How to define the output impedance of a gainstage?

[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;

R_in2 = 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.

R_in1 = 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).

R_in = 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;

R_out = R_L/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 R_L/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 R_L/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

[J M Fahey]

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

No, in this case you will have Rs in parallel with the much lower internal impedance of the cathode/source/emitter follower which is:
Z=1/transconductance

You can get transconductance from datasheets (generally tubes show it, some Fets do, some don't, and transistors hardly ever) but it can be deduced from the curves, if published.
So the typical tone stack driver cathode follower, with its classic 100/56/47K cathode resistor, does not have half that value but a much lower one, a couple kilo ohms.

EDIT: reading the above post, let me add that Re can be calculated.
I kow how to do it for a silicon transistor, never designed with Germaniums (thanks God).
There is *very* complex equation which can be reduced to a "magic number":
Re=26/Ie (in milliAmperes)
So a simple transistor gain stage, passing, say, 1 mA, even if it has its emitter connected direct to ground (common in old designs) or bypassed to ground with a big electrolytic, it still has 26 ohms in series with the emitter (internally).

[Rutger]

Thanks for the replies, it helps a lot!

@Fahey: I'm a little confused about this:

Rs in parallel with the much lower internal impedance

it (Re) still has 26 ohms in series with the emitter

It looks like a contradiction to me, or are 'Rs' and 'emitter' different things in your example?

About the source follower: I understand that in real life it's quite impossible to calculate the right Zout. So am I right by saying that in the above example Zout will just be 'very low', in the ballpark of say <100 Ohm? And when there is (for example) some other resistor between source and Vout, this will define the output impedance of the whole gain stage (in parallel with Rs)?

[J M Fahey]

No, no contradiction, just reread it until it clicks.
To say it using far more words and just for an NPN bipolar transistor, (won't fill the page calculating the same for Mos/Tubes/Fets, etc).
1) Suppose the transistor is biased by 2 x  100K resistors (one base to ground, the other base to +V , in this case I choose +20 V.)
The transistor "sees" 2 x 100K resistors in parallel, or 50K.
Now we connect a guitar pickup to that base (through a capacitor, of course, to avoid disturbing the +10V we have there).
To simplify, suppose the pickup impedance is 10K resistive.
The pickup is also a "signal generator" for the following stages.
Now the transistor base "sees" a 10K internal impedance generator (the pickup), in parallel with the biasing resistors (which amounted to 50K), so the total generator impedance seen by the base, is 50K//10K ~ (approximately) 8300 ohms.
***Now this is important***: since the transistor transmits the same voltage but multiplies current (because it has current gain), "Same voltage + larger current = smaller Impedance"
How smaller? It depends on current gain.
Suppose Beta/Hfe (current gain)=100.
"Reflected" impedance (impedance "seen" on the other side) is 100 times smaller.
8300 ohms/100=83 ohms.
Is tah all? Not yet.
I said wi still have some internal impedance in series with it.
Let's calculate it:
2) suppose the emitter sits at +10V (collector is +20 Volts) , and the emitter to ground resistor is 10K.
Current: 1mA.
*Internal* emitter resistance is 26/1mA=26 ohms.
Remember we consider the transistor a voltage generator (that's what the load sees: "something" that feeds it some voltage).
Internal impedance?: it has 2 parts, which add up because *they* are in series: "Reflected" impedance + Emitter impedance (Re)= 83 ohms + 26 ohms= 109 ohms.
You hinted at a close number, but by sheer chance, now you know where it comes from.
And of course said value changes *a lot* with design conditions (voltage/curren/biasing/gain/etc.) ; values are never random but come "from somewhere".

I forgot: those values I calculated are in series between them, but are in parallel with the resistor to ground.
Anyway 10K in parallel with 109 ohms changes that very little.
That's why I said earlier that 10k/2 was not the impedance by any means.

As you see, *sometimes* a couple seemingly simple questions are not answered here, or answered with a "short answer" without much detail, because going the full path can be too long ... and raise more new questions than it answers.

[Rutger]

Well, I don't mind a long explanation, as long as it makes things clear. And it does, thanks

I've read the so called 'contradiction' a couple of times and see where I went wrong.

Somehow I really need to understand what's going on in these circuits before I start to build something. I just don't want to simply copy/paste things and hope for the best.

[J M Fahey]

Don't worry.
Start by building simple stuff, and be proud/happy about it.
It's FUN.
Along that, try to understand it and read about it.
As I said before, I suggest *everybody* who gets into Electronics to grab any old Physics book, the one they usually teach you at School when you are around 16/18 years old.
The one that covers "Electricity and Magnetism".
It explains: current , voltage , battery , power , resistor , capacitor , conductor , insulator , magnet , transformer , switch , potentiometer , galvanometer (the basis of all "needle" measurement instruments).
No, no transistors, tubes or ICs yet, but these can be quickly understood later, since basically tubes and transistors are electronically variable resistors, after all.
TRAN SISTOR = TRANsference reSISTOR (Cool huh?)  :tu:

[phatt]

Wow, just read all that and now I'm even more bonkaz :lmao:

Thanks JMF and Roly,,Very well explained. :dbtu:

I was lucky enough to score a copy of *Art of Electronics* by Horowitz & Hill some years back and that was ever so helpful. <3)
Explaining most of the above comments
Yes there is the *internal R* (or is it Z?) of the device to consider.

Anyway after much reading I realized that building stuff was the only way,, so as Jaun has just noted,, build.
Bread board building is great because you can whip up circuits in no time and if it does not suit your purpose,, whip it out and make another one.

The best thing that can happen is ; Why won't it work??
Mistakes are the best learning tool because it forces you to follow through and read more.
needless to say I had to read a lot.  :lmao: 

Oh and when starting out don't try and build stuff you can't fix.
Phil.

[phatt]

@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 ?

Regards too FX send stuff; I think the voltage swing is likely an issue. I'll leave better minds to explain detail.
Some digital stuff can throw a fit if the input swing is too large from the efx send.

I've found some efx loop setups to be more of a problem than a help.
Phil.

[J M Fahey]

Yes, the big problem with Digital stuff is that it usually is designed by PC type guys who just got into Music and not the other way round.
They read somewhare that a guitar puts out , say, 200mV RMS and design along that.
While that input may receive well over 30 Volts peak to peak under certain circumstances.
Analog designers just add a series limiting resistor and a couple diodes pointing to +/-15V rails (check Crate and such) as to afoid utter destruction; any less than that just gets clipped and that's it.
But typical front end analog to digital converters get swamped and stupid, unless specifically designed to take that abuse.
Often they are not, and the distortion is not simply clipping, which we are used to, which worst case is made out of upper harmonics of what we are playing, but something very ugly and absolutely unrelated .

[Rutger]

@Fahey: yes you're right, building stuff is fun! I've done a few builds over the last year, without too much knowledge, and it worked out great But I'm at a stage now that I want to know what I'm doing and why, for me that's the other part of fun

Actually those questions I ask come from my attempt to design a simple fx-loop myself
Yes voltage swing is an issue as well, I see a lot of fx-loops that have a simple voltagedivider to cut down the voltageswing to a value that is suteable for effects. Diodes are another way to limit the output, maybe nice to combine that by using it as an overdrivestage as well. 

[J M Fahey]

Start by copying and building yourself small blocks taken from commercial amps; as, say, the loop circuit from a Valvestate or whatever.
Copying is not bad, on tyhe contrary, it's practicing and honing your skills.
And developing a mental relationship between what you see in a schematic, what you have actually built (even on a protoboard) and the actual sound.
*Now* I can look at a new, unknown schematic, and predict quite accurately whether the sound will be smooth/buzzy/muddy/whatever and imagine the sound in my head, ... but it took years of doing it.
Go step by step, build, test, *listen*.
That's the shortest and straightest path.