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Max input voltage with negative feedback in place?

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If there's a negative feedback loop connected to the cathode of a common cathode triode stage, that reduces the stage's sensitivity to input, and reduced its input impedance, correct?

But... Does this mean that the input stage triode's grid will start to draw current as the peak input voltage approaches its grid bias?

Or put differently, let's say I have a 12AX7 input stage biased with 1.5V grid-cathode; for instance, a 1.5k cathode resistor with Ia of 1mA.

Let's say I have 10dB NFB applied to the 12AX7 cathode from the output transformer secondary.

With the NFB in place, will the input stage 12AX7 start to draw grid current as input signal at its grid approaches 1.5V peak?

If yes, then I should change the operating points of the input stage so that the grid bias is more like 1.75V (grid-cathode) so that it can accept 1.5V peak signal at its grid. In my case, this will increase open loop distortion.

Hence the question.
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Not true actually. The negative feedback would increase, not decrease the input impedance. But only the part of the input impedance caused by the tube, but not the part caused the input resistor between grid and ground. The resistor is in shunt with the tube's input, and since the tubes input impedance is dam near infinite already, its increase caused by cathode nfb will be even more irrelevant compared to the grid resistor's. So, no real total impedance change. The voltage swing seen between the tubes grid-cathode, though, will be reduced, keeping it further from grid current zone. As the input (grid) goes positive, the nfb from the output pulls the cathode positive as well, so less difference between the elements.
 
Thanks for the reply. I'm just embarking on the negative feedback journey, and without much in the way of math skills. It's a challenge.

OK, so the input impedance goes up, and it was already high to begin with. The grid leak resistor defines the input impedance more than ever. Not a bad thing.

The voltage swing seen between the tubes grid-cathode, though, will be reduced, keeping it further from grid current zone. As the input (grid) goes positive, the nfb from the output pulls the cathode positive as well, so less difference between the elements.

So, if I have an (unbypassed) cathode biased triode with 1.5V at its cathode, and it has 10dB NFB applied (at the cathode), then I may find that I can apply 1.5V input signal at the grid without grid current flowing, because the feedback is pulling the cathode positive at the same time?

That might explain why in this circuit I've got going in spice (with 10dB NFB), I can apply 1.7V input to a 12AX7 with a 1.55V cathode bias and the input tube is not clipping.

Of course, spice can't simulate grid current effects very well. I was thinking it was giving me misleading results, but perhaps not?

??
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Thanks for the reply. I'm just embarking on the negative feedback journey, and without much in the way of math skills. It's a challenge.

OK, so the input impedance goes up, and it was already high to begin with. The grid leak resistor defines the input impedance more than ever. Not a bad thing.



So, if I have an (unbypassed) cathode biased triode with 1.5V at its cathode, and it has 10dB NFB applied (at the cathode), then I may find that I can apply 1.5V input signal at the grid without grid current flowing, because the feedback is pulling the cathode positive at the same time?

That might explain why in this circuit I've got going in spice (with 10dB NFB), I can apply 1.7V input to a 12AX7 with a 1.55V cathode bias and the input tube is not clipping.

Of course, spice can't simulate grid current effects very well. I was thinking it was giving me misleading results, but perhaps not?

??
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The level of voltage you can apply to the grid with NFB operating depends on the loop gain (open loop gain of the amplifier plus the NFB feedback factor) which is what you call the amount of feedback. With 10dB of feedback, the voltage fed back to your cathode is, by definition, three times bigger than the
grid/cathode voltage. So if you feed in a 3V signal there will be 2V at the cathode due to the NFB so the grid/cathode voltage is only 1V.

Most triodes start grid current at least 0.5V below the bias point. Most spice models are poor in this respect.

How do you know you have 10dB of feedback?

Can you post your LTspice schematic?

Cheers

Ian
 
The level of voltage you can apply to the grid with NFB operating depends on the loop gain (open loop gain of the amplifier plus the NFB feedback factor) which is what you call the amount of feedback. With 10dB of feedback, the voltage fed back to your cathode is, by definition, three times bigger than the
grid/cathode voltage. So if you feed in a 3V signal there will be 2V at the cathode due to the NFB so the grid/cathode voltage is only 1V.

Most triodes start grid current at least 0.5V below the bias point. Most spice models are poor in this respect.

How do you know you have 10dB of feedback?

Can you post your LTspice schematic?

Cheers

Ian


Thanks Ian. I adjusted the value of the series feedback resistor until the gain at the 4 ohm secondary tap with a 4 ohm load was reduced by 3X compared to open loop. -3X gain = -9.54dB (I should have called it -9dB of NFB)

Schematic attached. (This is an experiment. But it SPICEs well...)
 

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You will find in classic designs that a resistor in series with C8 will push the shelving effect of C8 to higher frequencies allowing you to roll off feedback in a higher freq,to have more feedback in the bass and mid-range and regions (transformer allowing ). More info here.
You do have a scope, right?
 
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Thanks, yes. The value needed for that resistor in series with C8 will need to be found experimentally (signal generator and o-scope). The simulation is showing overly good behavior, with a resonant peak at about 3MHz. I think that's much higher than it will be in real life.

I spent a lot of time getting the predicted open loop THD as low as possible, and in particular getting the 3rd harmonic as low as I could. I'm sure LTspice is miserable at predicting what the exact THD will be in real life, but I'm hoping it can predict what tube types and operating points will yield the lowest THD, relative to other tube types and operating points. It was a big surprise that this combination of 12AX7 and 12AU7 simulates so well. I fear it might be a case of 'too good to be true.' The only one way to know for sure is to build it. But I want to make sure I haven't designed an absolute laugher here, before I start wiring things up for real.

I paralleled the 12AU7 to get the output impedance down and transconductance up. I figure internal plate impedance will be at about 5k ohms for a paralleled pair of 12AU7s.

The 12AX7 is paralleled for no good reason. Well, maybe doubling the gm of the input tube is a good idea. Paralleling the 12AX7 might be a bad idea in the end, because of high Miller capacitance at the input. It simulates fine with a 100k grid leak resistor. -1dB at 20kHz, open loop.

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