SRPP DC working condition for Rk1=2Rk2

[slav]

Hi
I have question as in the title , how to establish DC working conditions knowing Rk1(upper valve cathode resistor) is 2 times bigger than vcathode resistor of lower valve? Under normal circumstances is quite easy (Rk1=Rk2) , voltage on the plate of lower valve is a half of voltage on the upper plate .
What is the reason to use such a conditions?
Thanks

[DF96]

To find the DC conditions you would need to use the characteristics from the data sheet. Raising Rk2 increases gain and reduces output impedance.

[slav]

Ok
Let's say I know voltage supplied to the upper anode ,I know the current flowing through both triodes is the same ,I have both Rk values , I have characteristics .What then?

[DF96]

Iterate? I'm just guessing, having never needed to do this.

[slav]

Iteration seems like long process , there must be at least some graphical method.
Anyone?

[kevinkr]

Someone here derived all of the SRPP equations some time ago including I think for cases of dissimilar RK1 and RK2 - not the case for the standard equations.

The easiest way would probably be to install something like LTSpice on your PC, find the appropriate tube libraries which should also be here (I've posted my LTSpice compatible library here in the past) and model your circuit in Spice.

Morgan Jone's book "Valve Amplifiers 3rd Edition" is an excellent general reference which I highly recommend, it does not however answer your question..

Note that the answer is not directly correlated to the ratio of the cathode resistors although that might get you some of the way there.

[DF96]

Calculating gain and output impedance for dissimilar cathode resistors can be done from equations, but I don't think that is what the OP is asking. He is interested in working point, which I think has to be done graphically or via simulation.

However, given the same current the grid-cathode voltages will be in ratio 1:2 so it is likely that anode-cathode voltages will be in a similar ratio. So roughly a third across the lower valve, and two thirds across the upper valve.

[kevinkr]

Originally Posted by DF96

However, given the same current the grid-cathode voltages will be in ratio 1:2 so it is likely that anode-cathode voltages will be in a similar ratio. So roughly a third across the lower valve, and two thirds across the upper valve.

This rule works ok with the 12AX7/ECC83 300V/850uA but seems not to be the case with the 6DJ8 at 180V/7mA where the difference is between 10 - 20%.. (68Vp lower tube/112V across upper tube, situation gets worse quickly at lower plate voltages like 150V, noting that much above 180V is pushing it with the 6DJ8 in this configuration) The higher transconductance types with steep Vg/Ip curves seem not to follow this rule that closely. The above simulations jibe well with prior bench experience, but may or not correlate well with any particular given tube. YMMV...

[Merlinb]

Originally Posted by kevinkr

The higher transconductance types with steep Vg/Ip curves seem not to follow this rule that closely.

The voltage across either valve is:
I*ra + mu*vgk
= I*(ra + mu*Rk)

Since the upper valve has twice the bias there will be an additional mu*Rk volts across it, compared with the lower triode (where Rk is the low triode's cathode resistor). So you are right, the voltage is not shared in a 2:1 ratio, unless I*ra is very small.

Incidentally, for identical triodes the maths is easy. The anode current is:
I = HT / (ra+muRk + ra + 2muRk)
= HT / (2ra+3muRk)
(where Rk is the lower cathode resistor and the upper cathode resistor is assumed to be twice as great)

Then just multiply by ra+muRk to find the voltage across the lower valve, or ra+2muRk for the upper valve.

[revintage]

Hey Slav,
Out of curiousity. Why do you want do this?