Bias question

[dhuebert]

Thinking about it, when I measure the cathode current through my 6L6 at full power (one ohm resistor from cathode to ground) I might be looking at 70 mA. What am I actually measuring? Is this a DC value or is it AC that my meter is misinterpreting as DC? If it is DC why does it go up as I increase the signal into the grid? Is it responding to the RMS value of the input signal? I have been wondering about this for a couple of years but could never remember to ask.

Don

[erichayes]

Hi All,

Don, as long as you're using a steady state signal (sine or square wave, for example) as your input, the cathode voltage you see across the resistor will be the DC component modulated by the sum of the AC component, which is symmetrical, which would equal zero, which means you're reading the DC voltage. It's only when you start throwing goofy waveforms (like music) at the amp that you have to take the AC component into consideration.

Assuming your amp is not single-ended, it's doing exactly what a push-pull amp running AB should do: draw more cathode current as output power is increased. Output power increases as grid drive increases, so everything's playing according to Hoyle. If you rebiased the amp to run class A, your zero output cathode current would be much higher, and not much different from your maximum output current.

[dhuebert]

Thanks for replying, but that exactly did not answer my question. Yes I am running AB and yes I am using a sine wave. So my net sum on the input is zero volts DC as you say, why then would the DC component at the cathode go up as signal amplitude is increased?

Don

Oh, wait...Since we are running AB the negative portion of the input signal drives the tube into cutoff, however the positive portion looks like a half wave rectified signal which has an effective DC value. As the signal amplitude is increased this DC value increases as well, moving the tube up the gm curve. If I remember correctly the RMS value of a half wave sine is 0.707 vp. Might be a good exercise to measure this and see if it fits.

[erichayes]

Hi All,

I think you're making it more complicated than it really is, Don. In class A, the tube is (theoretically) drawing the same current regardless of signal applied. In class AB and B, the current draw is proportional to input signal--the greater the signal voltage, the higher the current draw. Depending on the bias point, one or the other tube will be conducting more through a half cycle, except in class B, when one conducts and the other is cut off.

In one of the other threads, Shannon (and I) talk about running KT88s at over 60 mA at idling. If you were to look at the cathode current at that bias setting, you'd see it stay the same for a longer period as you increase the input level. That's because we're forcing them to operate closer to class A than they would be operating at, say, 50 mA quiescent. Conversely, if you were to set the bias so that, at full output (clipping), you just got rid of the crossover notch, you'd be operating in class B. When you subsequently remove signal, you'll find the quiescent cathode current is now significantly lower because the tubes aren't being called on to "work" with no signal applied.

What your meter is measuring, regardless of operating class, is DC, because the two halves of the AC waveform cancel each other out before the meter can resolve the AC component. If you were to lower the oscillator frequency to something subsonc, you'd see the needle vbrate or the dsplay to roll up and down. But if you switch your meter to AC volts, you'll find 0 VAC at idling increasing to some maximum value at full output. Better yet, hook up a 'scope to the cathode and switch between AC and DC coupling. That will show the effect on both the AC and DC components on the cathode resistor as the signal is varied.