Pentode CCS: Would somebody check my work?

[nullspace]

Hi All --

I've built some stuff from scratch before, but always using triodes. For my next project I was thinking of using a pentode CCS, and I was hoping to get some confirmation as to whether or not I know what I'm doing.

There will be 320v at the top of the circuit and 150v at the bottom; current draw should be ~20ma.

1) From looking at the data sheets, it looks like the anode will draw just under 18ma and the screen just over 2ma.
2) R1 -- So, 10k will put the screen about 20v lower than the anode.
3) C1 -- I frequently see 10uf here. Any other suggestions, or a formula to calculate the optimal value?
4) R2 -- Grid stopper. 1k okay here?
5) R3 -- 510ohms will drop about 10v at 20ma, which is what get looking at the loadlines; ~140v on the anode, -10vgk, and 20ma. Am I missing something here.
6) R4 -- 1k drops another 20v at 20ma. I've read that this resistor 'stiffens' the CCS. Thoughts?
7) C2 -- No idea what this capacitor does, but I generally see it in schematics, so I assume it belongs. Could someone enlighten me on what it does?

Thanks for any help I receive; I really appreciate it.

Regards,
John

[ray_moth]

I take it you're thinking of using the pentode CCS as a plate load for a triode? If so, take a look at Alan Kimmel's Mu Stage Philosphy article. It explains a lot about pentode CCS as a plate load.

Notes (based on my assumption of your intentions)

(a) R2 is not a grid stopper, it's a normal grid leak (or whatever it should be called) and its resistance should be around 0.5 Meg.

(b) The purpose of C2 is to pass the signal from the plate of the (triode) tube being loaded to the grid of the pentode. This makes the signal appear at the cathode of the pentode and has the effect of making the triode's load resistor R5 appear to be much larger that it actually is, giving a very high AC load to the triode and an almost horizontal load line. Gain will be close to the triode's mu and distortion will be very low. Output impedance is also low (best of both worlds!) if the output signal is taken from the cathode of the pentode. If OP impedance is not important to you, you can try taking the output signal from the triode's plate, to find out if it sounds any better.

(c) R5 should be connected directly to the cathode of the pentode, not via R3. Its value should be ~10k to 68k, depending on the mu of the triode.

(d) There should be a resistor from the junction of R2 & R3 to ground. Its value should be ~150/0.02 = 7.5k. This needs to be a high wattage resistor, say 10w to be safe (it will dissipate 3w).

[If your CCS is not intended as a plate load, please disregard the above!]

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Thanks for the help Ray. I had seen that article before, but it made a lot more sense the second-time around.

I'm hoping to use the CCS in place of the voltage dropping resistor on top of a VR tube (0D3) in much the same way Gary Pimm does in his projects. I've seen his circuits, but they're a bit more than I want to bite off for right now. I've used the resistor/OD3 circuit before, now I'd like to try a simple pentode CCS, then perhaps move on to one of GP's hybrid mosfet/pentode CCS.

Regards,
John

[ray_moth]

I use a pentode CCS in the tail of my 6SL7 long tail pair phase splitter. In other words, the CCS is in place of a resistor. For such use, you don't need C2 or R4 in your circuit. R2 then becomes a stopper, as you said before.

One problem of using a pentode for this purpose is that the screen voltage controls the current, so your CCS performance is entirely dependent on the stability of the screen voltage. You can cure that problem by using a high gain device that doesn't have a screen, i.e. SS.

[nullspace]

Thanks again Ray. I had gone back and looked at Kimmel's article again,and had concluded that I could use the CCS depicted in figure 3 of his writeup.

I clearly need to do some more reading on pentodes and operating points, because I was under the impression I could pick a plate voltage, grid-to-cathode voltage, and consequently a cathode current (just like a tride) and then the current would be split by a fixed ratio peculiar to the choice of pentode between the plate and screen. Pick a screen resistor to drop some voltage, and bob's your uncle. Any links you could offer me so I can continue my education on pentodes would be very much appreciated.

Regards,
John

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Thanks again Ray. I think I'm getting it now.

Here's a revised schematic. Gary Pimm recommends biasing the 6BQ5 at -5v, so I changed the cathode resistor to 250 ohms. I plugged in 47k for the screen resistor to start -- I'll measure the voltage drop across the cathode resistor, and if it's <5v then I'll decrease the size of screen resistor. If it's >5v, then I'll increase the resistor. And so on, until I'm getting a 5v drop and consequently 20ma. Does that sound about right?

Regards,
John.