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How to define the output impedance of a gainstage?

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Rutger:
Hi,

I have a hard time to understand what defines the value of the output impedance of a gainstage? Whatever it is build around an opamp or a FET, etc.

Mostly the inputimpedance of a gainstage is set by a resistor from input to ground. But I read different things about what sets the output impedance. I understand the outputimpedance of a FET/opamp is normally very low. But what happens when you put for example a pot right after the FET/opamp, or a simple voltagedivider or high-/lowpassfilter? What impedance will the next stage see then? And are there simple ways to calculate it?

Thanks!

phatt:

Rutger said;

"Mostly the input impedance of a gainstage is set by a resistor from input to ground."

No but it will always be somewhat lower than the resistor across the input.
(depends on input setup a bit)

Get your head around the input and mostly forget about Z output.

There is also the *internal imp of the device in question*
Many factors come into this and it will send you bonkaz.

Better people here to explain in depth details but if using opamps a simple buffer stage at the front will save having to go baCK to school for 4 years. :tu:

Phil.

Roly:
Here be Bonkaz. {gidday Phatt, long time no see  ;) }


Let's dispose of the easy case first - op-amps.  Generally speaking the output impedance of any op-amp worth the name can be taken as so close to zero it makes no never mind.  We normally assume that it is a pure voltage source - looking back into the output it is effectively a short circuit.

The input impedance into the non-inverting (In2, +) input will normally be the value of its resistor back to ground, (or reference point), R4 and in this case the series resistor R3;

Rin2 = R4+R3



...however when input is applied to the inverting input (In1, -) it is normally summed with the negative feedback from the output, R2, so the (-) input terminal is normally effectively ground, therefore the input impedance will be the value of the resistor in series with the input, R1.

Rin1 = R1

This is generally true of all op-amps, FET or Bi-Polar input.


A transistor, FET, or valve gain stage isn't quite as simple, and depends on the circuit arrangement, but we can make some generalisations.

The input impedance of a simple FET or valve stage will effectively be the value of the resistor used to tie the gate or grid back to ground, R3 below (note that this gets less true as the frequency rises, but is close enough for most audio applications).




The input impedance of a simple bi-polar transistor (BJT) stage is quite a different matter.  These devices are driven by current, not voltage, so they have a much lower intrinsic input impedance.  To a first approximation the input impedance will be about the value of the resistor used in series with the input.  If the input happens to be directly into the transistor base then the device input impedance will be approximately the impedance of the emitter circuit (Re) multiplied by the device current gain (hfe).

In this particular case however we also have the bias network R1 an R2 which are effectively in parallel across the transistor base impedance (the DC supply line is AC ground) so the actual input impedance will be the parallel combination of R1, R2, and (hfe * re).

Rin = R1||R2||(hfe*re) = 1/( 1/R1 + 1/R2 + 1/(hfe*re) )



In this case the emitter resistor Re (R3) is bypassed for AC.  For DC the input resistance at the base will still be about hfe * Re, but for AC the only effective emitter resistance is the bulk resistance of the transistor emitter itself, so the input impedance will be quite a bit lower; hfe * re.


The output impedance of a simple FET, BJT or valve stage can be estimated thus; the collector or anode voltage will normally be biased so that it rests somewhere about half the supply voltage (for maximum headroom), here 5V on a 9V supply.

We can then guesstimate the source impedance to be the parallel resistance of the load resistor, R4, and the transistor-plus-emitter resistance.  But to get about half supply the transistor etc must be about the same as the load R4, so to a first approximation we can guess the source impedance to be about half R4;

Rout = RL/2


If you want to follow any of these with a pot you can treat it simply as a combination of resistors (and apply ohms Law a lot).  This may upset the purists but it will normally get you quite close enough.


{strictly I've been very sloppy here by using "resistance" R and "impedance" Z as interchangeable, but resistance is a DC quantity while impedance is effective AC resistance}

HTH

Rutger:
@phatt: well, it's not that you cannot care, does it? When you design for example an fx-loop I think you need to think of the right Zout of your send-fx to get it to work properly.

Besides that, I like bonkaz  :cheesy:

@Roly: thanks for the theory and the equations :)

"If you want to follow any of these with a pot you can treat it simply as a combination of resistors (and apply ohms Law a lot)."
Do you mean that I need to treat it as a combination with, in your example, R4 (Rload)? So any following resistors I need to calculate as being in series/parallel with R4?


And when the gainstage is a bufferstage (sourcefollower), does the calculation of the Zout work about the same?



So will Zout in this case be: Zout=Rs/2 ?

Roly:
(As above) If we assume that the effective resistance of a class-A gain stage is about the same as its load, then the source resistance would be those two resistances in parallel, or about RL/2.

If we add a following pot for level control the output resistance from the wiper would be the resistance below the wiper to ground, in parallel with the upper part of the pot resistance in series with RL/2, however this is not normally a situation where we are interested in the stage output impedance because it is more normal to feed a tonestack (where we are interested in the drive resistance) directly from a gain stage (or buffer) and follow the tonestack with the volume pot.  The load impedance we present to the tonestack is, as Phil says, quite important.  In valve/tube amps a high load resistance/impedance is pretty easy to obtain, also with op-amps and FET's, but using BJT's we may need to resort to something like bootstrapping the base to get a sufficiently high input resistance/impedance.

No, in the case of followers the output resistance/impedance to a first approximation is the load resistance divided by the device gain, so it will normally be quite a bit lower than the value of the load resistor value alone would lead us to think.

For all of these there are closer approximations, and full definitions, but the problem starts to be that the fuller the definition the more device parameters you need, and you find that most datasheets don't mention them, transistor internal emitter resistance, re, for example.  Thankfully for most needs a reasonable ballpark figure is all that is required.

Full derivations can be found in Wikipedia under en.wikipedia.org/wiki/Common_collector and en.wikipedia.org/wiki/Common_drain.

HTH

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