Tie both cathodes together and bias with one LED?

[Wombaticus]

If the two stages are in series the signal is flipped 180* from one triode to the next. So the cathodes should be out of phase with each other. And that should be negative feedback. Of course you could always omit the issue altogether by fully bypassing the diode. If that's even needed. I've read that it isn't but I also know that diodes drop about a half a volt. I think that could show up as head voltage on the cathodes. So just parallel a 47uf cap with the diode and any feedback, positive or negative, is eliminated.

What you're missing is that the phase cancellation on the cathode leads to a *maximization* of the difference between grid and cathode on the first tube. So it is effectively positive feedback as suggested by Loudthud. To see that this must be the case one need only consider the degenerative feedback that occurs in the case of an unbypassed cathode resistor. The cathode following the grid in this case minimizes the grid-cathode potential and reduces gain.

Finally the DC voltage produced across the LED is irrelevant. What matters is the AC impedance, and this is surely very small. I *doubt* that there is much potential for deleterious positive feedback, although I'm willing to be proved wrong.

[printer2]

Common cathode resistor in an early Princeton.

[Chuck H]

What you're missing is that the phase cancellation on the cathode leads to a *maximization* of the difference between grid and cathode on the first tube. So it is effectively positive feedback as suggested by Loudthud. To see that this must be the case one need only consider the degenerative feedback that occurs in the case of an unbypassed cathode resistor. The cathode following the grid in this case minimizes the grid-cathode potential and reduces gain.

Either this is a poor explaination or I'm dumb as a rock (I'll accept either gracefully). What is the "difference between the grid and cathode"? The phase difference? AC differential? I realize tha the signal on the following cathode is greater than the first. So without bypassing the cathode of the first triode becomes an input, somewhat. And the signal from the cathode of the following triode is in phase with the grid of the first triode. Is THIS where this is going?

And, FWIW, a half a volt on the cathode should be enough to make a difference. Unless I'm wrong a diode will drop a half a volt no matter what the actual circuit voltage is.

Just bypass and be done.

[Chuck H]

Note the fully bypassed cathode resistor.

[Wombaticus]

Either this is a poor explaination or I'm dumb as a rock (I'll accept either gracefully). What is the "difference between the grid and cathode"? The phase difference? AC differential? I realize tha the signal on the following cathode is greater than the first. So without bypassing the cathode of the first triode becomes an input, somewhat. And the signal from the cathode of the following triode is in phase with the grid of the first triode. Is THIS where this is going?

And, FWIW, a half a volt on the cathode should be enough to make a difference. Unless I'm wrong a diode will drop a half a volt no matter what the actual circuit voltage is.

Just bypass and be done.

Let's go with poor explanation.

What I was referring to is the all-important potential difference between grid and cathode. And specifically this difference under signal conditions.

To explain, let's first contemplate an even more extreme example than the unbypassed cathode resistor in normal common cathode gain stage that I mentioned before. Think instead about what happens in the cathode follower, where there is essentially 100% degenerative feedback and gain is consequently u/(u+1), or just a bit less than unity. Even though there is constant DC bias voltage difference between the grid and cathode, when signal is injected at the grid, the cathode follows the grid very closely (this must be true, otherwise we wouldn't be able to take an output of nearly the same magnitude as the input from the cathode). In other words, as the grid swings up, the cathode ALSO swings up, and vice versa. Let's look at an example... imagine that the DC bias difference has the grid at 0V and the cathode at 5V. Now we put in a 2Vp-p signal. So on the upward swing the grid rises to +1V. But the cathode follows the grid and rises by, say, 0.9V. So it rises to 5.9V. Most significantly, the grid-cathode potential has only shifted from 5V to 4.9V -- a small change relative to the 2Vp-p signal we put in. If follows that the amount of current flowing through the tube also changes by a small amount.

Now, imagine that we were able to cancel out some of the signal appearing on the cathode. The signal on the grid will be unaffected, and will still rise to +1V. But imagine that the cathode now only rose 0.1V (i.e. was less able to follow the grid). Now the cathode will rise only to 5.1V, and the grid-cathode potential has now shifted from 5V at quiescence to 4.1V under the signal. It now follows that since this change is much larger than in the previous case that considerably more current is flowing through the tube than before.

The way that an un-bypassed cathode resistor reacts is exactly the same as in the cathode follower example, except, naturally, cathode follows grid to a rather lesser extent. The difference now is that there is a load resistance in the anode.. So the greater current through the tube in the "cancellation" case would correspond not only to more current through the tube, but also to a greater swing across the anode load. That is to say, more voltage gain. It stands that cancelling out signal appearing at the cathode of a gain stage results in increased gain. The out-of-phase signal injected at the cathode therefore has the net effect of acting as a POSITIVE feedback signal, even though it is nominally "canceling a signal". Does this make sense now?

And again, the voltage drop across an LED (it will be more like 1.6 or 1.7V for a standard red LED) is at DC... in other words, in the scenario outlined above, it is providing the bias voltage. The point I was making before is that it has very low AC impedance.. in other words, it's seen by AC as nearly a short circuit, and adding extra signal voltage won't significantly change the voltage appearing at the cathode. It's acting very similarly, in other words, to bypass capacitor. The only point of speculation really was whether or not this AC impedance was better or worse than what one might expect from a bypass cap at our frequencies of interest. That said, people do sometimes bypass the LED with a small film cap in places where signals are anticipated to be large, but this is really intended to shunt switching noise from the LED, as it may transiently cut out when driven really hard.

[loudthud]

I never understood feedback until I studied descrete power supply regulators and opamps. The notion that feedback "cancels" some of the input signal mislead me to think the feedback was "out of phase" with the input. Tube power amps didn't help because it's not clear what the phase relationship of the signal is after it goes through the OT.

Perhaps this will help. Take a 9V battery and connect the minus side to ground. If you touch the plus side to the grid of a tube common cathode stage, the plate voltage will go down. But if you touch the plus side to the cathode, the plate voltage will go up! So if feedback to the cathode in phase with the grid input, they cancel out. That's negative feedback. But if the feedback is out of phase with the input, they reinforce eachother, that's positive feedback.

[mooreamps]

I'm using a 6SL7 for my first two gain stages. If I go to a LED for the cathode biasing, can I just tie both cathodes together (like a 6SC7, only exernally connected) and use a single LED to set the bias voltage? As opposed to using a separate LED for each cathode.

Thanks!

In my experience, it does not work if you drive the gain stage too hard....

-g

[R.G.]

Welcome back, Gary. It's been 3.5 months.

Y'all play nice, now.

[R.G.]

Welcome back, Gary. It's been 3.5 months.

Y'all play nice, now.