Effect of feedback on transistor input impedance

[rumpfy]

Good on you Mr Al; As usual you bring a beautifully simple way of expressing the subject.
There was a couple of comments posted about the National circuit and the diagram seemed to come from some location and NOT a NS databook and did not include the comment. I happened to have the original Application Note and thought it may be of interest; so I opened up. Also, Mr. Nigel in post 9 asked about the bootstrapping to the drain and what was it for.
We 'did' the cathode follower in my engineering school in 1962, and I had not seen the adaptation to actually boot strap the grid (inphase) AND the anode (antiphase) to bootstrap out the anode to cathode impedance effects due to the anode to grid (drain to gate) capacitance. These effects are present in the FET as well. The whole mechanism is analogous to guarding in bridge arrangements. The beauty of the NS circuit is the 'guarding' of both the real part and the imaginary part of the input admittance.
My TEK 564 has the small RCA subminiature triodes as input buffers and the input admittance is just swamped with extra capacitance rather than the NS trick of bootstrapping it. The bandwidth of the TEK is only 10 mHz, so when you say the effect has some time delay, this puts an upper limit to the ultimate bandwidth achievable with this technique.

 

[Nigel Goodwin]

Hi,

I like the National link in this thread, but this idea is even simpler than that and also works with DC.

A simple example was say we have a 1 ohm (yes very low) input resistor in series with the input and we are measuring a 1 volt signal, and say the other end of the resistor goes to a low voltage near zero. That means with the 1v signal and 1 ohm resistor the measuring circuit would draw a full 1 amp because the input resistance (impedance) is 1 ohm.

But now pump the other end of the resistor up to 0.9 volts to make the same measurement. Now we have the 1v input minus the 0.9v bias so there's only 0.1v across the 1 ohm input resistor, so that is now only 0.1 amp when before it was 1 amp. So now the voltage being measured (1v) only sees a 0.1 amp current being drawn from it, so it appears to the 1v source as though it were feeding a 10 ohm resistor now because 1/10 equals 0.1 amps. So we've managed to increase the input resistance to 10 ohms using a 1 ohm resistor, and the increase of the "bias" from 0v to 0.9v comes from positive feedback.

But why stop there. Increasing the bias to 0.99v, we see only 0.01v drop across the resistor, thus the input resistance now looks like 100 ohms. Taking this farther, increasing the bias to 0.999999 volts and the input resistance then looks like 1 megohm, when we still have that 1 ohm resistor in series with the input.

The "bias" comes from linear positive feedback that is a portion of the measured value. With a 2v DC input test signal the bias would be automatically adjusted for twice the value it was with a 1v DC input. So with a 2v input the bias would end up being 1.999998 volts which would leave 0.000002v across the 1 ohm resistor which means 2ua current which means again 1 megohm apparent input resistance.

There's a slight catch here in that it takes a small amount of time for the feedback to react to the new input signal, so we would not really want to use a 1 ohm resistor but something much higher to start with.
Note also that we do not need infinite input impedance we just usually need it to be very high.

The simplest example of this is an op amp circuit with positive feedback. The positive feedback is always kept less than the input so it does not oscillate.
We also end up with a slight error in the measured value but it can be kept low or compensated for in other ways.

Give it a little further thought - you don't think you've just described a Wheatstone Bridge

 

[MrAl]

Hi Nigel and throbscottle

Nigel:
If you mean that we try to adjust the current through the resistor to be zero (or near that) then i guess i have to agree. I was trying to keep it as simple as possible.

throbscottle:
If you want to look at a two or three transistor circuit we could do that, maybe starting with a two transistor circuit that has a very clear cut way of setting the gain first, then carry over to the three transistor circuit.
I would suggest that you start with an op amp circuit though because calculating the resistances for a certain gain carries over to the transistor circuit with only a slight modification. Im not sure if you to do this or not, it's up to you.
I'll see if i can dig up that two transistor circuit we can start with, or maybe even start with a one transistor circuit unless you've already conquered that, just let me know.
Attached is the two transistor circuit. See if you can figure out why the voltage on the output is about 6v when the emitter of the first transistor is 1v. The input is also 1v as shown but there is also a 'bias' for the input of 0.5v and that is about equal to the base emitter drop of the first transistor.

Those kinds of circuits make use of negative feedback as i am sure you are well aware, while the input impedance increase came from a positive feedback. But just as we can INCREASE the input resistance with positive feedback we can also DECREASE the input resistance using negative feedback. By taking the negative feedback concept to the extreme, we can create a zero ohm current shunt. So that would be on the other end of the impedance scale.

General Readers:
I'd also line to mention the Kelvin Double Bridge circuit for measuring low resistances. This was probably more popular in ages past but some may still find it interesting and useful. These days we can get pretty good results using a DVM and a known current which is automatically a 4 wire system. One thing i do still like about it though is that we dont have to make a separate measurement of the current or have to use a set regulated current.

 

[crutschow]

...........................
Now my next question is, and something else that has always looked like a Dark Art to me, I know there are several types of feedback (current, voltage, series, parallel), but supposing I have for a very simple example an amp with 3 bipolar transistors (I'll stick with those because I understand them better) common emitter, biased very boringly with a p.d. to the base, and say I want it to have a closed loop gain of 100, how do I work out the value of a resistor taking negative feedback from the collector of the third tranny to the base of the first? Or supposing I have only one transistor, and I want a gain of 10, I can bias it with a resistor from collector to base, but how do I work out the value to get the required gain?

I'm assuming from the principle you explained above it will decrease the input impedance in this arrangement.

It's just one of those things I never understood, mainly because it gets very mathematical very quickly (and I only do slightly mathematical, sorry!)

If you assume a high open-loop gain from the three transistors then the gain can be calculated as you would with a op amp. It's the feedback resistor value divided by the input resistor value. The base of the input transistor is the summing junction. With a finite gain open-loop gain the closed-loop gain will be some less than that calculated value, depending upon the ratio of the closed-loop to open-loop gains.

The same thing applies to a single transistor but the calculated gain will tend to have more error unless you use a small closed-loop gain as compared to the open-loop gain. If you want a better calculation then you have to do the math that includes the open-loop gain in the calculation.

For both these configurations, the input impedance will be approximately the value of the input resistor.

 

[MrAl]

Hi,

I was hoping he would want to look at the op amp circuit first too, because that would be simpler. The transistor circuit could involve the original input resistance too so that would be slightly different than the op amp circuit and so might be harder to understand right off the bat.

We'll have to wait till he gets back here to find out what he wants to do.

 

[throbscottle]

Ok he's got back (Hey how come you are so sure he's a he anyway? He might be a she! )

He's digested (well, read or skimmed though at least, it's sinking in slowly) quite a lot of op-amp documentation so he's kind of in his own vaguely happy place there, but otoh if it helps explain the transistor case then all's well and good

Right, I found out for the single transistor case that the gain is given by dividing the collector load resistor by the emitter resistor - actually I think I might actually have learnt that in college but forgotten since I never needed to use it. I've seen how the AC gain can be controlled using a resistor in series with the emitter bypass cap.

Crutschow - I thought it was way more complicated than that. I thought the principle must be same, but it had never occurred to me that the input might be through a resistor. So that's a known value, but how do you derive the element contributed by the transistor's base?

I found the calculation for an opamp taking into account open loop gain, so can I use the same formula for a transistor amp with the same kind of f/b, or are the there some more variables?

Ok so I remember reading a book in nineteen eighty something which showed voltage and current feedback in combinations of series and parallel, and I didn't understand it. Now, afaict, we're looking at voltage feedback - oh heck, now do you consider it to be series or parallel? If this describes it's relationship to the input signal rather than the amp itself then it is parallel, yes? And for the single transistor the emitter resistor creates current feedback in series with the signal?

Ok now my head is going round and round again...

 

[audioguru]

A transistor:
If you connect a resistor in series with the emitter to ground then its signal has the same AC voltage and same phase as the base. The signal at the collector is out-of-phase with the emitter signal so the emitter signal cancels some of the signal level at the collector (negative feedback).
Since the emitter resistor is in series with the base-emitter junction then it increases the input impedance a lot.

If you connect a resistor from the collector to the base then it biases the base with DC and adds negative feedback which reduces the gain. Since it connects directly to the input it cancels some of the base signal and reduces its input impedance a lot. Usually a series resistor feeds the input signal to the base to increase the total input impedance. This input resistor (and/or the output impedance of the signal source) becomes part of the gain calculations.

 

[MrAl]

Hi again,

Well i guess i did not know if you were a he or she but assumed you were a he and that you would update that status if need be <chuckle>

So anyway, i see you are getting more now about this and that's certainly good to see. It's not that hard once you get to do a few. One thing to keep in mind that makes it easier is that for a lot of these transistor circuits we use a lot of approximations to understand how it works, and how to modify the circuit. For your example of the collector and emitter resistor that is about the same thing. The collector resistor value divided by the emitter resistor value gives the approximate voltage gain, and we dont care if it is exact or not. Just like the two transistor circuit i provided in this thread, if you consider the 5k resistor to be the feedback resistor, if the output voltage has to drive that resistor and the 1k is in series, then the voltage divider action is 1k/(1k+5k) which means 1/6 of the output gets put back into the emitter circuit, and with the input being a total of 1v plus 0.5v, and 0.5v is dropped across the base emitter (approximate again), that means that the circuit is satisfied when the output is 6v. So in this case is it very much like the op amp, where the non inverting gain is 5k/1k+1 which equals 6. So the gain is 6. Again that is approximate though, and can be anything from 5 to 7 roughly. The way i set it up though it came out closer to 6 like 5.8 or something like that.

But i showed that circuit because it is a good example and is a little more independent of the other transistor requirements that you now found out about like the base emitter resistance. Note again that in that circuit the feedback goes to the emitter of the first transistor, not the base. So we bypass any complications due to the base itself more or less. If we did go to the base, we'd have to figure out how much we would loose as base current and that would affect the overall gain too. So it would be almost like an op amp with input impedance (resistance in our case) that goes to ground. So that would steal some feedback and so we'd have to adjust the feedback so that we can get more of a feedback signal than without that loss.
If you intend to provided feedback to the base though, then you must also have some input resistance in series with the base otherwise you'd be counting on the previous circuit output impedance, which can be done too.
The base emitter is a diode however, not a resistor, but to simplify things we can think if it as a resistor that has a set value somewhere in the middle of where it will actually be during use.

You started with a one transistor stage and that's probably a good idea. You can see how the feedback from the collector (common emitter stage) affects the input and so limits the gain. The idea is to keep the gain lower than the lowest beta the transistor will ever have, and for fast response at least 10 times lower is a good idea, although you should be able to go higher with audio circuits.

Of course spice is a friend here too and can help you get started as you try different things. I dont know though if you care to download a free simulator or not and learn how that works too.

For now you should probably concentrate on voltage feedback. True current feedback is used in some op amp circuits and in power supplies but that's a little different. Of course voltage feedback in a transistor circuit though also requires current. Current through the resistor changes the voltage. The emitter resistor provides automatic voltage feedback because the base emitter diode is also in series with the input so when the emitter voltage rises the base emitter voltage differential falls and so the current into the base falls and so the transistor conducts less. That's negative feedback.