Dear folks,
I want to use a SRPP with 6N1P to drive a low impedance ( ~ 10K ). This nice Triode has mu=35, rp = 4400
After reading
The Tube CAD Journal,SRPP Decoded
http://http://valvewizard2.webs.com/SRPP_Blencowe.pdf
http://www.euronet.nl/~mgw/background/tubes/uk_tubesinfo_1.html
I´m a little bit confused about how to determine Zout. The optimum value for Rak seemed to be 750Ohms at Rload=10K for fully bypassed Rk.
Using formular at tubecad I calculate 632Ohm
Using formular at valvewizard I calculate 725Ohm
And using it from euronet.nl I calculate 2250Ohm
If I measure the Output impedance using 1Khz signal i calculate 4889Ohm
What is the "right" formular for a fully bypassed SRPP??
Thanks a lot for any help..
Karsten
I want to use a SRPP with 6N1P to drive a low impedance ( ~ 10K ). This nice Triode has mu=35, rp = 4400
After reading
The Tube CAD Journal,SRPP Decoded
http://http://valvewizard2.webs.com/SRPP_Blencowe.pdf
http://www.euronet.nl/~mgw/background/tubes/uk_tubesinfo_1.html
I´m a little bit confused about how to determine Zout. The optimum value for Rak seemed to be 750Ohms at Rload=10K for fully bypassed Rk.
Using formular at tubecad I calculate 632Ohm
Using formular at valvewizard I calculate 725Ohm
And using it from euronet.nl I calculate 2250Ohm
If I measure the Output impedance using 1Khz signal i calculate 4889Ohm
What is the "right" formular for a fully bypassed SRPP??
Thanks a lot for any help..
Karsten
You may be confusing output impedance with optimum load. They are not the same thing. To add confusion, different people seem to give different formulas. I have been doing an analysis of the SRPP which one day I might publish. My results are:
output impedance Zout is
(Ra1+(mu+1)Rk1 + Rk2)Ra2/(Ra1+(mu+1)Rk1+Ra2+(mu+1)Rk2)
where Ra is anode impedance, Rk is cathode resistor (which may be zero if bypassed), suffix 1 refers to lower valve and 2 to upper valve. For identical valves and lower cathode resistor bypassed (and smallish Rk), this simplifies to (Ra^2)/(2Ra+(mu+1)Rk); in the typical case this will be about Ra/3.
The optimum load, Zopt, set by requiring equal currents from each valve, is approximately
(mu Rk2 - Ra2)/2 -- note that this is often quite small (could even be negative!) unless a large value for Rk2 (upper cathode resistor) is used. This is best done using a mu follower, which is a closely related circuit with different bias arrangements.
Assuming bypassed lower cathode, Rk2=750, Ra=4400, mu=35 I get
Zout = 633 (full formula) or 541 (simple formula)
Zopt = 10925
Bear in mind that you need to use the actual Ra and mu at the chosen bias point, not the quoted values for the valve in general. My guess is that 750 ohms at the cathode will bias the valve back, so Ra will be larger and mu smaller, but I am not familiar with the valve. Check the data sheet.
output impedance Zout is
(Ra1+(mu+1)Rk1 + Rk2)Ra2/(Ra1+(mu+1)Rk1+Ra2+(mu+1)Rk2)
where Ra is anode impedance, Rk is cathode resistor (which may be zero if bypassed), suffix 1 refers to lower valve and 2 to upper valve. For identical valves and lower cathode resistor bypassed (and smallish Rk), this simplifies to (Ra^2)/(2Ra+(mu+1)Rk); in the typical case this will be about Ra/3.
The optimum load, Zopt, set by requiring equal currents from each valve, is approximately
(mu Rk2 - Ra2)/2 -- note that this is often quite small (could even be negative!) unless a large value for Rk2 (upper cathode resistor) is used. This is best done using a mu follower, which is a closely related circuit with different bias arrangements.
Assuming bypassed lower cathode, Rk2=750, Ra=4400, mu=35 I get
Zout = 633 (full formula) or 541 (simple formula)
Zopt = 10925
Bear in mind that you need to use the actual Ra and mu at the chosen bias point, not the quoted values for the valve in general. My guess is that 750 ohms at the cathode will bias the valve back, so Ra will be larger and mu smaller, but I am not familiar with the valve. Check the data sheet.
Dear DF96,
Excelent! Thank you so much for providing proved formulars. I will try to reproduce this in the next evenings ( Thanks a lot again to give me some reasons to vanish in my basement )
So the hope is real now to drive a Zload of ~10K using a SRPP stage.
Deep bow about your knowledge..
regards
Karsten
Excelent! Thank you so much for providing proved formulars. I will try to reproduce this in the next evenings ( Thanks a lot again to give me some reasons to vanish in my basement )
So the hope is real now to drive a Zload of ~10K using a SRPP stage.
Deep bow about your knowledge..
regards
Karsten
I'm not sure, I haven't seen that one - do you have a reference? My results are slightly different from Blencowe (and Broskie, IIRC). Also, Morgan Jones gives a formula which is different from mine but his numerical result agrees with mine.
So, Karsten, please don't bow too low - I'm smart, but not infallible!
If anyone else wishes to have a go themselves, I will summarise my analysis. The SRPP starts as a common cathode amp with an active load. The complication is that the output is taken from the top cathode instead of the bottom anode. This means that the open circuit voltage gain is reduced a little because of the signal voltage drop across the upper cathode resistor. The short circuit current output can be worked out from the lower valve simply driving the upper cathode resistor, then this voltage is multiplied by gm. The ratio between open circuit voltage gain and short circuit transconductance is the output impedance. (Note that this is circuit transconductace, not valve transconductance).
The optimum load is approximately given by both valves contributing equal signal currents. There is a bit of slack here, which may be why different people get slightly different formulas. The differences are small, though, the distortion has a broad minimum and valve characteristics will not be precisely known so any formula should only be taken as a rough guide.
So, Karsten, please don't bow too low - I'm smart, but not infallible!
If anyone else wishes to have a go themselves, I will summarise my analysis. The SRPP starts as a common cathode amp with an active load. The complication is that the output is taken from the top cathode instead of the bottom anode. This means that the open circuit voltage gain is reduced a little because of the signal voltage drop across the upper cathode resistor. The short circuit current output can be worked out from the lower valve simply driving the upper cathode resistor, then this voltage is multiplied by gm. The ratio between open circuit voltage gain and short circuit transconductance is the output impedance. (Note that this is circuit transconductace, not valve transconductance).
The optimum load is approximately given by both valves contributing equal signal currents. There is a bit of slack here, which may be why different people get slightly different formulas. The differences are small, though, the distortion has a broad minimum and valve characteristics will not be precisely known so any formula should only be taken as a rough guide.
Split the SRPP resistor in half and tap your output at the middle.
Will make both triodes work equally into the load. Zout increases,
this is true. But Zopt decreases to lowest possible for that SRPP,
as the slew limits become symmetrical.
Of course you must include cathode follower impedance as part
of "the SRPP resistor" when deciding where the true middle is...
Pretend the top cathode is loaded with a CCS, and what would
that impedance be? Then add SRPP resistor, and divide by two.
Thats the point where you can best drive a challenging load...
Strangely enough, plate resistance of the lower triode doesn't
figure into finding the middle. It does however, figure into Zout.
(Plate resistance + Cathode resistance + SRPP resistor) /2
Will make both triodes work equally into the load. Zout increases,
this is true. But Zopt decreases to lowest possible for that SRPP,
as the slew limits become symmetrical.
Of course you must include cathode follower impedance as part
of "the SRPP resistor" when deciding where the true middle is...
Pretend the top cathode is loaded with a CCS, and what would
that impedance be? Then add SRPP resistor, and divide by two.
Thats the point where you can best drive a challenging load...
Strangely enough, plate resistance of the lower triode doesn't
figure into finding the middle. It does however, figure into Zout.
(Plate resistance + Cathode resistance + SRPP resistor) /2
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I have had another look, and I still get the same result:
output impedance Zout (DF96) is
(Ra1+(mu+1)Rk1 + Rk2)Ra2/(Ra1+(mu+1)Rk1+Ra2+(mu+1)Rk2)
If I assume upper and lower values are the same (i.e. unbypassed lower) then this becomes
(Ra+(mu+2)Rk)Ra/(2Ra+2(mu+1)Rk)
If the lower resistor is bypassed then I get
(Ra+Rk)Ra/(2Ra+(mu+1)Rk)
Blencowe has
(Ra+(mu+2)Rk)(Ra+2Rk)/(2Ra+2(mu+2)Rk)
but I am unclear whether this is with lower bypass or not (probably unbypassed). He has a few extra Rk in both numerator and denominator. In most cases the numerical difference will be small.
We both get the same result for optimum load: (mu Rk - Ra)/2
Blencowe mentions adding an anode resistor (equal to Rk or 2Rk) to the upper valve for better balance. I am unclear why. He says that omitting it makes little difference. It may be that his Zout formula assumes that this extra resistor is present; this might explain the extra Rk which appears.
output impedance Zout (DF96) is
(Ra1+(mu+1)Rk1 + Rk2)Ra2/(Ra1+(mu+1)Rk1+Ra2+(mu+1)Rk2)
If I assume upper and lower values are the same (i.e. unbypassed lower) then this becomes
(Ra+(mu+2)Rk)Ra/(2Ra+2(mu+1)Rk)
If the lower resistor is bypassed then I get
(Ra+Rk)Ra/(2Ra+(mu+1)Rk)
Blencowe has
(Ra+(mu+2)Rk)(Ra+2Rk)/(2Ra+2(mu+2)Rk)
but I am unclear whether this is with lower bypass or not (probably unbypassed). He has a few extra Rk in both numerator and denominator. In most cases the numerical difference will be small.
We both get the same result for optimum load: (mu Rk - Ra)/2
Blencowe mentions adding an anode resistor (equal to Rk or 2Rk) to the upper valve for better balance. I am unclear why. He says that omitting it makes little difference. It may be that his Zout formula assumes that this extra resistor is present; this might explain the extra Rk which appears.
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