Second design - SE EL84, cascode pre

[tubeswell]

Therefore, bypassing the lower triode’s cathode resistor does not have as much of an effect on bandwidth, as fully bypassing the upper triode’s grid does.

Can you re-word that?  I can't make any sense out of it as it stands.

If you were to bypass either only the cathode of the input triode, or the grid of the output triode, then the latter would have more effect on gain and bandwidth.

[2deaf]

If you were to bypass either only the cathode of the input triode, or the grid of the output triode, then the latter would have more effect on gain and bandwidth.

Thanks.

[shooter]

from observed

that was my working hypothesis 
so, math says in ideal 1/2pirc, but the real world's never ideal 

thanks
dave

[jjasilli]

Shooter, that's the wrong conclusion.  For a standard passive RC-filter, cutoff frequency, the frequency f for which the gain is 3dB lower than the gain in the pass band, is:  f = 1/ (2*pi*r*c).  This is carved in stone.


But a tube is a reactive, not a passive device.  It does not behave like a fixed resistor, especially for AC operation.  So, the passive formula does not fully apply to bypassing a tube element with a cap.


Added complexity:  Merlin states that the performance of actual cascode circuits may not conform to design modeling. 


So 2deaf is building & measuring to find out what's actually happening.

[shooter]

This is carved in stone.

got the tablets 
you're still calculating a passive RC with Rk/Ck, you just need more math to calculate tube dynamics, then put the answers in a yatzee shaker n pour out 

OR
as 2Deaf shows, n my preferred method, observe using math to insure you're watching the correct channel 

[tubeswell]

cap connects as shown in attached.

That certainly makes a lot more sense, but why 470nF?  Looks like it will take a long time to charge when B+ is first applied during which time the gain will be reduced due to the lower "screen" voltage.  It seems to me that 10nF would be plenty for a flat response.

I missed this question before.

Longer charging time also equals longer discharging time, which equals shunting lower frequencies.


Also, everything in the amp takes a while to charge up at startup, so there is nothing unusual about that

[2deaf]

Also, everything in the amp takes a while to charge up at startup, so there is nothing unusual about that

Not to mention the 234 seconds to adjust your shorts and such.

[2deaf]

Yeah,so...I looked at equation (2) in Chapter 12, Section 2 of RDH4 for a minute, maybe a minute and a half.  My motivation crisis has become so severe that I'm not even going to try setting A'/A to 0.707 and rearranging the equation to solve for frequency.  So I decided to run with what PRR said in #42.  Seems to work for triode gain stages, but what about cascodes?  I re-worked the cut-off frequency equation a little.

f_c' = 1/(2*pi*R_k' *C)   Where f_c' is the -3dB cut-off frequency and R_k' is (R_k x r<sub>k) / (Rk + rk).  A slight re-arrangement of that equation gave me:

fc' = (Rk + rk) / (2*pi*Rk*rk*C)

I determined the internal cathode resistance for the different operating conditions and plugged in the numbers.  Gives a real good fc' for the 1K and 2K cathode resistors and close-enough results with the 560r cathode resistor.

(edit: I have apparently angered the font size gods)</sub>

[shooter]

I like it font fail n all 
despite my mathobia I enjoy seeing it played out.

[2deaf]

I started looking at Merlin's Fixed Biasing the Upper Triode on Valve Wizard's Cascode pages.  I just can't see how C_g2 is going to have any effect on the frequency response when there is no audio current in the fixed bias circuit.  What am I missing?

[PRR]

I started looking at Merlin's Fixed Biasing the Upper Triode on Valve Wizard's Cascode pages.  I just can't see how C_g2 is going to have any effect on the frequency response when there is no audio current in the fixed bias circuit.

Do you mean this page, image, text?
http://www.valvewizard.co.uk/cascode.html

"If we choose a value of 560k for R1 and .... R2 ....= 104k ....85k.... This value is used to find a suitable decoupling capacitor. So, for a low roll-off of 20Hz...

I think his fingers got ahead of his mind. The grid is so high impedance that 85k is "zero" out to 20kHz (assume 100pFd Miller C).

But this point "should" be bypassed or it will suck-up crap from adjacent parts and the room; also power supply crap. The "required" value depends on how much crap is around and what your final-test specs are. My gut says it could maybe be less than 0.1uFd, but that's a common part and a smaller value would not save dollars. If you build by the millions, pennies count, better think carefully. But in DIY, 0.1uFd seems a fine value, even if the explanation there is misleading.

[2deaf]

Do you mean this page, image, text?
http://www.valvewizard.co.uk/cascode.html

"If we choose a value of 560k for R1 and .... R2 ....= 104k ....85k.... This value is used to find a suitable decoupling capacitor. So, for a low roll-off of 20Hz...

Yes, that's the one.

0.1uFd seems a fine value, even if the explanation there is misleading.

I figured I had misinterpreted it.  Once again, thanks.

[tubeswell]

Cg2 anchors the g2 voltage under signal conditions (according to the frequency point that the capacitance of the cap can usefully ‘fix’). Without Cg2, the grid would fluctuate slightly, as current through the upper triode changes, because the shunting function of the g2 voltage divider isn’t perfect enough to override the effect of changes in g2 grid current. It is analogous to bypassing (or decoupling) the screen supply in a pentode.

[2deaf]

Cg2 anchors the g2 voltage under signal conditions (according to the frequency point that the capacitance of the cap can usefully ‘fix’). Without Cg2, the grid would fluctuate slightly, as current through the upper triode changes, because the shunting function of the g2 voltage divider isn’t perfect enough to override the effect of changes in g2 grid current.

Merlin says, "The upper grid does not draw any current, so we require only a voltage reference . . . "  This has been found to be true and C_g2 has no effect whatsoever right up to the point of grid-current clipping.  There are actually two grid-current clipping events and C_g2 affects the onset point of both of them.  It only has a subtle effect on the first onset point with a much larger effect on the second onset point.

It is analogous to bypassing (or decoupling) the screen supply in a pentode.

Not really.  We fully bypass the pentode screen in order to keep gain constant and we fully bypass the cascode "screen" in order to prevent unwanted noise at g₂.

[shooter]

um. I'm the pointy hat kid in back;

We fully bypass the pentode screen in order to keep gain constant and we fully bypass the cascode "screen" in order to prevent unwanted noise at g2.

so far I'm holding on, but the "unwanted" crap/noise at the fake g2 - does that hold true in a real pentode where the cap "keeps gain constant", is it also keeping crap out as a bonus?? 

[2deaf]

... but the "unwanted" crap/noise at the fake g2 - does that hold true in a real pentode where the cap "keeps gain constant", is it also keeping crap out as a bonus??

I don't actually buy into the noise at g₂ thing, I'm just playing along.  The gain for a signal injected at g₂ of the cascode is very low as compared to the gain for a signal injected at g₁.  The same is true for a pentode where the gain using the screen grid is way lower than the gain using the control grid.  The lack of amplification for the noise when injected at the "screen" in either case really diminishes the concern about the noise.  But, yeah, I'm sure the real pentode is reaping a bonus by having the bypass capacitor lowering the crap level.

[tubeswell]

It is analogous to bypassing (or decoupling) the screen supply in a pentode.

Not really.  We fully bypass the pentode screen in order to keep gain constant and we fully bypass the cascode "screen" in order to prevent unwanted noise at g₂.

Well yes, ... noise that would otherwise be caused by tiny fluctuations in grid current. If you hold g2 at a constant voltage, and raise and lower the cathode voltage, you will respectively raise and lower the tube current. Most of that current will be sourced through the plate. A smaller amount will be sourced through the mechanism that is used to fix Vg2. The voltage divider used for fixing the g2 bias is a crude shunt stabiliser, which ‘drowns out’ most, but not all, of the (few nano-amps of forward and reverse) g2 current. The cap stabilises Vg2 for the bit that isn’t drowned out, stopping noise.

[Gnobuddy]

so, math says in ideal 1/2pirc, but the real world's never ideal 

There was a very smart man on the Aussie Guitar Gearheads forum, who had a signature line I really liked: "If theory and practice don't agree, you haven't used enough theory."

That simple 1/(2 pi R C) formula works perfectly for a single RC high pass or low pass filter. In the real world, if you use a 1% resistor and a 1% cap, and put them between two op-amp buffers, the real-world frequency response will be within a couple of percent of the theoretical 1/(2 pi RC) formula, over the entire frequency range which the opamps handle well.

But a single resistor and a single cap isn't what we have at the cathode of a triode gain stage. We don't get a simple high pass filter response. Instead, we have a shelving filter, which has *two* corner frequencies - flat at very low frequencies, flat at much higher frequencies, and with a sloping "ramp" connecting those two flat regions. This is because there is more than one resistance involved: not just the external cathode resistor, Rk, but also other invisible resistances inside the tube.

What matters isn't Rk, then, but rather (as PRR said earlier), it is the resistance "seen" at the cathode if you measured it. This is a combination of the external cathode resistor Rk, the internal cathode resistance (same as 1/gm, sometimes called rk), and in the case of a triode, also on external *anode* resistance Ra (aka Rp), the internal anode resistance ra (aka rp), and the mu of the tube!

With a 12AX7, for instance, if we take the nominal mu of 1600 microamps/volt, then the internal cathode resistance is (1/mu), or about 625 ohms. This is not ten times bigger than the typical 1500 ohm external cathode resistance, nor is it ten times smaller. That means we can't just ignore one resistor when calculating corner frequencies caused by the cathode bypass cap: we really have to include both resistances, and that means a considerably more complicated formula than just plain old 1/(2 pi R C).

These days it isn't hard to use a computer to simulate the frequency response, but if we're talking about guitar gain stages which have already been built thousands of times before, why bother? I only bother calculating if I'm using an oddball tube that nobody seems to have used for a guitar amp before, which might need rather different external resistors and cap values to produce the desired frequency response.

And it's really nice to have an actual frequency response measurement, if only to confirm that we have, in fact, used enough theory...so just recently I splurged some fifty bucks ($CAD) on a Syscomp CGM-101 ( https://www.syscompdesign.com/product/cgm-101/ ).

That must have been a closeout deal, as the CGM-101 seems to be gone for good. But it's pretty much exactly what I wanted - eleven bits of vertical resolution but a small 200 kHz bandwidth, useless for today's fast digital circuits, but pretty much what we want for audio measurements. 11 bits of vertical resolution is much better than the somewhat-affordable digital 'scopes like my Rigol, which only has a rather pitiful 8 bits (256 steps) of vertical resolution, but has far greater bandwidth (which is useless for audio.)

I cannot vouch for it's accuracy, but there is an online cathode bypass cap calculator here: https://www.ampbooks.com/mobile/amplifier-calculators/cathode-capacitor/calculator/

The attached image shows the result of running that calculator on a half-12AX7 with a 1uF cathode bypass cap, and the usual 1.5k Rk and 100k Ra (aka Rp). As you can see, it's neither a high pass nor a low pass filter, but instead, a shelving filter with two corner frequencies.

-Gnobuddy

[shooter]

you just need more math

what's the 'ol lyric, different strokes for different folk
I like this formula;

        (time/fun) * $ = theft of others + (individual tweaks)
-------------------------------------------------------------------------------------
                                   musical happiness

[Gnobuddy]

I like this formula;

        (time/fun) * $ = theft of others + (individual tweaks)
-------------------------------------------------------------------------------------
                                   musical happiness

Agree very much about the musical happiness. Just this last Monday (a holiday here in British Columbia), I spent a day of musical happiness at an outdoor jam, held under big shady trees on a beautiful hobby farm in the city in which I live. I met several good musicians, got to make music with several of them, enjoyed the food (potluck) we'd all brought to share, played with a cute dog, and even got to pet a friendly foal who came right up to the fence to have his forehead scratched. It was a wonderful day!

Lately, musical happiness is the motivation for most of the math I'm actually bothered to do - it's almost always in the quest to understand guitar sound better!

-Gnobuddy

[shooter]

 
It should be the inverse of:

       (time/fun) * $ = theft of others + (individual tweaks)
-------------------------------------------------------------------------------------
                                   musical happiness

otherwise the more musical happiness the smaller the outcome 

[2deaf]

But a single resistor and a single cap isn't what we have at the cathode of a triode gain stage. We don't get a simple high pass filter response. Instead, we have a shelving filter, which has *two* corner frequencies - flat at very low frequencies, flat at much higher frequencies, and with a sloping "ramp" connecting those two flat regions. This is because there is more than one resistance involved: not just the external cathode resistor, Rk, but also other invisible resistances inside the tube.

A shelved response is not the same as a shelving filter.  A high-pass filter has a shelved response above the cutoff frequency, but it is not a shelf filter.  A low shelf filter also has a shelved response above the cutoff frequency, but what happens below the cutoff frequency isn't necessarily the same as a high-pass filter.  A low shelf filter affects the frequencies below the cutoff frequency by either cutting them or boosting them.  So the term "high-pass filter" tells you what is going to happen to the frequency response and the term "low shelf filter" only tells you where the frequency response is going to deviate from flat.

[2deaf]

With a 12AX7, for instance, if we take the nominal mu of 1600 microamps/volt, then the internal cathode resistance is (1/mu), or about 625 ohms. This is not ten times bigger than the typical 1500 ohm external cathode resistance, nor is it ten times smaller. That means we can't just ignore one resistor when calculating corner frequencies caused by the cathode bypass cap: we really have to include both resistances, and that means a considerably more complicated formula than just plain old 1/(2 pi R C).

mu is not 1600uA/V, transconductance (g_m ) is.  1/g_m is not the internal cathode resistance of a common-cathode triode gain stage, but it is an approximation of the internal cathode resistance for a cathode follower.  The internal cathode resistance for the common-cathode triode gain stage is (R_p + r_p ) / (u+1).

One time I calculated the parallel impedance of R_k and C_k for a bunch of frequencies and plugged that value into the gain formula in place of R_k .  I converted the results to dB using the fully-bypassed gain as the reference point and graphed it.  I then took an actual 12AX7 and observed the actual gain at those frequencies and graphed that in dB, also.  Those two graphs are similar enough that it could easily be concluded that the 12AX7 is responding to the parallel impedance of C_k and R_k in the same manner as replacing that impedance with a lone resistor of the same value.     

[shooter]

One time I calculated

I thought I got it, but I didn't   


...... 
the 4th read I got 

at those frequencies

fwiw when I get really bored I look at this math, then spend 20 minutes hunting up the graph around here that already did the math, I just can't find that one 

[Gnobuddy]

mu is not 1600uA/V, transconductance (g_m ) is.

Correct. Sorry, that was a minor typo on my part.

  1/g_m is not the internal cathode resistance of a common-cathode triode gain stage

Oh yes, it is (approximately)! More on this in a moment.

but it is an approximation of the internal cathode resistance for a cathode follower.

It is, because it is exactly the same thing we're talking about! The impedance seen from the outside "looking into" the cathode is exactly the same thing as the output impedance of a cathode follower, except insofar as it's modified a bit by the presence of an external anode resistance (Ra). (More on that in a moment, too.)

The internal cathode resistance for the common-cathode triode gain stage is (R_p + r_p ) / (u+1).

Let's think that through for a minute. There are two terms, Rp/(mu+1), and rp/(mu+1). In high-mu triodes like the pair in a 12AX7, the "+1" in  (mu+1) is negligible compared to mu itself, and far less than statistical parameter variations, so we can drop it.

So we now have Rp/mu + rp/mu.

And what is the second term, rp/mu?

We know mu = gm x rp. Divide both sides of the equation by rp, and you get (mu/rp) = gm.

Now take the reciprocal of both sides, and you have (rp/mu)=1/gm.

Ta-da! You say pot-eh-to, I say pot-ah-to; you say (rp/u), I say (1/gm). They are the same thing! 

I did mention that the external anode resistance Ra has an effect, but didn't include it in the approximate formula. I should have, as it is sometimes comparable to the (1/gm) term.

One time I calculated the parallel impedance of R_k and C_k for a bunch of frequencies and plugged that value into the gain formula in place of R_k .  I converted the results to dB using the fully-bypassed gain as the reference point and graphed it.  I then took an actual 12AX7 and observed the actual gain at those frequencies and graphed that in dB, also.  Those two graphs are similar enough that it could easily be concluded that the 12AX7 is responding to the parallel impedance of C_k and R_k in the same manner as replacing that impedance with a lone resistor of the same value.   

You overlooked an error of at least a factor of two. It's easy to overlook a factor of two error on a log-log graph, because of the way a logarithmic axis compresses bigger numbers. On top of that, what tolerance was your capacitor? If it was an electrolytic, it's not unusual to have a -50%/+100% tolerance. That's another huge error.

Try the same experiment again, this time with a 15k cathode resistor going to a negative supply rail (say -15V so the triode still biases up to roughly 1 mA cathode current), and a 1% film cap for the bypass cap, and see what your measurements say. The glaring error in your formula should be clearly visible this time.

This particular error - focusing on the external resistor rather than the combination of external and internal - also plagues many books on transistor electronics. There you frequently see the formula f = 1/(2 pi Re Ce), which is completely wrong; with transistors circuits, the internal emitter resistance is usually far, far smaller than the external resistance, and dominates the frequency response.

-Gnobuddy

[Gnobuddy]

...shelved response...shelving filter...high-pass filter has a shelved response...it is not a shelf filter....A low shelf filter...has a shelved response...

I'm sorry man, you're using a bunch of terms that don't exist in any legitimate electronics textbook, and they make no sense at all to me. "Low shelf filter", for example, makes no sense whatsoever, and has no engineering definition.

But call it by whatever name you want (let's say you want to call it a Pruggle-Flerd filter), the end result is exactly the same: the frequency response caused by a cathode bypass cap is always a flat region at low frequencies, a flat region at high frequencies, and a rising ramp in between.

This is NOT the same as the frequency response of a high-pass filter. The attached image shows the frequency response of an ideal high-pass filter. As you can see, it does NOT flatten out at low frequencies, but falls forever. There is no lower shelf. (A practical implementation of an RC filter will eventually flatten out when it hits the noise floor of your measurement equipment, hopefully at least 90 dB down from the passband region. Very different from a triode with a bypassed cathode cap, which will typically flatten out just a few decibels down from the upper passband region.)

There is more information about the meaning of "high-pass filter" here: https://www.electronics-tutorials.ws/filter/filter_3.html

I hope that helps, and Happy Pruggle-Flerd Filter day to you! 

-Gnobuddy

[2deaf]

I'm sorry man, you're using a bunch of terms that don't exist in any legitimate electronics textbook, and they make no sense at all to me. "Low shelf filter", for example, makes no sense whatsoever, and has no engineering definition.

I'm making up terms that don't exist?  I was quoting you.

Anybody that has done audio editing on a computer to any extent knows what a low shelf filter is.  But you had me doubting myself, so I ran a search on it.  Turns out a low shelf filter is exactly what I said.  I guess I still have a few memory cells left.

[2deaf]

Let's think that through for a minute. There are two terms, Rp/(mu+1), and rp/(mu+1). In high-mu triodes like the pair in a 12AX7, the "+1" in  (mu+1) is negligible compared to mu itself, and far less than statistical parameter variations, so we can drop it.

So we now have Rp/mu + rp/mu.

And what is the second term, rp/mu?

We know mu = gm x rp. Divide both sides of the equation by rp, and you get (mu/rp) = gm.

Now take the reciprocal of both sides, and you have (rp/mu)=1/gm.

Ta-da! You say pot-eh-to, I say pot-ah-to; you say (rp/u), I say (1/gm). They are the same thing!

I did mention that the external anode resistance Ra has an effect, but didn't include it in the approximate formula. I should have, as it is sometimes comparable to the (1/gm) term.

Right out of the gate, the expression involves tomatoes, not potatoes.

You can't just drop R_p out of the equation.  Even when the R_p in the equation is replaced by R_p in parallel with the typical 1M grid leak resistor of the following stage, it is still larger than r_p.  1/g_m is the internal cathode resistance of a cathode follower because R_p is zero and it drops out of the equation.

Isn't the last sentence an admission that you are wrong?

You overlooked an error of at least a factor of two. It's easy to overlook a factor of two error on a log-log graph, because of the way a logarithmic axis compresses bigger numbers. On top of that, what tolerance was your capacitor? If it was an electrolytic, it's not unusual to have a -50%/+100% tolerance. That's another huge error.

Oh no, I didn't.

Listen son, this isn't my first rodeo.  I selected a film 680nF cap that tested out real close to 680nF and a 1.5K resistor that was dead-on 1.5K.  I have a large collection of 12AX7's on hand, so I selected one that had characteristics that were similar to published values.  There is no place to make an error of at least a factor of two when using the formula for the gain of an unbypassed triode.  R_k appears in the formula and r_k doesn't, nor does the parallel combination of R_k and r_k if that is what you are implying.  I have it under good authority that r_k is already factored into the formula for unbypassed gain. 

[Gnobuddy]

Listen son, this isn't my first rodeo.

My daddy's bigger than your daddy.

Have a nice day!

-Gnobuddy

[Gnobuddy]

So how do we calculate the proper bypass cap value?

I say that the formula f = 1/(2 pi Rk Ck) is NOT correct for a cathode bypass cap, and we need a different formula. Arguing about the formula wasn't helping to clarify the issue, or find a solution. So I took a different tack, and used the LTSpice circuit simulator to demonstrate the problem with the formula.

In time honoured mathematical methodology, let's start out by supposing that the formula actually is correct. In that case, as long as we keep the product (Rk x Ck) the same, we should get identical frequency responses, because 1/(2 pi Rk Ck) would be exactly the same.

So, for example, I could use 1.5k and 22uF in one gain stage, then double the resistor to 3k and halve the capacitor to 11uF in a second gain stage. In both cases, the product Rk times Ck is the same (1.5k x 22 uF = 33 milliseconds;  3k x 11 uF = 33 milliseconds; both are the same.) So if our formula is correct, both gain stages should have identical frequency responses.

The attached image shows an LTSpice simulation of two triode gain stages. Both triodes are identical (half a 12AX7), and in the artificial world of the simulator, they are truly identical - there are no normal manufacturing tolerances! The anode resistor and grid bias resistor are also the same for both stages at 100k and 1 megohm respectively.

In the first of the two gains stages, I've used 1.5k and 4.7uF for Rk and Ck. For the second gain stage, I've used ten times the resistance (15k), and one-tenth the capacitance (2.2uF). This guarantees that the product Rk x Ck is identical in both cases, at 7.05 milliseconds. To keep both triodes operating at exactly the same point on the characteristic curves, I've connected a negative supply voltage to the far end of the 15k cathode resistor, and adjusted the voltage so that both triodes show exactly the same current (1.36 mA ) in the simulation.

Both triodes are powered from the same 400V B+ supply. Both anode load resistors are identical (100k). Both anode currents are identical (1.36 mA). This means both anode voltages are also identical (264V in the simulation). Both cathode voltages are identical (+2V in the simulation). Everything is identical in both stages, including the product Rk x Ck. The only difference is that one stage has a ten-times-bigger Rk, and a ten-times-smaller Ck.

So, if our formula is correct, both gain stages will have identical frequency responses.

But look at the result of the simulation. Both stages have identical gain above 3 kHz. But at low frequencies, the frequency responses are not the same, but very different. The 1.5k/4.7uF stage crosses the 14 dB line at 60 Hz. The 15k/0.47uF stage crosses the same 14dB line at 400 Hz!  Even though Rk x Ck is identical in both cases, the 15k / 0.47 uF stage is obviously screaming for a bigger capacitor.

Conclusion: The formula f = 1/(2 pi Rk Ck) is incorrect for cathode bypass capacitors. It predicts a much-too-small value of capacitor in circuits where the cathode resistor is large.

Even when the value of Rk is not unusually large, the formula is wrong - it always predicts a better bass response than you will actually get with that capacitor. But the formula is not as badly wrong when Rk is relatively small, as it is when Rk is much bigger.

(An aside: I set the source voltage to 0.1 volt in the simulator, so the graphs are 20 dB lower than you'd see if I had used 1 volt input. The output voltage above 3 kHz is about +15 dB, but the voltage gain at that frequency is 20 dB bigger, at about 35 dB.)

I'm attaching a screenshot of the simulation. If anyone wants the actual LTSpice simulation file (.asc), or the 12AX7 model (.txt), please let me know, and I'll try and upload them here, if the forum permits those file types. That way, you can run the simulation yourself, and verify what it shows.

-Gnobuddy

[2deaf]

Read this:

[shooter]

 
Thanks 2Deaf, my ADHD likes the readers digest version

[Joe P]

Hello again, folks. Haven't gotten much further (financial reasons), but I'm stuck thinking about the PSU bit. Following the advice given, the plan is as follows:
1. Build a 1974x psu, and make it work
2. Build Epi Vjr. stage by stage and make each work
3. Mod - for this I'm thinking I want to add one of the cascode stages from the über-deluxe posted in this thread, a James TS, and convert the second Vjr stage to a cath follower.

Now, I'd like to have a load for each PSU node, so I can take measurements on each before building the respective stage. However, I'm at a loss as to how to figure out what those loads should be... assuming this schematic for the Vjr: http://mercurymagnetics.com/images/pdf/schematics/wiring/E-VJ-schem1a.pdf and going by this statement from PRR

A happy SE stage acts-like roughly the OT primary impedance. Use 7K resistor for Champ with 7K OT.

in this thread: https://el34world.com/Forum/index.php?topic=21256.0 - I figure B1 would reasonably be 4.7k @ 25W (that's the closest standard value I've found)? What about B2?
And for B3 I guess I could just do the math for 1mA, considering Merlin's article on PS filtering.

And then for future conundrums, what would be the current draw for a 12ax7 cathode follower? 1mA again? And an 6DJ8/6922/E88CC cascode? This datasheet: http://tdsl.duncanamps.com/dcigna/tubes/sheets/amperex/6922-2g.gif gives a "typical characteric" plate current of 15 mA, should I go with that?

Or is this all overkill? Should I just whack a ~5k 25W on the recto tube and be happy?

Lots of questions here, sorry about that...

[shooter]

I "test" my PSU at the 1st tap virtually every time, I've got a bunch of 2K N 5K 50W'rs
here's how I do it;
I start with other ppls design 
grab the tube datasheet
have a "target" 1st tap voltage I'm aiming for.
in this datasheet 250vdc with 1 tube shows average ~~ 48mA @ plate + 5.5mA at G2

so 250vdc/ .04A - ~ 6200 ohms
I gator-clip it to the 1st tap, have my meter clipped across the R and power-up.  I quickly grab the Vdc, power down, do math, if I'm happy, I'll powerup and check for things getting warm.  If it makes it 5min and no disaster I assume the downstream taps will be ok and start my build

[shooter]

things got kinda sideways with the cool kids 

My thinking is, first get the Valve jr/1974 up and running

1. Build a 1974x psu, and make it work

on the recto tube
the schematic you linked has a SS bridge
can you clarify for me or link to your version schematic?

thanks

[PRR]

> me> A happy SE stage acts-like roughly the OT primary impedance.

That's for transformer loading.

Resistor loading, in ignorance, assume the whole stage acts-like twice the big resistor in the stage (usually plate, but in CF the cathode load).

Using Fender 12AX7-100k-1.5k biasing, it comes to more like 3+ times the plate resistor (so 330k); but "twice" is still a fair guess.

> 6DJ8/6922/E88CC ...a "typical characteric" plate current of 15 mA

That is a TV Tuner tube. It is optimized for 100MHz where everything sucks and noise "snow" is a real issue, and real-high current is the main thing.

We never run these tubes that hot for audio. Do you have a plan? Is it a 47k plate load? Then even if (both halves) the tube were dead-short it could only suck 300V/47k = 6mA. And that won't work. More likely at least half the 300V is dropped in tube(s), so half in the 47k, 3mA. Probably less in cascode.

[Joe P]

Wow, that's two weeks that went "poof"... Anyways,

can you clarify for me or link to your version schematic?

yup, it's the psu from this: https://www.thetubestore.com/lib/thetubestore/schematics/Marshall/Marshall-18-Watt-Schem-Schematic.pdf:

Do you have a plan? Is it a 47k plate load?

I snagged the cascode (attached) wholesale from the over-the-top Deluxe Reverb posted previously in this thread: https://sites.google.com/site/stringsandfrets/Home/deluxe-plus. It's bootstrapped by the following CF, upper plate resistor is 100k, lower 68k.

Though that also confuses me, according to Merlin you should be

splitting the previous stage's anode resistor into two equal parts

...

[PRR]

> two equal parts

68k is 100k, pretty near. There's a bunch of trade-offs which get worse if they are very different, but they do not have to be exactly equal.

Your latest image shows C? at 50V. There's four main parts here: resistor resistor tube tube. On the "roughly equal" theory, each gets ~about~ 1/4 of supply. If supply is 300V, this puts 75V on your 50V part.

I also (again?) suspect there is a high-value grid resistor missing. As drawn, the 820r+1uFd suck-out all the treble current.

[Joe P]

> two equal parts

68k is 100k, pretty near. There's a bunch of trade-offs which get worse if they are very different, but they do not have to be exactly equal.

Your latest image shows C? at 50V. There's four main parts here: resistor resistor tube tube. On the "roughly equal" theory, each gets ~about~ 1/4 of supply. If supply is 300V, this puts 75V on your 50V part.

Oh, on the original schematic, there was no voltage rating on that cap, I'll up it then, thanks.

I also (again?) suspect there is a high-value grid resistor missing. As drawn, the 820r+1uFd suck-out all the treble current.

Bad screenshot/bad idea to put the values inside the symbol... That "R" is actually a "k"

[PRR]

> Bad screenshot/bad idea

Sorry. My blind.

[PRR]

Read this:

Read this:

[murrayatuptown]

Good info in here!

The stringsandfrets cascode site is Ulrich Neumann's, not Mr. Aiken's (forgot his first name, but I think he's not Jim Aiken, mon...bad joke). Ulrich redid his website. I couldn't get the earlier link to open but am not sure I see a difference - the one below works for me.

https://sites.google.com/site/stringsandfrets/deluxe-plus

I have been looking at the 1956 (Wireless World?) Hedge amp schematic that uses  a cascode LTP with voltage divider bias on the upper triode and wondering how poorly it would work to grid-leak bias the halves of the LTP.

Murray