Does the anode resistance of a triode change throughout a waveform?

[anson]

Hi folks,


I was just reading Merlin's article on triode gain stages, and have a question about the anode resistance of a triode gain stage. Basically the article says that the anode resistance can be calculated from the slope of the grid curve, or by this:

anode R = delta V / delta I

where delta V / delta I are the slope of the grid curve around the locality of a given bias point.

http://www.freewebs.com/valvewizard1...Gain_Stage.pdf

page 22

But a grid signal doesn't just hang around the bias point--it increases and decreases in voltage around it, and at some time during its excursion it might go down to -4v or so, where the grid curves are nearly half what they are for a center biased stage of say 1.5v. So does that mean that while the waveform travels momentarily to -4v, the anode resistance at that time would be increased because of greater delta V/delta I value around -4v?

I was wondering about this because if the anode R changes throughout a wave form, then it will have an effect on the high frequency roll off point of treble control circuits that rely on a given value of output impedance.

Thanks!

[Alex R]

This is a speculative rather than an authoritative reply, but I would have expected many of the parameters in a tube to change whilst it follows a signal wave up and down. Interested to hear a more authoritative answer.

[Steve Conner]

Yes, it does. All tube parameters are linear approximations valid only for small deviations about an operating point. Voltage swings in tube amps are anything but small.

As an extreme example, when the tube is cut off and passing no current, the anode resistance is obviously infinite, because a small change in plate voltage will cause no change in plate current, since there isn't any.

I have heard it said that the cathode follower driving the Marshall tone stack cuts off completely on negative signal swings, because of the stored charge in the bass capacitor, which must surely have a "large-signal" effect on the bass response in the way you suggested.

Generally, the parameters all vary as the three-halves or five-halves power (or was it two-fifths or two thirds? ) of the plate current, plus other deviations from that law (such as cutoff) that are specific to the details of the tube construction, and give different brands of tubes their different tones.

[J M Fahey]

Yes it does.
Even more important, the curve non-symmetry means that that triode won't amplify the same both halves of a sine wave, at *all* frequencies, generating even harmonics .... but you already knew this.
Besides that, the ability to source/sink current is not the same.
In the cathode-follower example, the 100K resistor to ground can only sink what a 100K resistor would (duh), if the DC point is, say, 200V, it is passing and can sink 2mA tops, (on the negative half of the sinewave) while on the positive half it can source whatever the triode can provide: at least 3 or 4mA for a 12AX7 and easily 10/12mA for a 12AU7.
Into a relatively low load impedance (the tone control) the waveform will be *very* unsymmetrical, which contributes to the *real* Mojo.

[kg]

i eat mojo for breakfast!

can't get enough...

[Alex R]

Yes it does.
Even more important, the curve non-symmetry means that that triode won't amplify the same both halves of a sine wave, at *all* frequencies, generating even harmonics .... but you already knew this.
Besides that, the ability to source/sink current is not the same.
In the cathode-follower example, the 100K resistor to ground can only sink what a 100K resistor would (duh), if the DC point is, say, 200V, it is passing and can sink 2mA tops, (on the negative half of the sinewave) while on the positive half it can source whatever the triode can provide: at least 3 or 4mA for a 12AX7 and easily 10/12mA for a 12AU7.
Into a relatively low load impedance (the tone control) the waveform will be *very* unsymmetrical, which contributes to the *real* Mojo.

The first stage of a Vibroking is a cathode follower. It is there to drive the three-knob reverb that is built into the amp. I was once asked to wire a VK preamp minus reverb into a wrecked old Bassman and did so without the cathode follower. But it didn;t sound right till I put the cathode follower in - it gave it a zingy touch-sensitiveness. I have often wondered why, and now wonder whether this is the explanation.

[anson]

As an extreme example, when the tube is cut off and passing no current, the anode resistance is obviously infinite, because a small change in plate voltage will cause no change in plate current, since there isn't any.

That's really interesting. So that means that when the valve is in cut off, the plate resistance is huge, so if our anode resistor is 100k ohms, then for that moment the output impedance will be about 100k ohms.

And then for the rest of the waveform, it seems that since the load line is relativelly linear we could say that the output impedance is delta V/delta I, which is about 70k ohms. Then the output resistance will be the 70k ohms plate resistance in parallel with 100k anode resistor. That's 41k ohms. Assuming fully bypassed Rk.


That's a change in output impedace by about a factor of two. If I'm connecting a "snubbing" capacitor from ground to the junction of the plate and the 100k anode resistor, then the higher order harmonics from the "soft clipping" portion of the wave will be rolled off more due to the higher output impedance, and the higher order harmonics from the grid clipping will be less rolled off due to the decreased output impedance. ...If I'm understanding things correctly that is.

Like Alex R says, that might explain the different tone characteristics I noticed by putting tome controls and frequency shaping circuits in different places.

Thanks for the feedback!