Calculating input impedance for single-device stages WITH feedback

[earthtonesaudio]

...How do you do it?

Horowitz and Hill are vague on the subject, discussing transistor feedback in general terms, and op-amp feedback in more specific terms, but skipping specific info about transistors.


How about the Zvex SHO circuit?  It is a single-transistor stage with negative feedback from drain to gate (as well as degenerative source feedback, but that's not what I'm interested in).  If you have it set for unity gain, and assume that the drain is a voltage source (which it isn't, but just pretend for a minute), then the input Z is 10M in parallel with 5M (because the upper resistor has twice the current flowing through it, or about 3.3M.  At a gain of 10 you get 10M in parallel with 0.91M, because a +1 signal at the input will result in a -10 signal at the output, so there is 1-(-10) times the current flowing through the upper resistor as there is through the lower one.
In this idealized case the input impedance is 10M in parallel with [10M/(1+gain)].
Of course the drain is not a voltage source at all, and rather than supply any current necessary to make the required voltage, it will sag under current load, reducing the amount of negative feedback and increasing the overall gain.  I can understand this intuitively, but I can't find any "hard data" to back it up, much less make any sort of reasonable calculations on the subject.

Help please.    Thanks!

[R.G.]

Getting to the answer takes some doing, and getting to the right answer takes some formal principles and math. Not a lot, but some. The Art of Electronics, for all the weight of the book, is a very broad brush. It does not go into great detail in many areas.

You're actually doing pretty well. In looking at the SHO:
- one can ignore the MOSFET gate at "low" frequencies. I put low in parentheses because (a) the MOSFET input is effectively infinite at DC but (b) the gate-source capacitance is not negligibly small but ( c) the source resistor makes the effective value of the gate-source capacitor negligible to a much higher frequency.
- yes, the bottom 10M is  in parallel with the input
- yes, the feedback 10M is effectively reduced by the voltage gain of the stage from input to output
- yes, the drain is not a voltage source; however, it is a current source running through a resistor, the very definition of a Norton equivalent voltage source, which is the equivalent of a Thevenin source of equal voltage with the same resistance in series with it; this makes the drain a voltage source in series with 5.1k, which means that compared to a 10M resistor, you can neglect the 5.1K and it *is* a voltage source and won't sag noticeably.

The formal way to figure the input impedance is to write up the mesh equations for the network and solve them, which is complicated by the need to insert the transconductance of the MOSFET.  The practical way is to do what I think of as on the spot engineering - applying the principle of superposition in linear systems and then doing divide and conquer, tossing out at every step things which can be viewed as not mattering much, to get to a quick answer. You're doing this last, probably as a result of both your reading and being corrupted from the One True Way by reading my posts on similar matters. 

Lemme do some respond-in detail:

It is a single-transistor stage with negative feedback from drain to gate (as well as degenerative source feedback, but that's not what I'm interested in).

But you need to be interested in the source feedback, not so much here, but in general for this kind of thing. That's because the device input impedance has a big effect, and the source impedance changes it hugely, in general. Here it just changes the frequency where you have to stop ignoring the gate-source capacitance, and its ultimate effect. I put this issue under the need to understand, at least some, the applicable model of the device you're considering at the moment. In this case, the source feedback tells you to ignore the gate impedance to a higher frequency, which is good, because you can mentally just cut the wire from the 10Ms to the gate and not worry about the gate for this computation. It simplifies things.

If you have it set for unity gain, and assume that the drain is a voltage source (which it isn't, but just pretend for a minute),

No need to pretend. As I mentioned, it IS a Norton source, and that can be transformed into an equivalent Thevenin source to find the source impedance, and then neglected because it's trivial compared to the 10M.

then the input Z is 10M in parallel with 5M (because the upper resistor has twice the current flowing through it, or about 3.3M.  At a gain of 10 you get 10M in parallel with 0.91M, because a +1 signal at the input will result in a -10 signal at the output, so there is 1-(-10) times the current flowing through the upper resistor as there is through the lower one. In this idealized case the input impedance is 10M in parallel with [10M/(1+gain)].

Yep. Feedback 101. It's why the feedback cap in the Vox wah circuit looks variable. You're varying the gain, not the capacitor.

Of course the drain is not a voltage source at all, and rather than supply any current necessary to make the required voltage, it will sag under current load, reducing the amount of negative feedback and increasing the overall gain. 

The part you're missing is that you doubt yourself because you can't (yet) do the Thevenin/Norton transformation and have a formal basis for ignoring the source impedance of the drain. It does sag; but so little that you can ignore it, at least compared to the 10M it drives.

I can understand this intuitively, but I can't find any "hard data" to back it up, much less make any sort of reasonable calculations on the subject.

You've built the intuition but without the formal principles and having your hands metaphorically slapped by crusty older hardware design engineers/professors who call the simplification principles to your mind.

I personally view a lot of engineering design in the light of what Bruce Lee said about learning martial arts techniques. He said that he was forever learning various techniques so he could forget them. He wanted to learn the principles so he could "forget" them - apply them intuitively, without thinking "Aha! My opponent is doing a face punch. This calls for age-uke." It's another version of the idea in "The Gambler"; you gotta know when to hold 'em, know when to fold 'em. And you have to know, by experience, when to pull out the node/loop equations and when you can ignore them and do a quick Thevenin/Norton chain, RC filter/feedback reduction in your head and come to an approximate answer on the spot. Computers are good at doing the exacting detail. The great value of human insight is knowing what matters and what can be ignored (even approximately, mostly) for the moment.

You're getting there. You just lack some experience in knowing you're getting there, and perhaps the classroom work in knowing that you can ignore what you ignore.

[earthtonesaudio]

Thanks, R.G.

I had a little "aha!" myself while reading your reply.  Not so much about the actual technique of visualizing the drain as a voltage source with a resistor in series, but the fact that I actually CAN visualize it that way and get a fairly close answer.  Once you said that, it made sense.