, can this current fuse the screen grid ?
Tube dissipation ratings may be "averaged over a complete audio cycle".
So if the rating is 8 Watts, and the dissipation is 16W for a short time, and 4W most of the time, you are safe.
Because screen current is highly non-linear, and no simple formula, it has to be plotted from the curves. This is very tedious.
It is better to take the data-sheet conditions for a single-ended amp. A long time ago, some junior engineer worked over a hot tube for hours to find "good" conditions, so his company could sell more tubes.
Yes, factory data-sheets for many large tubes do not show single-ended conditions. Why would you want a huge SE amplifier, when a push-pull could make more power with less cost?? Both in tubes and in the output transformer. So there will have to be Smoke Testing.
A DC meter watching the screen current (or voltage across a small 100r resistor to the screen), time screen voltage, tells you the screen power "averaged over a complete audio cycle".
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> aim to coexist KT88, GU50, EL34 and 6L6 with the same impedance, voltage and current employed.
Then you need voltage and current, and *power dissipation* (and maybe impedance) which all tubes can work well at.
A good 6550 can stand 42 Watts dissipation (KT88 should be similar).
6L6GC is rated 35W, but I've had trouble at 30W. (A genuine 6L6 from 1938 is 19W; genuine old 6L6G is the same.)
EL34 says 25W but we all know they stand 30W.
GU50 is rated 40W.
So we have 30W and 40W tubes. You *can* run the KT88 or GU50 at 30W dissipation. Then all tubes will make nearly the same power output (about 12W per tube).
6L6GC has the lowest plate voltage rating, 500V.
It is usually NOT necessary to use HIGH voltage in a single-ended amplifier. Power output is limited by dissipation, not voltage losses.
Also the "optimum" screen voltage should -drop- for higher plate voltages. (High plate voltage means low plate current, so we do not need a high screen voltage, and do not want a high screen voltage because it is bad for the screen and we will need more control grid voltage to balance the high screen voltage.)
So I suggest these values for ANY of those tubes:
500V 60mA 8K3 load
450V 67mA 6K7 load
400V 75mA 5K3 load
350V 86mA 4K1 load
300V 100mA 3K load
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Tube dissipation ratings may be "averaged over a complete audio cycle".
So if the rating is 8 Watts, and the dissipation is 16W for a short time, and 4W most of the time, you are safe.
Because screen current is highly non-linear, and no simple formula, it has to be plotted from the curves. This is very tedious.
Tank you so much! This is precisely what I wanted to know. It was always doubtful this concept for me.
....Why would you want a huge SE amplifier, when a push-pull could make more power with less cost?? Both in tubes and in the output transformer.
Because I love the sound of Single Ended. Anyway, this project was start with intention to make 2 version with the same Preamp. Obviously, second SON was / is thinked with Push Pull.
...It is usually NOT necessary to use HIGH voltage in a single-ended amplifier. Power output is limited by dissipation, not voltage losses.
Also the "optimum" screen voltage should -drop- for higher plate voltages. (High plate voltage means low plate current, so we do not need a high screen voltage, and do not want a high screen voltage because it is bad for the screen and we will need more control grid voltage to balance the high screen voltage.)
So I suggest these values for ANY of those tubes:
500V 60mA 8K3 load
450V 67mA 6K7 load
400V 75mA 5K3 load
350V 86mA 4K1 load
300V 100mA 3K load
Ok. I should value it very seriously and I'll test with graphics and curves.
...But... Perhaps I did not make myself clear in my first speech ( the language is a limit for me, and I think that can bring confusion ):
With my power transformer I cannot go further about 440 V. So, I presume, the first operating condition suggested it's not possible for me. But I'll try the other!
@ HotBluePlates
...I have a vision for the complete amp:
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Kagliostro will test it with a load very adapted ! :l2: :l2: :l2:
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> Because I love the sound of Single Ended.
Yes, I understand.
However "MOST" designers, back in the time when vacuum-tubes were common, used SE only for "little" amplifiers, to minimize the number of parts to buy and assemble.
A one-6L6 amplifier is about the same cost as a two 6V6 amplifier. While you need a second power tube plus a phase-splitter, the output transformer is much less money (no unbalanced DC), the power output may be a little higher, and the distortion number is "better".
So very few tubes have suggested operating conditions for SE over about 10 Watts.
One exception, though an odd one. The HK-257 has a typical condition for 30 Watts output from one tube in Class A.
http://www.mif.pg.gda.pl/homepages/frank/sheets/114/h/HK257B.pdf (http://www.mif.pg.gda.pl/homepages/frank/sheets/114/h/HK257B.pdf) middle-right of page 3
Note that the same tube can be run at 1000V or 500V with very little change of output power, but a large change of load impedance.
Note that two such tubes gives, not twice, but TEN times the output power.
And the sheet says "Audio and Television". Television was new in 1946. The TV video signal has to be impressed on the radio wave. In 1946 this could be a big SE tube modulating the carrier signal-- that avoids having a Modulation Transformer which is difficult with video. However a 30W modulator could only manage a 50W-100W total transmitted power output; enough for testing in 1946 but by 1948 they were using much bigger transmitters and different modulation systems. So "big SE" faded away again.
These big SE amps are fun to think about. A couple of guys here actually built one on the breadboard. One is a professional house/factory Electrician, knows the dangers and how to reduce the risk of electrocution and fire. However I think "Real Life" got in the way of finishing the work.
http://el34world.com/Forum/index.php?topic=10262.msg93751 (http://el34world.com/Forum/index.php?topic=10262.msg93751)
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Hi PRR :thumbsup: :thumbsup: :thumbsup:
just a curiosity, which is the socket type name ?
it is similar to UX7 (that fits 1625)
(https://tubezone.co.uk/published/publicdata/ECOMMERCE/attachments/SC/products_pictures/SOCUX7-LGok_enl.JPG)
but seems much more larger
(http://i.ebayimg.com/t/Pair-New-Ceramic-Large-7-pin-Vacuum-Tube-Sockets-for-813-4E27-4-125B-etc-/00/s/MzAwWDU4NA==/$(KGrHqNHJDkE-N)DGl8wBPoguqs9Yg~~60_12.JPG)
Thanks
Franco
EDIT: Is an U7G socket ?
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PRR, I will take account of your professional experience and your warning about this idea for a big single ended. Maybe, one day, I'll tell you "Oh yeah, you were right.", but now I want try this "obsession".
...If we have to speak at it from the perspective of "cost" and "the number of parts to buy and assemble", you are completely right. In fact the whole project looks like madness, but I like this and I make it possible whatever it takes! :icon_biggrin: Ok, I'm just a crazy man.
In view of the potential dangers that it involves, I will try, if at all possible, to make sure everything is going well.
Your suggestion about the HK-257 sounds very good for another project (...maybe with a microwave transformer's ??? ) in Singele ended. I'll think about it. Seriously. It's very interesting!
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Visto che la la HK-257 va anche con anodica di 500v e rende più o meno allo stesso livello (25W invece di 30W), forse è meglio evitare 1000v di anodica, inoltre con 500v ti basta un TU da 2.6k invece con 1000v ti servirebbe un TU da 12k
Since the HK-257 also goes by anodic 500v and makes more or less at the same level (25W instead of 30W), perhaps it is best to avoid 1000v B+, 500v also requires just a 2.6k OT, instead 1000v requires a 12k OT
K
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Just to throw more crazy ideas into the mix, if you want high power SE with modest plate voltage, look at the 13E1 (http://frank.pocnet.net/sheets/025/1/13E1.pdf). Rule of thumb - 1 x 13E1 ~ 3 x KT88/6550 or 4 x EL34! But make sure your PT can supply the high current.
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... The HK-257 has a typical condition for 30 Watts output from one tube in Class A.
...
Note that two such tubes gives, not twice, but TEN times the output power.
You implied this, but casual readers may overlook it... When you note that 2 tubes gives 10x output power, the data sheet gives "Load Resistance, plate to plate" and also describes the condition as "2 Tubes, Overbiased."
The first note tells you this is push-pull, not single-ended anymore.
The second note is very-old-school terminology for "Class AB" rather than Class A. "Overbiased" means "more than class A bias" to keep the plates from overheating by allowing the tubes to cool off by shutting off part of the signal cycle. The increase in supply voltage from 500v or 1000v to 1.5kV is your other clue about this being a class AB condition.
500v also requires just a 2.6k OT, instead 1000v requires a 12k OT
Once you realize that load impedance is selected on the basis of supply voltage available and output power needed (not because the tube directly dictated a needed load) then it becomes clear why the load impedance increases with increased voltage.
Play with the equations Power = Voltage * Current and Voltage = Current * Resistance to prove this for yourself. You can mix the equations to derive Power = Voltage2/Resistance to make it even more obvious. (CAVEAT: "Voltage" in this case really means RMS signal at the plate, not the supply voltage; however, the RMS signal voltage is always less than the supply voltage.)
For example:
500v Supply
25w = Voltage2/2600Ω
65000 = Voltage2
Voltage = ~255v RMS
RMS Voltage = 51% of supply
1000v Supply
30w = Voltage2/12 000Ω
360 000 = Voltage2
Voltage = ~600v RMS
RMS Voltage = 60% of supply
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Many thanks HotBluePlates
The second note is very-old-school terminology for "Class AB" rather than Class A. "Overbiased" means "more than class A bias"
I'm glad I learned a new thing :smiley:
The other stuff you added is really explained in a very easy to understand way
Play with the equations Power = Voltage * Current and Voltage = Current * Resistance to prove this for yourself. You can mix the equations to derive Power = Voltage2/Resistance to make it even more obvious. (CAVEAT: "Voltage" in this case really means RMS signal at the plate, not the supply voltage; however, the RMS signal voltage is always less than the supply voltage.)
For example:
500v Supply
25w = Voltage2/2600Ω
65000 = Voltage2
Voltage = ~255v RMS
RMS Voltage = 51% of supply
1000v Supply
30w = Voltage2/12 000Ω
360 000 = Voltage2
Voltage = ~600v RMS
RMS Voltage = 60% of supply
Franco
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> Voltage = ~255v RMS
> RMS Voltage = 51% of supply
> Voltage = ~600v RMS
> RMS Voltage = 60% of supply
More math but perhaps more insight:
Look at Peak voltage.
Ideally Peak voltage is the supply voltage. But the tube gets in the way.
255V RMS is 361V Peak, or 72% of 500V supply. Not great, not horrible.
600V RMS is 848V Peak, or 85% of 1000V supply. Not great, but better.
On the flip side, look at lost voltage. The 500V case loses 139V (at at-least 139mA Peak). The 1000V case loses 151v (at at-least 71mA Peak).
Which suggests they left some power on the table. The diode-line is about 45V at 100mA. It should be able to pull nearly 90% of B+ as peak output. The 72%-85% figures are probably for "low distortion" (however they don't say how low is low). In G-amp service you could probably get more Watts (and chest-impact) before it changed from "flavor" to "fuzz".
Yes, factory data for a 2-tube condition is always push-pull. Only fools (like us) would throw in more tubes without taking advantage of "better" ways to use them.
Yes, the "overbias" case perhaps should not be called "Class A". But H&K were radio-heads, mostly working class C with occasional hard-B (idle near cut-off). Maybe anything softer looked like A to them. The fact it is heavy AB is clear from the DC current change, >1:3 idle to full.
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Thanks PRR
as always a very good explanation :thumbsup:
Franco