This is a diagram of a pentode power amplifier: enter image description here

This amplifier is used to control an electro-pneumatic actuator. Let me describe it a little bit:

  • the input voltage (the control signal) is V3 (-30V < V3 < +30 V)
  • the output is at two pins of C8 (or two anodes of P3 and P4);
  • There are two stages: stage 1 includes P1 and P2, stage 2 includes P3 and P4. When the input V3 changes, the voltage difference between two anodes of P1 and P2 changes correspondingly. The signal is sent to stage 2 through voltage dividers R6-R7 and R8-R9.
  • Z1 and Z2 represent two coils inside the electro-pneumatic actuator. When the actuator moves, it constantly feeds a singal back, represented by V4.
  • V1 and V2 are constant voltage sources, V1 = -150V ; V2 = +150V ;

The whole purpose of this amplifier is to keep V4 = V3. In the other word, for example, if we set V3 = 10V, then the actuator will move so that its feedback V4 = 10V.

The pentodes used in the diagram are 6Ж5Б-B, which has following properties: enter image description here

I would like to know how this circuit works. I put this on LTSpice but it doesn't work because LTSpice does not have pentode models, just icons. I put some pentode models that I downloaded from Internet but it still doesn't work, one of the reasons is the models are not 6Ж5Б-B.

So I decide to understand this by hand-calculation using the characteristic curves of 6Ж5Б-B. For example, what are the current in the anodes of P1 and P2 when V3 = 10V? Even more, what are the purpose of C3 ; R12 and R13 ; R29 and C9 ; C8...?

  • 3
    \$\begingroup\$ absolutely no expert here, but it looks a bit like a differential input pair (p1 & p2) and a differential power stage (p3 & p4). So basically a fully differential op-amp of sorts. \$\endgroup\$
    – tobalt
    Oct 15, 2021 at 9:03
  • \$\begingroup\$ Try micro-cap - it has a 6K5B pentode as a model although your part number appears to be Russian maybe? \$\endgroup\$
    – Andy aka
    Oct 15, 2021 at 9:16
  • \$\begingroup\$ @tobalt Could you draw an equivalent ciruit using a fully differential op-amp? I think the circuit will have an op-amp and serveral resistors as we ignore the pentodes. \$\endgroup\$
    – Dat
    Oct 15, 2021 at 9:23
  • \$\begingroup\$ @Andyaka great, is 6K5B equivalent to the characteristics of 6Ж5Б-B ? \$\endgroup\$
    – Dat
    Oct 15, 2021 at 9:24
  • 1
    \$\begingroup\$ I bet Mr. Carlson would know. Hopefully this link gets his attention. \$\endgroup\$
    – rdtsc
    Oct 15, 2021 at 11:51

1 Answer 1


You can probably substitute just about any pentode model (with anode current rating at least 32mA) and have it work reasonably well. Valve circuits are usually quite tolerant of parameter variations; by design, since characteristics drift as the valves age.

So go ahead and try the 6K5B. (And something else, maybe EL84 for luck, and compare the results) (EDIT : my Brimar book lists the 6K5G as a triode, with 1mA anode current, but the 6K6G is a power pentode with 32mA anode current so a pretty good match. Other possibilities : 6V6, 38mA max anode current, 6L6 even more)

Some basic analysis : the first stage is a long tailed pair, acting both as a gain stage and a phase splitter. Note the high values of R1,R3,R4 : anode current must be in the range 0 to 200 uA, suggesting Vg-k for each will be around -4.5V.

With Vin +ve, P1 will be turned ON, and P2 (with its grounded grid) will be OFF : following their anode voltages to the output stage, P3 will be OFF and P4 ON, pulling Z1 low and providing NFB via R12,R25. So far, so good... Then the actuator will presumably drive V4 until P1 grid matches P2 grid (0V). Just like an opamp.

So... in balance, R4 and the cathodes must be around +4.5V to +5V (-Vg), giving us the cathode current (155V/0.82Meg = 190uA), split between P1,P2 cathodes, 95 uA each.

Anode load is 1Meg to 150V (R1) in parallel with 3Meg (R6,R7) to -150V, or by Thevenin, 0.75Meg to 75V. But the cathode current is split between anode and screen grid (and the screen grid voltage is set by R2,R5 to about 25V max, further reduced by screen grid currents). So if we assume that half of the cathode current becomes anode current; say 50uA, the anode voltage would be 37.5V below the 75V Thevenin source, i.e. 37.5V. Pretty low.

Then we can establish Vg of P3,P4 at -37.5V, and their cathode voltage somewhere around -35V, giving 115V across R10,R24 (4.36K) or around 25mA shared between P3,P4 to drive the actuator.

Once the basic amplifier is understood and working, the negative feedback by R12,R13,R25 appears to control the P term (if you view it as a PID controller), R29,C9 provide some sort of I term, and C3 across R8 may provide a D term. These should allow tuning the controller for the (unknown) characteristics of the actuator.

  • \$\begingroup\$ I was wondering how the first-stage long-tailed pair (P1+P2) splits the phase = inverts phase for P2's anode/output, considering that P2's gate is grounded. A bit of a "back to school" moment for me... Turns out that the coupling happens via the shared cathode node in the circuit. As P1 pulls up (P1 anode current increases), the cathode node gets higher relative to GND, which makes P2 close a bit = P2 anode current decreases. Hence the phase inversion. The power stage is clear. To me as an op-amp kid, the feedback is interesting. I'd hook this up via the P2's gate (wrong polarity though). \$\endgroup\$
    – frr
    Oct 15, 2021 at 13:54
  • \$\begingroup\$ @frr that's right : the key is the cathode current is held "constant" - ideally by a constant current source, but R4 is close enough for valve work, so as P1 adds current, it subtracts from P2. \$\endgroup\$
    – user16324
    Oct 15, 2021 at 14:05
  • \$\begingroup\$ @user_1818839 Just a quick note, first scan. The screen voltage on U1 and U2 is nominally about +21 V, which means the plate voltage cannot be less than this. (Electrons would flood the screen instead of the plate, otherwise, especially so with the suppressor sitting at cathode voltage.) So the long tailed pair cannot have a plate current of more than about 120 uA as a first-pass guesstimate. (U3 and U4 have self-biased screens.) \$\endgroup\$
    – jonk
    Oct 15, 2021 at 20:20
  • \$\begingroup\$ @jonk We seem to be in close agreement on screen grid voltage. However, anode voltage certainly can fall below sSG voltage (and yes, that does increase SG current). Your estimate of 120uA in both anodes is IMO quite close to my assumption of 50uA in each; the remainder of the cathode current going to the screen grids. \$\endgroup\$
    – user16324
    Oct 15, 2021 at 20:35
  • \$\begingroup\$ @user_1818839 Yes. That was just a first pass as just a quick, once-over. Of course, there actually will be some screen current. But the impedance is only about 78k there. So we are diddling over a few volts difference, I think. I didn't bother with the offered charts to try and figure out the DC bias point, which is also impacted by the DC biasing via R12+R13 divider and necessary feedback through R25 as the cathodes also have to do a bit of self-bias through R4. Last time I designed a tube circuit was 1969, though. Long time, I wasn't good at it even then, and much forgotten. \$\endgroup\$
    – jonk
    Oct 15, 2021 at 20:41

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