Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: c.stoffel on May 18, 2016, 05:19:37 pm
-
Hello everybody! I built this circuit inside an old amp and it's sounding good but there's something I'm just a little bit worried about. I'm using fixed bias and there's no standby switch, nor a dedicated bias tap on the power transformer. I'm also using solid state rectification for the HV, so the B+ comes up really quick, almost instantly, but the bias voltage takes something like 5 or 6 seconds to get where it should be (thanks I believe to that 220k resistor after the diode on the bias supply...).
Some points:
1. I want to keep the front plate intact so I don't want to add a standby switch
2. Maybe cathode bias could solve my concerns but fixed bias worked much better in this amp because it has bias wiggler tremolo (applied to point A on the schematic), and I liked the tone much more once I converted it from cathode bias so...
3. Would a dedicated bias tap on the power transformer make the bias voltage go up as quick as the HV, due to a lower RC time constant? Maybe I could buy another power transformer with a bias tap, but it's an expense I'd prefer to avoid if possible.
I know there's a lot of debate on standby switches, and lots of things have been demystified, but I couldn't find anybody talking about this specific situation, so the question is: should I be concerned or just leave it alone and play the thing? :icon_biggrin:
Thanks!
Christian
-
I am wondering why the bias circuit takes that long to get to operating voltage.
It is also fed by a SS diode so it should be getting it's voltage almost as quickly as your HV circuit.
Maybe your e-caps in the bias circuit aren't up to snuff, and leaking? :dontknow:
Your output tubes should be taking about the same amount of time for their heaters to free electrons for full conduction.
If your - bias is where it should be when the tubes warm up, it should be fine.
Only concern I might have is when the tubes are already hot and you turn it back on and the - bias takes that long to come up.
Then you could have excessive current flow in the output tubes until they get their proper bias. :think1:
-
Then again, maybe because that 220K resistor comes before the diode on the Plexi's fixed bias circuit. :dontknow:
-
You can get a cheap/small 12v filament transformer from Radio Shack. Connect the 12v secondary to your existing 6.3v filament string. This will give you about 60VAC on the primary that can be used for the AC source for the bias supply, allowing you to use a much smaller (faster TC) resistor.
Or, you can run the amp cathode biased. Tremolo will wiggle the zero volts on the grids of the power tubes just as good as it will wiggle a negative bias voltage. I have a couple amps that do this.
Or, you can just play it like you stole it! :icon_biggrin:
-
That 220K is current limiting the charge path for the bias supply too much. Cut it by 80-90%, change to 22K to 47K. 20 uf is kind of light filtering for a bias supply, you may need more ufds. Bias uses almost zero current.
:sad2: :w2: :worthy1: :worthy1: :worthy1: :worthy1: :worthy1: :worthy1: :BangHead: :BangHead: :cussing: :cussing: :icon_biggrin:
enough already.
-
Wouldn't reducing that resistor increase the - V ?
I thought HBP explained to me that that resistor is there to knock down the AC before the HW rectifying diode.
And that adjusting the value of that will change your - bias range.
I still need to figure out this part of my quad 6V6 Plexi circuit.
Putting that R before the diode might reduce or eliminate the delay?
You don't like the blinky emo thingys, 11?
I usually just put those in to add some color to the message. LOL
-
If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?
-
That 220K is current limiting the charge path for the bias supply too much. Cut it by 80-90%, change to 22K to 47K. 20 uf is kind of light filtering for a bias supply, you may need more ufds. Bias uses almost zero current.
There's plenty of Marshalls out there that use that exact bias circuit. The fact that the bias circuit is such a light load means that you can get by with small caps.
-
If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?
No.
Without the - V on the grid and the cathode connected to Gnd,
the tube will have it's accelerator pedal pushed to the floor.
Cathode bias puts the cathode at a + V as opposed to the grid.
-
If the bias voltage takes longer to "rise", meaning to get to -50 or whatever, doesn't that just mean the tubes aren't conducting current till that happens anyway?
No. Just the opposite. The closer the bias voltage is to zero, the harder the tubes burn. 'Course, the tubes still gotta wait for the filament to heat the cathode before any current will flow.
It's an ECL82. "No reason to get excited." The thief, he kindly spoke.
-
The dwg below shows how a Cap charges when voltage is applied through a resistor.
The cap is considered "fully charged" after 5 * RC in seconds but is darn close after 4 RC, with R in ohms, C in farads. Or R in megohms and C in microfarads.
You have 220K * 20 uf, effectively. That is .22 Meg * 20 ufd in "RC units" thus each RC time unit (aka "seconds") is 4.4 seconds and you need at least 4 of them to get close to fully charging your bias caps. This is why your bias is taking so much time to charge no matter how many Marshalls use the circuit.
(http://i172.photobucket.com/albums/w32/ttm4/RC_02_zpsox3sm2cs.png) (http://s172.photobucket.com/user/ttm4/media/RC_02_zpsox3sm2cs.png.html)
-
So if we move the resistor to before the diode,
then we have the cap connected with 0 R and the delay goes away, right?
-
The dwg below shows how a Cap charges when voltage is applied through a resistor.
The cap is considered "fully charged" after 5 * RC in seconds but is darn close after 4 RC, with R in ohms, C in farads. Or R in megohms and C in microfarads.
You have 220K * 20 uf, effectively. That is .22 Meg * 20 ufd in "RC units" thus each RC time unit (aka "seconds") is 4.4 seconds and you need at least 4 of them to get close to fully charging your bias caps. This is why your bias is taking so much time to charge no matter how many Marshalls use the circuit.
I totally agree. I think the OP does too.
-
"So if we move the resistor to before the diode,then we have the cap connected with 0 R and the delay goes away, right?"
Not really, the diode does not much affect the RC time constant. It is the SIZE (value) of the resistor that the cap charging current has to flow through. Whether before or after the diode, every charging electron has to flow through that 220K (in this case) current-limiting, charge-limiting resistor. The delay would "go away" if we eliminated the R completely. Not completely, of course, caps require some time to charge, but for all real world purposes, losing the Resistor would take the delay down to well under a second.
Since the ultimate aim of the bias rectifier is to produce "in the neighborhood" of 50 volts (forget the polarity) and a half wave rectifier off half a 300 volt winding should produce .7 * 300 = 210 volts which we then have to whack down to 50, we would like to be able to use 100 volt caps instead of 250 or 300 volt filter caps. So, by inserting the resistor either in front of the diode OR immediately after the diode, hardly matters, we can chop the voltage that ends up on the caps AND perform a bit of current limiting just to be nice to our parts. So somewhere between "zero delay" and "too much delay" is a 1-2-3 second time constant. I tend to use about 15K in that spot, but I *also* like to be able to develop enough (negative) bias volts so that I can completely or almost completely shut off the output. IMO it's handy for troubleshooting the preamp section if you have a hum problem which can be time consuming, you don't have to cook your output tubes while you chase it down. I like to be able to get as much as (negative) 65-75 volts on the grids and it also allows for component drift. The first time one applies power to a new build is what/when? Test the heaters, see if they light up? Check unloaded B+ or, better, throw a resistor calculated to pull say 40-50 mils across the main power supply just to be sure it's working and so you don't blow the main filter caps with too-high volts? Check bias? You bet. I like the idea of knowing that I have way too high negative volts (yes, sounds odd) to shut the output section hard off, just in case something fails 20 seconds after I turn it all on for the first time. The 22K I suggested would yield .4 seconds * 4 or 5 which is 2-3 seconds which should be fine.
Now we get to the voltage divider network, everything after the last cap in the chain. All you want to do here is to 1: provide enough adjustment range and 2: make it so that you CAN'T dial the bias so low that you redplate the output tubes. I don't know the proper bias volts for the given tubes, but for 6L6 where we want about 45-52 volts, I like to be prevented from going under say 30-35 volts. That's governed by the R UNDER the adjust pot.
-
"No reason to get excited." The thief, he kindly spoke.
Probably my favorite lyric ever.
-
If we move the resistor to the other side of the diode,
then the 1st filter cap is connected directly to it's charging source, no resistance.
As you have explained, the second filter cap is still connected thru a resistor to it's charging source, so some delay remains.
But moving that resistor should greatly reduce the time it takes for the bias voltage to ramp up. No?
-
If we move the resistor to the other side of the diode,
then the 1st filter cap is connected directly to it's charging source, no resistance.
In an RC series circuit, it makes no difference what order the components are in as far as charging and discharging goes. The resistor could be between the + side of the bias capacitor and ground and the time constant would still be the same. Moving the resistor to the other side of the diode doesn't change the fact that it is in series with the capacitor.
-
I don't think so.
If the resistor is on the other side of the diode it is on the AC side.
The pulsing DC coming from the diode is connected directly to the e-cap.
No resistance remains on the DC side of the diode. :w2:
-
Very well.
-
Go back to the drawing, the schematic. The power source is a battery, a pure DC source that is on 100% of the time. The diode is a check valve which prevents any anything from happening when the diode is reverse biased. The rectifier circuit is thus a battery if and only if the diode is forward biased. So you have your battery but you have it only 50% of the time. Actually, 49.9975% of the time, and even less as charging goes on because the positive going part of the very first cycle of AC delivered to the diode anode can only forward bias the diode once it is over .7 volts positive relative to the cathode (bar end) and the cathode is climbing as the cap(s) charge. If the cap is charged to say 40 DC volts, then the diode anode has to reach 40.7 volts to conduct and continue passing current to charge the cap. So on the upswing of cycle #1 you lose .7/300 while the voltage is climbing and the diode again loses forward bias when that voltage falls below being .7 volt positive relative to the cathode. As the cap charges, there is charging activity less and less of the time because the diode is fwd biased less and less of each cycle and we assume that all the cycles are the same duration, thus take the same amount of time. Blah blah blah.
The circuit in effect is cut, chopped, interrupted half or more of the time. "Duty cycle". The circuit would not act like a battery at all, in any way, if the diode was not there, and the cap would act much like a short. Where the charge-limiting resistor is located in the chain doesn't matter one bit, just like where the switch is located doesn't matter. It could be two resistors, one on either side of the cap. Or three resistors. Their summed total resistance would be the operative number.
There is no circuit when the diode is non conducting. Charging the cap bears the requirement that the power source is DC. It's only DC half the time....or less.
The second cap, yes, is located at a greater resistance away from the point we think of as the DC source, and thus it does charge at very slightly higher delay from the first cap, but it is almost undetectible because when the first cap is not charging during the off part of the diode's conduction, the first cap is charging the second cap.
-
I get that the HW diode is only on 50% of the time.
So in the diagram, we are effectively turning the switch on and off 60 times a second.
With the resistor on the other side of the diode, there still is no resistance between the switch and the cap.
There is a very easy way to prove my theory wrong. :think1:
Move the resistor to the other side of the diode.
You guys are saying that it makes no difference and it will take the same amount of time.
If it takes exactly the same amount of time for the bias voltage to ramp up, I am obviously wrong.
But if the bias voltage now comes up much quicker, I must be right. :l2:
-
You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time. I've never done that and I wouldn't do that because the time involved is not significant. The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect. Kinda like that quote from Mr. Zimmerman.
-
There is a very easy way to prove my theory wrong.
Move the resistor to the other side of the diode.
You guys are saying that it makes no difference and it will take the same amount of time.
If it takes exactly the same amount of time for the bias voltage to ramp up, I am obviously wrong.
But if the bias voltage now comes up much quicker, I must be right.
Try it.
-
This is not my amp we are talking about.
I just cited the difference of where the resistor is placed in this amp,
and where the resistor is on the Plexi circuit I recently built.
In effect I have already tried it, in that I didn't notice this delay in my Plexi's circuit.
I could have just overlooked it, so I flipped over my Plexi and connected my meter's leads.
Hit the power switch and immediately got - V that took less than 2 seconds to reach it's set point. :dontknow:
That is less than half the time he said it is taking his amp to ramp up to his bias voltage.
Something must be different, and the only thing I see is where that resistor is in the circuit. :w2:
-
My Plexi 6V6 uses that exact circuit, except I used 25µF caps rather than 10µF caps. It takes an average of 14.8 seconds for the bias caps to fully charge. I used a Simpson 260 and Galaxy S5 stopwatch. The bias voltage just barely beats my cold GZ34 rectifier! If the GZ34 is still warm, B+ comes up faster than the bias. I think I'll change my caps to 10µF.
-
My Plexi 6V6 uses that exact circuit, except I used 25µF caps rather than 10µF caps. It takes an average of 14.8 seconds for the bias caps to fully charge. I used a Simpson 260 and Galaxy S5 stopwatch. The bias voltage just barely beats my cold GZ34 rectifier! If the GZ34 is still warm, B+ comes up faster than the bias. I think I'll change my caps to 10µF.
I didn't use a stopwatch. I am using the 10uF caps, and a lower B+ voltage (not that the B+ should matter). Your bias circuit took way longer than either of ours to charge up, but not as long as Eleventeen's posted formula suggests yours should take.
I don't know what is going on here, with 3 vastly different results, and none of them are exactly the result the formula suggests.
Usually, the formulas in the books can be verified by real world circuit results. :dontknow:
-
Couldn't he also use an off-stdby-on switch for a power switch? I know he says he doesn't want a standby switch, but to me it would be a very simple solution.
-
He could do that. I don't think his situation is a problem.
-
Exactly when a capacitor is fully charged is too subjective. You need to time it to something like the 80% point or even 63.2%. If setting up an experiment to determine the effect of resistor placement, use larger values for the resistor and/or capacitor so that the times are longer and easier to measure.
-
Thanks everybody, lots of interesting responses.
Well, I tried the 220k resistor before the diode but the charging time of the bias supply didn't change at all. So we can conclude the RC time constant is the same with both arrangements.
I think I'm going to try cathode bias again because the preamp topology changed a lot since I experimented with it and didn't like the tone of the amp that much. That would be the simplest solution. But wouldn't a hot cathode biased push-pull amp be less sensible to bias wiggler tremolo as the cathode current counteracts the effect a bit, making for a less pronounced tremolo effect? Well, I guess I'll try it and see.
A small transformer backwards from the 6.3V sounds like a good idea, if I don't like the results of cathode bias. Food for thought!
The idea of using 350V capacitors on the bias supply is nice too, cause the 220k resistor could be removed and the time constant would drop tremendously, right? I never realized that using lower voltage capacitors on the bias supply was a matter of economics, but it makes sense!
The off-stdby-on switch is a cool idea too, except that I want to keep the front plate intact (just for visuals) and the original power switch is, well, different... take a look at the attachment.
Talking about bias supplies, has anybody seen how it is done on the Vox AC50/4? That's a weird circuit, I'm not sure how it works!
I'm gonna try cathode bias and see if it sounds good!
Thanks!
Christian
-
Your bias circuit took way longer than either of ours to charge up, but not as long as Eleventeen's posted formula suggests yours should take.
Well I stopped the timer when the voltage was within 1 volt of fully charged. It takes a long time to juice up the last little bit. It really takes 58 seconds to reach absolute full charge, such that the meter reading ceases to increase. That's pretty close to the 5RC rule. .22 x 25 x 2 x 5 = 55 seconds. Your short charge time is the one I question.
-
But wouldn't a hot cathode biased push-pull amp be less sensible to bias wiggler tremolo as the cathode current counteracts the effect a bit, making for a less pronounced tremolo effect?
The cathode current is stabilized by the cathode bypass resistor. Your bias wiggler should work very well with the ECL82s. Here are a couple of my circuits that work very well. The Maggie is ECL86s (similar to your tubes) and the Ampeg is 6V6s (a bit harder to wiggle than the ECL86s but still a very strong tremolo).
http://sluckeyamps.com/lil_maggie/Magnatone_M2.pdf (http://sluckeyamps.com/lil_maggie/Magnatone_M2.pdf)
http://sluckeyamps.com/RCA/Ampeg_J12B.pdf (http://sluckeyamps.com/RCA/Ampeg_J12B.pdf)
-
You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time. I've never done that and I wouldn't do that because the time involved is not significant. The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect. Kinda like that quote from Mr. Zimmerman.
The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path. Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used. This saves not only money but lots of space.
-
You can take the 220K resistor out of the charging path and use 350V capacitors and then use a big voltage divider after that to get a real fast charge time. I've never done that and I wouldn't do that because the time involved is not significant. The bias from voltage doublers and full-wave bridges with no center tap using that capacitor to get bias voltage take an incredible amount of time to charge apparently with no ill effect. Kinda like that quote from Mr. Zimmerman.
The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path. Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used. This saves not only money but lots of space.
This is a concise and clear explanation of what I thought I understood. Knocking it down early and using smaller caps.
I don't mind being wrong about something like this.
It is what techs like me do. We put out our theory based on what we think we know.
Then we devise a test that clearly either proves or disproves our theory.
If proven wrong, it is back to the drawing board to formulate a different theory and another test for that theory.
What does bother me is not understanding why I was wrong on this.
So far I have not heard a clear and compelling reason as to why my logic was faulty.
Some techs don't bother to try to figure out why their theory was wrong.
This stuff kind of eats at me even after I have fixed the problem.
Why was my theory, which I still think logically makes sense, proven wrong by testing?
Where is the gap in my understanding that this situation is falling through?
How can I fill in that gap?
What makes me happy is that I seem to have the fastest charging bias circuit.
I don't know why, but I do know that it is not taking my circuit near as long as yours to stabilize at it's set point.
-
The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path. Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used. This saves not only money but lots of space.
The charging path would be from a winding tap, through a diode, across a capacitor, then to another winding tap (or the other way if you like conventional current flow). The voltage divider is then placed in parallel with the charging path. This arrangement will charge up so fast that you will have trouble measuring it. Putting a 220K resistor in the charge path will allow you to use smaller, cheaper capacitors, but the trade-off is going to be time. I don't think the time involved here is any problem. How much could a tube possibly red-plate in a couple of seconds?
-
I don't mind being wrong about something like this. . . What does bother me is not understanding why I was wrong on this.
I'm thinking it doesn't matter where the resistor(s) goes: E.g. (other circuits with rectification): 1) in a B+ circuit a dropping resistor cold go between CT & ground, instead of in the B+ rail; 2) in an LED circuit, a current limiting resistor could go before or after the LED; in either position it limits current & drops voltage.
Maybe another reason bias voltage delay is not a problem is that there's a similar delay in the duty cycle of the B+ filter caps???
The voltage divider has to be somewhere; and it seems that wherever it is, it's is still in the charging path. Hence it's better to knock the bias supply voltage down early, in order to feed the caps a smaller voltage -- so physically smaller caps can be used. This saves not only money but lots of space.
The charging path would be from a winding tap, through a diode, across a capacitor, then to another winding tap (or the other way if you like conventional current flow). The voltage divider is then placed in parallel with the charging path. This arrangement will charge up so fast that you will have trouble measuring it. Putting a 220K resistor in the charge path will allow you to use smaller, cheaper capacitors, but the trade-off is going to be time. I don't think the time involved here is any problem. How much could a tube possibly red-plate in a couple of seconds?
I'm not clear on this. A diagram would be helpful.
Perhaps the Vox circuit -- posted by Paul, with the stacked caps before the bias diode -- is a variation on this theme.
-
He could do that. I don't think his situation is a problem.
I agree, I don't think this is a problem.
With all our circuits, we seem to have some - V on our grids almost immediately.
With the Cathode tied to Gnd, and at least some - V on the grid,
that should help keep some of the newly awakening electrons from rushing to the Plate.
If our bias is rising in relation to the B+ filter caps also charging and the rising Plate voltage,
then we really don't have a condition where our tubes accelerator is mashed to the floor.
I try to protect my vintage tubes as best as I can, so I do understand his concern when he noticed this.
But without any evidence of the tubes red plating, even for only a second or two,
then I don't think this is a real problem.
-
Well today I tried cathode bias in this amp again and I really don't like it. I wanted to like it but I can't! I've experimented with it before and now I remember why I changed to fixed bias... The clean tone is nice but the overdriven tone in cathode bias is just... I don't know, there's a fizz on top of the notes and it's really ugly. It gets too bright in a non-musical way, totally uninspiring to play. And I've experimented with:
- different values for cathode resistor (tried around 70% plate dissipation and about 90% too to see if the tone would improve)
- different values for cathode bypass capacitor (10uF, 22uF, 100uF, 220uF, 1000uF)
- even tried using separate cathode resistors/bypass cap for each power tube.
The only situation where it improved a bit was using a common cathode resistor and NO bypass cap. Anybody has a clue on why? Maybe that's just the nature of the ECL82/6BM8? Even when I tried copying some successful power amp topologies like the 56T (6BM8 version) and the Little Wing I've never been able to get those good tones. The tubes I'm using are Winged C 6F3P, are these any different in practice from the other ECL82's?
Using fixed bias is a completely different story. The overdriven tones sound really good, and no sign of that fizziness...
I found out that Laney uses a separate winding on the power transformer to get the bias voltage for some of their amplifiers. It's the same situation as mine: solid state rectification, and no standby switch. The CUB10 and CUB12 use the same power transformer, and searching I found that it has a 225V winding and a 20V tap for bias. Since the 10 uses 6V6s and the 12 uses EL84s they have slightly bias arrangements... maybe a separate bias transformer would be my best bet?
I have no problems messing with this circuit cause it was all rebuilt already, it's just the aesthetics of the amp that I want to keep original. My concern is that I want this amp to be reliable, that's why I'm worried about the bias taking so long to go up.
-
I'm not clear on this. A diagram would be helpful.
-
If the resistor is on the other side of the diode it is on the AC side.
The pulsing DC coming from the diode is connected directly to the e-cap.
Think about the current going through the resistor and the diode. It is the same. The resistor and the diode both have the same pulsing DC no matter which side the resistor is on. There is no AC on either side of the diode. It may look like there is if you hook your 'scope to it, but the probe makes a circuit with AC that isn't there otherwise.
-
The Cub 10 has a FW rectifier feeding the bias circuit.
If you are still concerned about this bias voltage delay,
maybe making yours FW would reduce the caps charge time?
What do you guys think, time reduced or not?
-
I'm not clear on this. A diagram would be helpful.
But according to eleventeen, the charging time should remain the same. Que passa?
-
What do you guys think, time reduced or not?
I doubt it. There is no getting around the RC constant.
If you make a full-wave bias circuit by putting a 220K resistor and a diode on the other end of the winding, the DC voltage will be significantly more negative than with a half-wave circuit. The full-wave version will have significantly less ripple than half-wave.
How about if we use 47K resistors between the windings and the diodes and then use a lower resistance from bias to ground to get the desired range? The circuit will draw more current, but still negligible. The 47K's will allow the capacitors to charge up much faster. The lower resistance to ground will increase ripple, but the lower ripple with full-wave will counter that.
-
But according to eleventeen, the charging time should remain the same. Que passa?
Orale vato, where did he say that?
-
What do you guys think, time reduced or not?
I doubt it. There is no getting around the RC constant.
If you make a full-wave bias circuit by putting a 220K resistor and a diode on the other end of the winding, the DC voltage will be significantly more negative than with a half-wave circuit. The full-wave version will have significantly less ripple than half-wave.
Why would we get significantly more negative voltage?
The way I look at it, we would just not be flipping the on off switch 60 times a second. The second diode is going to give us the same voltage pulsing DC output as the first one gave us. It will only fill in those blank spots from the 1st diode by giving us another pulse when the 1st one is shut off.
The way HBP had explained this circuit to me before, reducing that resistor is what would give us a significantly more negative voltage.
That seems to make sense to me. :dontknow:
-
Why would we get significantly more negative voltage?
The way I look at it, we would just not be flipping the on off switch 60 times a second. The second diode is going to give us the same voltage pulsing DC output as the first one gave us. It will only fill in those blank spots from the 1st diode by giving us another pulse when the 1st one is shut off.
Yes and that's why. The cap charging refresh rate is now double. As it fills in those "blank spots" the caps do not discharge as much.
So it keeps the cap(s) charged up higher.
-
Yes, but they can't charge up more than the voltage level we are supplying them. :w2:
We don't get double the voltage from a FW rectifier than we do from a HW.
The way I see it, the caps will just charge up twice as fast. :dontknow:
-
The way I see it, the caps will just charge up twice as fast.
Right, but your not seeing the peak charge from the rectified dcv, your seeing the average dcv. The caps hold the charge up to an average which is what we see when we take a B+ reading.
The longer the time between the cap getting charged the more the cap has to release it's charge to hold up the B+ average dcv.
So the cap discharges. As this happens, over and over again, the result is a lower dcv average. It can't keep up the dcv, so the B+ dcv average drops.
But double the charge feed to the cap and there's less discharge of the cap between charges, so the dcv average stays higher with less ripple.
-
"We don't get double the voltage from a FW rectifier than we do from a HW.
It depends whether you are talking about a FW *bridge* or a "two diode" FW rectifier like the type used on the main B+ supply like a Fender Twin, where the two HV legs are fed to two diode anodes (or two series strings of 3 diodes each) and the cathodes (bar ends) are tied together. BOTH types are called "FW". The type using the 4 diodes in the (as usually but not always drawn) diamond is called a "FW bridge".
But the "2 diode" rectifier configuration requires a CT to work. Now I know you are going to say "but the tranny DOES have a CT"....well, yes it does but for the bias supply, the way we are stealing a single HV lead through a single diode is as a half wave rectifier and we are making that circuit between the CT which is grounded, using only half the HV winding, and completely independent of how we are using BOTH halves of the HV winding elsewhere in the amp. So, looking only at that half of the HV winding, across that half-winding, we DO NOT have a CT. Thus we/you can not thinking we have a "2 diode" FW rectifier and thus, if you say "FW" you MUST be referring to a bridge rectifier, and sure enough, a bridge WILL put out double the volts of a half wave. But the final conundrum is that you can not use a FW bridge for this purpose because the "ground" of the bridge rectifier would fight with the "ground" of the other rectifier. Half of the output of a bridge that you tried to use for this purpose would be shorted out. You can't use EITHER type of FW rectifier for the bias without resolving the ground conflict when you are stealing one HV leg to feed the rectifier. You HAVE TO use the half wave.
"The way I see it, the caps will just charge up twice as fast. "
The way the electrons see it, they won't.
-
So we can't just hook up another - diode with the center tap connected to ground, I get that.
If we disconnect the HV CT, and then connect the - diode, then we would have a FW bridge with the + going to B+ and the - going to the bias.
Equal and opposite now as referenced to Gnd.
We get double the voltage from the FW bridge by using the - as the GND reference.
The only problem I see with using a FW bridge by disconnecting the CT, is the path the current will now be force to take to GND.
I'm not sure if that would cause all our B+ current to flow back through bias circuit now or not. :dontknow:
-
It would disable the main rectifier. 6L6's will make no output with -50 on their grids and zero on their plates. That is a perfect certainty. But we could have a lengthy discussion about it, I guess.
-
So we can't just hook up another - diode with the center tap connected to ground, I get that.
Yes we can. We put it on the other hi voltage wind secondary leg.
If we disconnect the HV CT, and then connect the - diode, then we would have a FW bridge with the + going to B+ and the - going to the bias.
No FWB has 4 diodes.
We get double the voltage from the FW bridge by using the - as the GND reference.
No that's not why.
-
6L6s :w2:
This was the circuit he posted about, ECL82 output with PS shown.
Maybe we could discuss this circuit?
-
6L6s :w2:
This was the circuit he posted about, ECL82 output with PS shown.
Doesn't matter what power tube, disconnect the CT no electrons can be brought into the PT secondary wind for current to flow out.
Just like disconnecting a tubes K to ground, no output electrons because there's no input electrons.
-
... I don't mind being wrong about something like this. ...
I think I'll blissfully stay out of this one. But regarding "RC" the "R" is anything/everything in the path between the "C's" two leads.
So winding -> diode -> cap minus -> cap plus -> R -> ground -> winding will be the same effect on time-to-charge as the stock arrangement (which is what I think 2deaf was saying much earlier).
This was the circuit he posted about, ECL82 output with PS shown.
Maybe we could discuss this circuit?
Your drawing, in Reply #2 is functionally-same as c.stoffel's drawing in the original post. If it charges too slowly for his liking, it is because of the 220kΩ resistor (the "R" in "RC"). Sluckey suggested getting an independent winding for bias via an added transformer, of lower voltage, so the "R" could be smaller and speed up the charging, while still maintaining the same voltage output for the overall circuit.
Remember, the definition of "ampere" is a rate of electron flow (1 coulomb past a point per second equals 1 ampere). Resistance slows current, so for same voltage you get less A's. If current is slowed by big-R, then cap charging (which is dumping a bunch of electrons on one of the cap's plates) also slows.
-
6L6s :w2:
This was the circuit he posted about, ECL82 output with PS shown.
Doesn't matter what power tube, disconnect the CT no electrons can be brought into the PT secondary wind for current to flow out.
Really Willabe?
I may not be an electrical engineer or a tube amp designer, but I'm not an idiot.
I was a damn good repair tech, back in the day.
I'm still quite rusty, but things are starting to come back to me.
You've come into this discussion and posted a few things my Army training tells me are just plain wrong.
I haven't attacked you about it, but come on, give me a break.
I'm just trying to have a civilized discussion about these concepts to stimulate my memory.
-
... I don't mind being wrong about something like this. ...
I think I'll blissfully stay out of this one. But regarding "RC" the "R" is anything/everything in the path between the "C's" two leads.
So winding -> diode -> cap minus -> cap plus -> R -> ground -> winding will be the same effect on time-to-charge as the stock arrangement (which is what I think 2deaf was saying much earlier).
I had been hoping you would have weighed in here by now.
You seem to have a way of distilling complex subjects down to something I can understand.
-
I also think Sluckey's idea about adding another transformer is probably the easiest and best solution. :worthy1:
-
I modified my earlier post to add some extra information.
If you have a specific question, I'll try to help. I think everyone else was giving good, correct answers, but perhaps not in a way that turned the lightbulb on for you.
-
I recall in the 6V6 Plexi thread you had indicated that adjusting that resistor would change the voltage level supplied to the bias circuit.
I'm still trying to figure this circuit out completely,
so I can make a reasonable guess as to what resistor I need to put in to my 120 VAC tap to supply my bias board with 25-30 VAC.
-
... I'm still trying to figure this circuit out completely, so I can make a reasonable guess as to what resistor I need to put in to my 120 VAC tap to supply my bias board with 25-30 VAC.
I don't know. I'm sure there's a formula out there somewhere, but I don't know what it is. And that's because while you'd think "Use Ohm's Law," the diode makes the process non-linear and current pulses non-continuous, which makes the math more involved. So what I typically do is guess way higher than I think it should take, and trim the resistance down on test until I get the voltage I want.
I do notice your board specifies an input of 65v. I'd probably be looking for a small transformer which provides that 65v or less to feed the board.
-
Why would we get significantly more negative voltage?
If the HW and FW versions were both charging a purely capacitive load (no resistor to ground), they would both charge up to the same DC voltage which would be slightly less than the peak AC voltage driving the circuit. If you introduce a resistive load in addition to the capacitive load (a resistor to ground), the FW version will have a higher DC voltage than the HW version. If you increase the resistive load (decrease the value of the resistor to ground), the ratio of HW DC voltage to FW DC voltage decreases.
-
There are 3 distinct issues with the ckt you just posted. One is that the input 220K is largish and thus it slows down charging the bias filter caps to their intended steady-state value because the RC time constant of the 220K > 20 uf (effective) is over 4 seconds and you need about 4 of those periods to get the caps near their final desired voltage. To solve this, lower the 220K to some 5-digit value, like 15K, 22K, 33K, 47K.
The second issue is that the way the "adjust" pot is connected to the output of the bias supply, adjusting that pot changes the apparent load on the entire supply and possibly changes the bias supply output. This (meaning, the undesirable effect of this) is exacerbated by the over-large 220K feed resistor. We don't want that. We want the bias supply to see a fixed load and thus (as much as possible for an unreg supply) to provide a fixed voltage. That load should be a stack of three resistors. In idealized form, the middle R is a pot and we take the bias from the wiper. As we adjust the wiper, the supply sees the constant load of the 3 stacked resistors and thus its output does not change. For bias, we are taking only the most miniscule current, we just sticking a charged electrode into the tube. We just "pick off" a voltage of our choice over the range of that pot, but unlike the dwg you show, we do not change the overall load the supply sees. The lowest resistor is the "preventer" that keeps you from being able to adjust the bias so low that the output tubes overcurrent and redplate. That's the third issue. The adjust pot (as wired) changes the load on the whole supply and yes, it will affect the bias delivered to the 470K/470K junction but in a goofy way.
Now you will see bias supplies with the wiper of the adjust pot connected to one end of the adjust pot, making a "rheostat". Fine. Two things about that. One is, that should the wiper fail which is a thing that happens once in a great while, the tubes would lose their neg bias and go right away to overcurrent. So if we connect that pot as a rheo, then the worst that can happen with such a failure is that the bias voltage snaps to the voltage available at one end of that "pot". That means, we would prefer to connect the wiper to the more negative end of the pot so that if it fails, it fails to under versus over current.
And finally, to do the rheo, there needs to be the "preventer" resistor underneath it.
As an extra bonus, a possible 4th problem is that the twin 470K resistors feeding the output tube grids are probably too large. On most amps, those are 220Ks. This may not matter.
-
Boy, I had this drawing all ready to post and then I read eleventeen's post. I think I did some stuff that he said not to do.
If you are going to take the bias from the wiper, you should connect a large resistor from the wiper to the more negative side of the pot. That way, if the wiper fails open, the tube bias reverts to the most negative voltage available from the supply.
I tied the wiper to the more negative side of the pot. The change in the load is only 5K, so I'm not losing any sleep. Should the wiper fail open, the bias reverts to the most negative voltage available from the supply.
This circuit will probably use ten times as much current as the one in the OP. Still only a few ma's and no sleep deprivation. If we went down to 15K, I might get a little concerned about the current consumption.
The ripple should be very small due to the full-wave rectification.
The actual values might need to be adjusted a little if a circuit like this was ever used, but they should be pretty close.
This circuit will charge up lickety-split.
-
There is a 220K resistor between the bias supply and the 470k resistors in the OP drawing. That needs to go away so that the audio signal coming from the PI will have a direct path to ground at the junction of the 470k's.
-
My Plexi 6V6 uses that exact circuit, except I used 25µF caps rather than 10µF caps. It takes an average of 14.8 seconds for the bias caps to fully charge. I used a Simpson 260 and Galaxy S5 stopwatch. The bias voltage just barely beats my cold GZ34 rectifier! If the GZ34 is still warm, B+ comes up faster than the bias. I think I'll change my caps to 10µF.
I changed the caps to 10µF. Now the voltage ramps up to -30v in only 4.25 seconds. Still takes about 15 seconds to reach full charge of my operating bias which is -36v. I'm not nearly as excited now! :icon_biggrin:
-
There are 3 distinct issues with the ckt you just posted. One is that the input 220K is largish and thus it slows down charging the bias filter caps to their intended steady-state value because the RC time constant of the 220K > 20 uf (effective) is over 4 seconds and you need about 4 of those periods to get the caps near their final desired voltage. To solve this, lower the 220K to some 5-digit value, like 15K, 22K, 33K, 47K.
The second issue is that the way the "adjust" pot is connected to the output of the bias supply, adjusting that pot changes the apparent load on the entire supply and possibly changes the bias supply output. This (meaning, the undesirable effect of this) is exacerbated by the over-large 220K feed resistor. We don't want that. We want the bias supply to see a fixed load and thus (as much as possible for an unreg supply) to provide a fixed voltage. That load should be a stack of three resistors. In idealized form, the middle R is a pot and we take the bias from the wiper. As we adjust the wiper, the supply sees the constant load of the 3 stacked resistors and thus its output does not change. For bias, we are taking only the most miniscule current, we just sticking a charged electrode into the tube. We just "pick off" a voltage of our choice over the range of that pot, but unlike the dwg you show, we do not change the overall load the supply sees. The lowest resistor is the "preventer" that keeps you from being able to adjust the bias so low that the output tubes overcurrent and redplate. That's the third issue. The adjust pot (as wired) changes the load on the whole supply and yes, it will affect the bias delivered to the 470K/470K junction but in a goofy way.
Now you will see bias supplies with the wiper of the adjust pot connected to one end of the adjust pot, making a "rheostat". Fine. Two things about that. One is, that should the wiper fail which is a thing that happens once in a great while, the tubes would lose their neg bias and go right away to overcurrent. So if we connect that pot as a rheo, then the worst that can happen with such a failure is that the bias voltage snaps to the voltage available at one end of that "pot". That means, we would prefer to connect the wiper to the more negative end of the pot so that if it fails, it fails to under versus over current.
And finally, to do the rheo, there needs to be the "preventer" resistor underneath it.
As an extra bonus, a possible 4th problem is that the twin 470K resistors feeding the output tube grids are probably too large. On most amps, those are 220Ks. This may not matter.
Cool, looks like this would make a better and safer bias supply. And I think you're correct about the 470K resistors feeding the output tube grids. The ECL82 datasheet specifies a max. value of 2M for cathode bias and 1M for fixed bias, and the 6F3P (which I'm using now) specifies 1M for cathode bias and 500K for fixed bias. I think I'm gonna change them to 220K and double the coupling caps.
There is a 220K resistor between the bias supply and the 470k resistors in the OP drawing. That needs to go away so that the audio signal coming from the PI will have a direct path to ground at the junction of the 470k's.
Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? Take a look at the Fender 5G9 and the Gibson GA-5T (this one also doesn't have a depth control). By the way, the value of this resistor and the others on the bias supply should be considered as part of the resistance from the output tube grids to ground, right?
-
... Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? ...
You could look at it that way if you're applying tremolo.
The bias circuit will charge just as fast with/without it. That resistor (and the bias feed resistors you're talking about dropping from 470kΩ to 220kΩ), as well as the coupling caps feeding the output tube grids, will determine how fast the output tubes recover from an overload input signal. Too slow a response leads to farty blocking distortion. Too small looks like a heavy load on the phase inverter outputs, making it hard for the phase inverter to deliver a clean output signal of the required size.
p2pamps has a tagline: "Everything affects everything." This is an example; don't try to design "the best ever," but instead use existing amps as a guide for what might be "good enough" in terms of all the tradeoffs, and see if the amp needs adjustments after it's built.
-
Well, this amp doesn't have a depth control for the tremolo, so that 220K resistor is there to act like a 'fixed' depth control, otherwise as I see it the tremolo couldn't be applied to that point, am I wrong? Take a look at the Fender 5G9 and the Gibson GA-5T (this one also doesn't have a depth control). By the way, the value of this resistor and the others on the bias supply should be considered as part of the resistance from the output tube grids to ground, right?
I went back and read your original post with the note about Point "A". Disregard my comments on the 220K resistor.
-
I appreciate you all having patience with me and my hard head sometimes.
I need to read over most of your comments many times for it to sink in for me.
Your explanations help me fill in my big gaps I am missing after a long dormant period. :worthy1:
RC charging, ahhh can you say series circuit, Paul?
Sometimes I go off in the weeds, and can't see the forest from the trees.
KISS, Paul, KISS. Keep It Simple Stupid.
I'm really trying to get a strong grasp on all these tube amp PS and bias circuit fundamentals.
Because as was just said "Everything affects Everything" and this is the foundation for it all.
I used to think that e-caps just filtered DC and that's it.
Now, I'm coming to realize that the uF of filtering can influence and tune (sort of) the response of the tubes performance.
"Everything affects Everything", KISS, Paul, KISS.
There is probably damn good reason they used certain cap and resistor configurations.
I'm really getting a little anxious, as I'm about to finalize the modifications to the Plexi circuit to accommodate my quad 6V6s.
And start wiring things in place. I don't want to wreck my vintage stuff with one of my stupid mistakes.
I want to use circuit designs that protect my components while helping them perform at their best.
And that is not as easy as it sounds because, "Everything affects Everything", KISS, Paul, KISS. :BangHead:
-
I went back and read your original post with the note about Point "A". Disregard my comments on the 220K resistor.
That's alright! I'm learning a lot from this discussion.
Another amp I found that uses fixed bias, SS rectification and no standby switch is the Fender Pro Junior. It takes the bias voltage from the HV winding but with a capacitively coupled bias power supply. Is this kind of circuit faster to charge than the one I've been using with a large resistor from HV? Or maybe Fender just don't care if the bias takes a little to go up?
-
Yes, it will charge faster, practically immediately, because there is no series current limiting resistor that ALSO, as it limits current, creates an RC time constant as we've discussed. That .047 cap is not really necessary.
If at any time you wish to seriously increase any remaining bafflement you may have on this topic...go look at a Bogen power supply, where the main B+ comes from a half-wave voltage doubler.
http://el34world.com/charts/Schematics/files/bogen/bogen_chb100.pdf (http://el34world.com/charts/Schematics/files/bogen/bogen_chb100.pdf)
Once again, we are stealing the bias from one of the HV lines that creates the B+ for the amp. In THIS configuration the "blocking" cap that feeds the "backwards" bias rectifier diode ABSOLUTELY MUST be present. And it generally has to be a FAT capacitor, like a .22.
Why do we need that cap? Because at the junction of the stacked diodes that make up the doubler, we have all kinds of pulsating DC flying around, and that would get into the poor, innocent bias supply and ruin its day.
-
Still talking about my amp, does anybody have an idea on why I get such ugly tones I mentioned earlier when cranking the volume using cathode bias (shared or single resistor), but it gets better when I remove the cathode bypass cap(s) and using fixed bias it changes completely with no sign of that nasty tones? And it happens using exactly the same circuit, just changing the bias method.
-
From my rudimentary understanding, the bypass cap increases gain in the tube.
You may be getting too much gain when cranked up that is distorting your signal in a bad way.
Fixed bias doesn't have any bypass cap, as far as I know, so that even when you go cathode biased but no bypass cap the tubes response is similar? :dontknow:
-
The Pro Jr. bias is the old Ampeg thing that Marshall also used in my 900 and Bogen used a variation of. The capacitor and resistor before the diode forms an AC circuit with a real ugly signal. The diode then rectifies this AC into negative DC. These things typically take a lot of time to charge up.
Not only is C12 necessary, but its value is critical. If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced. It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.
The value of R28 is also critical. Values significantly greater than or less than 56K will both reduce the AVOT bias voltage.
Increasing the value of R29 will increase the AVOT bias voltage and decreasing the value of R29 will decrease the AVOT bias voltage.
Unlike many things in guitar amps, these bias circuits have components that must be the value shown and in the positions shown or the circuit won't work. It is an extreme case of everything affecting everything.
The Pro Jr. is a little odd in that it doesn't have a load resistor after the second filter capacitor so that C13 and C14 will charge up to the same voltage. R30 forms a series RC circuit which will slow the charge rate. For some reason I seem to feel that the whole thing is rather wimpy and that contributes to the long charge time.
-
The voltage doubler in the Bogen CHB100 is full-wave at B+ with respect to ground. It is half-wave at the junction of the two 40uf capacitors. The bias circuit is the same idea as the Pro Jr. one. This time a voltage divider is used in the AC portion, but otherwise it is the same.
If you try to take the bias from the other end of the winding, it won't work because there is a straight-wire path in parallel with the bias circuit that is much more appealing to electrons. With a FWB without a center-tap, you can take the bias from either leg.
-
As an Army repair tech we were trained to try to break circuits down into their smallest logical units for troubleshooting.
So for power supplies we would tend to say we have an AC module and a DC module.
AC was pretty much everything up to the rectifier, DC was everything after.
I've noticed this is causing me to miss the bigger picture sometimes.
We were never allowed to modify or improve any circuits design.
So that method was helpful in quickly isolating and identifying faulty components.
But in an "Everything affects Everything" environment I seem to have blinders on now.
I'm hoping I can widen my view now to see how the PS works as a whole unit.
Thank you for the detailed explanations that illuminate the bigger picture for me.
Even if I seem to still be in the dark.
I will reread your detailed explanations until I finally see the light.
Best regards,
-
Not only is C12 necessary, but its value is critical. If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced. It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.
Yes, I'm thinking that the cap input circuit [cap > shunt resistor > diode] is serving double duty as both a voltage divider & a filter, though I don't have my head completely around this yet. See: http://forum.allaboutcircuits.com/threads/resistor-role-in-a-high-pass-filter.47449/ (http://forum.allaboutcircuits.com/threads/resistor-role-in-a-high-pass-filter.47449/) & https://en.wikipedia.org/wiki/Voltage_divider (https://en.wikipedia.org/wiki/Voltage_divider)
-
Look at page 6 of this pdf...
http://sluckeyamps.com/misc/Amp_Scrapbook.pdf (http://sluckeyamps.com/misc/Amp_Scrapbook.pdf)
-
Not only is C12 necessary, but its value is critical. If you change it from .047uf to .022uf, the absolute value of the (AVOT) bias voltage will be dramatically reduced. It you increase C12 to 0.22uf, the AVOT bias voltage will be increased.
Yes, I'm thinking that the cap input circuit [cap > shunt resistor > diode] is serving double duty as both a voltage divider ...
Concur with 2deaf & Jjasilli. The cap and the resistor-to-ground which follow it form a voltage divider for a.c. The cap's portion of that divider is its capacitive reactance (in ohms) at the frequency of the pulsating d.c. applied from the PT/bridge. To recall an earlier discussion, this voltage division can only happen if a.c. is applied to the cap/resistor, because the diode is after them.
Sluckey's info also shows how the baseline is shifted to allow an apparent negative voltage to be rectified by the diode.
-
Look at page 6 of this pdf...
http://sluckeyamps.com/misc/Amp_Scrapbook.pdf (http://sluckeyamps.com/misc/Amp_Scrapbook.pdf)
Very cool! A picture is worth 1000 words :thumbsup:
-
Look at page 6 of this pdf
Sluckey, you need to post when *updates* are available :laugh:
(doc's updated, thx)
-
Look at page 6 of this pdf
Sluckey, you need to post when *updates* are available :laugh:
(doc's updated, thx)
Yes, thank you for that resource Sluckey! :worthy1:
OK, just one more question for the slow learners like me.
We try to knock the voltage down to use lower voltage e-caps, saving money and space.
I already had cheap 10uF 400V ecaps from China I used in my bias circuit.
The formula did not indicate that this would make any difference.
Does using 400V ecaps where 100V are indicated have anything to do with charging time? :dontknow:
-
have anything to do with charging time
volts determine when it explodes, farads determine how long it takes to charge
-
Caps are "analog devices". They have to be physically larger to handle more of something, be it Volts or Farads. That is, they need more of the material they are made from; and hence must be bigger in physical size.
Caps are made of different materials a such as ceramic, silver mica, electrolytic, poly, etc. This affects physical size from type to type. E.g., a ceramic cap may be physically smaller than a poly cap of the same value.
-
OK, so I didn't think the 400V would make a difference.
Thank you for confirming that.
I didn't accurately time my charging, and I probably just considered them fully charged when they got to that 80+% and significantly slowed their charge rate.
Knowing that this amount of - grid V would prevent the tubes from red plating.
I wasn't concerned that my vintage output tubes would be damaged.
In the future if I'm concerned about protecting fixed bias vintage output tubes,
I will look to have a separate tap that has the least amount of R in the charging of the bias caps as possible.
Thank you all for helping me understand these concepts better. :worthy1:
-
I don't get the original point or question. If billions of amps are already working great with the bias supply circuits that have already been designed in the past, it would seem much more logical not to re-invent the wheel and just use what has always been used. If i blow a tube a little sooner than its expected life span I just order a new one, its not a problem they are not that expensive. LOL :icon_biggrin:
-
I think a quest for a deeper understanding of circuits is driving this thread. It is disconcerting to come to the realization that bias voltage is not instantaneous. It does not appear that anyone is trying to reinvent the bias circuit.
-
I am interested in understanding this. Given this info that the bias capacitors take some time to charge ...would it make any difference in an amp tht has a standby switch? The bias caps would be charged and tubes heated, before you hit the standby.
And for those who prefer not to have a standby switch, then it is better to have smaller capacitors with a lower voltage rating that charge faster?
-
Yes. And from a purely practical perspective I agree with both your posts. When all is said & done, the B+ supply is also delayed because of the duty cycle of its R-C networks, and filaments take time to heat up. So bias delay is not fatal. It's easy enough to plagiarize an existing bias circuit; but it's also good to have a deeper understanding.
-
The Cub 10 has a FW rectifier feeding the bias circuit.
What do you guys think, time reduced or not?
Yes, connect another diode from the other leg of the transformer winding and the rise time will increase, but the final voltage will remain the same. Quick, simple, done.
You can also connect a diode between the grid and cathode of each power tube to reduce how hard they conduct as the coupling caps charge, before the bias supply chokes everything off.
-
The Cub 10 has a FW rectifier feeding the bias circuit.
What do you guys think, time reduced or not?
Yes, connect another diode from the other leg of the transformer winding and the rise time will increase, but the final voltage will remain the same.
The only time that a HW and FW bias supply will result in the final voltage remaining the same is when there is a capacitive load, but no resistive load. The bias supply being discussed here has a resistive load, so the FW version will result in a more negative voltage than the HW version. This is not just my opinion. A quick, simple experiment will verify my statements.
The time it takes to charge the capacitive load to 63.2% of the maximum voltage is going to be real similar if not the same whether you use FW or HW. Don't take my word for (nobody was going to, anyway), set up an experiment. Like I said before, use larger values to increase the time so that it is easier to measure.
-
In what will undoubtedly not the be the final word on the topic of whether a cap is required between the HV feed and the input to the HW rectifier supplying bias, here is a page from the 1963 RCA receiving tube manual showing a 7027-fired push-pull amplifier with the cathodes of the 7027s grounded and a perfectly straight-ahead bias supply fed via NO CAP to the cathode of the bias-rectifying diode. I have not experimented with whether that diode can be fed DIRECTLY from the HV feed, it is entirely possible that might not work the same. But the drawing clearly shows that only a voltage-dropping resistor is needed. I myself always place a resistor in this position for the reasons I have stated, it allows lower working voltage caps to be used in the bias supply and provides current limiting into the bias supply.
(http://i172.photobucket.com/albums/w32/ttm4/RCA_50_zpsbltnrmbk.png) (http://s172.photobucket.com/user/ttm4/media/RCA_50_zpsbltnrmbk.png.html)
-
a perfectly straight-ahead bias supply fed via NO CAP to the cathode of the bias-rectifying diode.
The AC voltage feeding that bias rectifier has a zero volt baseline. IOW, the sinewave swings positive and negative. There is no need for a coupling cap to shift the baseline in this case.
-
To say what Sluckey said, in a different way: The circuit just posted has a full-wave rectifier (which a grounded center-tap on the high-voltage winding). The previous circuit which lead to discussion of the cap feeding the bias circuit uses a full-wave bridge rectifier, with no center-tap for the high-voltage winding.