How is output impedance calculated?

[Mark Hammer]

One of the things I've never really understood well is how output impedance of a circuit is determined.  I see instances where there may be a large value terminating resistor or large value output pot, yet the specs will say something like "output impedance 1k".  Then I'll see other instances where the stated output impedance value is precisely the value of the terminating resistor to ground just before the output jack.  Now, I recognize that just because it's written somewhere on the net doesn't mean it's true, but ignoring instances where someone posted incorrect calculations, there is still a major gap for me between what I see hanging off the ends of circuits and what is stated.  The fact that commercial schematics which indicate impedances also don't show a correspondance between visible component values and impedance specs further prompts me to ask this question.

Consider the following.  You have an op-amp with a 10uf cap on the output and a 100k pot after that.  When turned fully clockwise, the pot behaves like it is a 100k resistor to ground.    The schematic may say a different output impedance than that, but I'm assuming that, for whatever role the pot plays, there are other components to factor in as well.  If the pot is rotated to its middle resistance point, you now have a 50k resistance in series with the op-amp output, and a 50k resistance to ground at the output.

So, uh, what has just happened to the device's output impedance, or is it unchanged?  I am really and truly confused and in the dark here.

Could someone please address this confusion?  Some attention to the case of an output hanging off the emitter or collector of a bipolar, different types of op-amps, whether there is a cap involved or not, and so, would be deeply appreciated.

Thanks in advance.

[bioroids]

Hi Mark!

I think (may be wrong) the resistance to ground on the last stage doesn't play a major role on setting the output impedance (it can be viewed as part of the input impedance of the next stage).

A resistor in series (like a half turned pot) adds to the output impedance.

I believe it has something to do with the amount of current the stage can source, so, on a BJT amplifier for example, the lower the collector resistor, the lower output impedance. Don't ask me for equations though...

This is all I know about that (and may be wrong)...

Luck

Miguel

[R.G.]

Welcome to the black box. Say hello to Mr. Ohm, Mr. Thevenin, and Mr. Norton, as they lurk in every black box.

Georg Ohm informs us that output impedance is a measure of how much current the described device can put out under load. Over and done, now Mr. Ohm can go home. But we still have to figure out what he meant.

Consider a Perfect Voltage Source (PVS), with a voltage of 10.000000000Vdc. If we put a 1K resistor across it, we get 10ma of current flow, and the voltage is still 10.000000000Vdc. If we put a 10 milliohm copper wire across it, the source can supply the 1K amperes through the copper wire, and it never sags by even a microvolt. A perfect voltage source can supply literally any amount of current and its voltage never sags. That is a zero ohm output impedance, by definition.

Things change if we put some resistor in series with the PVS. Let's put a 1 ohm resistor in series, and now we connect some loads on the output. A 1K resistor pulls 10.00000V/1001 ohms, or 9.991 milliamperes, which most of us could not tell from 10ma. The 10milliohm load, though, now gets 10 amps, and the voltage across the 10 milliohm load is now 100mV, not 10.00000V.

This is the model that Thevenin proposed: any voltage source can be modelled as a perfect voltage source in series with some impedance. Further, we can determine the impedance by shorting the composite voltage source and its output impedance and measuring the current. The open circuit voltage divided by the short circuit current will always be equal to the impedance inside. And that is one way to calculate output impedance - the open circuit voltage divided by the short circuit current.

Every year, EE sophmores are given a black box with a battery and some other parts inside, and asked to figure out what's in the box without actually opening it. Those that do not have a working acquaintance with Thevenin's theorm fail - or crib the results from the group egghead.

You can have an impedance in a couple of ways. One is to have a series resistor as the last thing before the output terminal of the box. This insures that the output impedance will not be less than the resistor. Or you can have some internal facet of your voltage source, like an internal resistance in a transistor, or a bonding wire on the IC die, or the resistivity of the silicon, and those all contribute to impedances that prevent an active voltage source from putting out an infinite current.

Mr. Norton is waiting.

Mr. Norton said that Thevenin had it only half right. What if what's actually in the box is not a PVS, but a Perfect Current Source (PCS)? That is, an imaginary component that will supply any voltage, even an infinite one, to cause its rated current to flow. This kind of thinking does not come easily to us voltage bigots, but string along.

Just as there are no PVS's, there are no PCS's either. There's always some impedance. Imagine a 10A current source. What is the output impedance? Um... OK, it's open circuit voltage divided by short circuit current.  Open  circuit voltage is ... OK, infinite, and short circuit current is 10A. The output impedance is infinite. Correct. Mr. Norton hands out cigars.

"But!" he expostulates, "what if there's a 1 0hm resistor across the current source? What is the open circuit voltage then?"

OK, no problem, it's 10A times 1 ohm, 10V.

"Good. And what is the short circuit current?"

...um... it's 10A. So the output impedance is 10V/10A and that's one ohm!

Norton and Thevenin pointed out that voltage sources and current sources are two faces of the same thing. A PVS with a 1 0hm series resistor is indistinguishable from a PCS and a 1 ohm parallel resistor. And that realization is what separates the clever, attentive sophomore EEs from the true curve setters. You can transpose between Norton and Thevenin sources any time you like.

"..um.. R.G., that's all well and good, but what's that got to do with calculating output impedances?" I can hear you say.

It's a tool. This is one of those areas where you have to know the tools and a few simplifying assumptions.  Strictly speaking, to learn the output impedance of an amplifier, I'd put in a signal, measure the no-load voltage, then short it and measure the short circuit current, and a quick trip to my friendly neighborhood calculator gives me the answer.

The problem is, of course that the amplifier might not survive the test. Short circuits are tough conditions for solid state amps, and open circuits can be deadly to tube ones. So we do a less strict test. If you do the math, you can put an impedance equal to the internal output impedance on the output of an amplifier, and you will find that the open circuit output voltage is cut in half. So you measure the open circuit voltage, load the amp and twiddle the load until the voltage is half, and then measure your resistor. It's now equal in magnitude to the internal impedance.

If you sat through all that babbling, you have probably begun to see a glimmer of hope of working this out. Then we start talking about feedback.

Feedback can cause outputs to have apparently greater or smaller output impedances than they really have. It does this by throwing away excess gain to fake out a ''virtual" impedance. If, for instance, you have an opamp that without feedback has a 100 ohm output. If it also has a gain of 100,000 and is used in a circuit that only uses at a gain of 10, it has a gain of 10,000 left over to fake perfection. I could go get out the derivation, but the result is that the output impedance is reduced by the excess loop gain, or to about 100 ohms/10,000 or 1 milliohm. This is a reasonable approximation of a real case. Voltage output feedback amplifiers have quite low output impedances, approaching zero, as long as they are within the raw voltage and currents of the device. This is a good approximation to know. A feedback current source has an output impedance approaching infinity - until the device runs out of voltage or current.

Where that leaves us is with approximating feedback amps, especially opamps, with perfect sources. Note that if we put a resistor in series outside the feedback loop, the output impedance is just that resistor. If we put it inside the feedback loop, the effect is divided by the excess gain, and has essentially no effect other than to narrow the span overwhich the amp appears perfect.

So opamps are perfect sources but with sudden limits. If you put a resistor in parallel with a PVS, it does - nothing - to the impedance, as the source can supply it and any other load. It's only series resistors that affect PVSs. If you put a resistor in series with a PCS, it also does nothing, but a resistor in parallel with a PCS eats some of the PCS's current and it's no longer available outside, so it lowers the composite output impedance.

So, we're almost to the place we can do some work.

Your question:

Consider the following. You have an op-amp with a 10uf cap on the output and a 100k pot after that. When turned fully clockwise, the pot behaves like it is a 100k resistor to ground. The schematic may say a different output impedance than that, but I'm assuming that, for whatever role the pot plays, there are other components to factor in as well. If the pot is rotated to its middle resistance point, you now have a 50k resistance in series with the op-amp output, and a 50k resistance to ground at the output.

So, uh, what has just happened to the device's output impedance, or is it unchanged? I am really and truly confused and in the dark here.

The opamp we assume is a voltage gain amp, and so it's so near a PVS that it doesn't matter. It's driving a 10uF cap and a 100K resistor to ground. The cap is large enough that we'll assume that we're way away from the frequency where the cap matters, so we replace it with a short circuit. Now, what impedance does a load on the wiper of the pot see?

We know two of the answers. When the pot is full up, it's a resistor in parallel with a voltage source, so the pot does not matter, and the impedance is just the impedance of the feedback opamp, almost zero as long as we don't ask the opamp to supply more voltage and/or current than it really can.

Likewise, when the wiper is at ground, the impedance is zero, as the wiper is shorted to ground. No signal, but hey, we're worried about impedances here.

What happens in the middle? You already have the tools.

The impedance seen by the load on the wiper varies up from zero on either end to some maximum and then goes down again. The maximum is when the two resistances on each side of the wiper are equal, of course.

We can calculate it. The setup looks like a PVS driving two 50K resistors in series and we want to measure the open circuit voltage and the short circuit current, then divide. The open circuit voltage at the junction of the two resistors is half the voltage source (Ohm again, for resistor dividers), and the short circuit current is the voltage source divided by 50K. Dividing those two we get - 50K.

What happens if we put a 1K in series with the wiper? Then the impedance we see varies from a low of 1K (that we stuffed in series) to 51K.

OK, what about a 50K parallel to ground? Same - you figure out what the open circuit voltage and short circuit currents are at each rotation, divide, and that's the impedance you get. By the way, that's the math behind tapering resistors.

Notice that the opamp fell out early. If we had a transistor amp with an output impedance of 10K, that 10K would be in series with the pot's impedance.

A common one you won't see much - the output impedance of a common emitter transistor stage at the collector. This one is almost exactly the collector resistor because the collector of a transistor looks like a constant current source mostly, as does the drain of FETs.

The emitter of a bipolar is a lowish impedance, a few ohms. Likewise the source of a FET.

You can convert a Thevenin voltage source to a Norton current source by making the Norton source be the short circuit current of the Norton source, and the parallel impedance be the series impedance of the Thevenin source - and vice versa. By converting back and forth and using ohm's law you can reduce complicated circuits down to where you can assign an impedance based on circuit parts - usually.

I'm sure that's at least partially confusing. Say which part is dense, and I'll try to expand.

[Mark Hammer]

You're hornin' in on my territory, Keen.  Posting length is my TRADEMARK!!  :lol:  :lol:

Seriously, as delightfully thorough and well-delivered a reply as I had hoped for, and from the source I hoped to receive it from.

Much to cogitate over, for myself and everyone else.

Many thanks,
Mark

[WGTP]

Thanks, I knew there was a reason I wasn't an EE.

[brrt]

Wow R.G.! Great stuff! Practical explanations of these basic EE theories really expand my interest and appreciation of electronic circuits.
What i'm wondering now: can the input impedance be calculated likewise? Also with Norton and Thevenin conversions (theoretically)? And what is a practical method of measuring the input impedance?

Brrt

[brett]

RG, you are a master of explaining the detail behind the rule-of-thumb.

the output impedance of a common emitter transistor stage at the collector. This one is almost exactly the collector resistor because the collector of a transistor looks like a constant current source mostly, as does the drain of FETs.

The emitter of a bipolar is a lowish impedance, a few ohms. Likewise the source of a FET.

I presume op-amps have emitter-follower outputs or similar, since they are low output impedance devices?

In any case, these rules of thumb have got me thru 99% of design that needed an understanding of output impedance (which usually means keeping it low, which in turn means not using a common emitter amplifier and using a high value resistor on the collector).

For complex devices that I don't understand, I simply measure output impedance by putting a large value resistor (e.g. 1M) across its output and measuring the output voltage.  Replacing the resistor with a pot and adjusting it until the output drops to half indicates that the resistance of the pot and the output impedance are equal.  (which is what RG also said above, I suppose)

cheers

[R.G.]

You're hornin' in on my territory, Keen. Posting length is my TRADEMARK!!  

I ... um ... learned from a Master.  

Seriously - I got tired typing. Yell when (not if) you find the places where I glossed over things in the interest of time. Thevenin/Norton and conversion as well as black box lab is about four class periods and two labs, so I didn't do it nearly the justice it deserves. But your curiousity is well placed - this stuff is incredibly practical in messing about with electronics. Once you get the hang of it, you can approximate input and output impedances very quickly.

What i'm wondering now: can the input impedance be calculated likewise? Also with Norton and Thevenin conversions (theoretically)?

Input impedances are tougher to thumbnail because there is usually an active device input staring you in the face when you get through calculating series and parallel resistances. About all I try to do is to ballpark input impedances unless I really have something critical.

F'rinstance: the input of a normal bipolar transistor is approximately the AC current gain times the sum of the Shockley resistance and the unbypassed emitter resistance. Unbypassed emitter resistor is easy, but Shockley resistance??

The Shockley resistance is a quirk of bipolar devices that makes the input act like there is an internal resistor inside the emitter of a value of 26mV/Ie (in milliamps). So a device with Hfe=100 and a collector current of 1ma with its emitter grounded will have an apparent input impedance of 100*(26mv/1ma) = 2.6K.

Last I looked, no one has explained the Shockley resistance authoritatively. Shockley just codified what he observed and others have since then found that he was about right.

So designers try to make it not matter. Things like a 1K resistor in the emitter will run the apparent input impedance up to 100K. Pulls your gain down, but at least you can calculate the input impedance of the transistor. The bias resistors on most bipolars pull the input impedance down by being effectively in parallel with the input. A very common bias setup like 150K/15K biasing makes for about a 13K impedance to just those resistors, and then the transistor input is in parallel with that.  There are bootstrapping tricks that will make bipolars look like a much higher impedance, but it's still tricky. Best to not go there if it's critical unless you just have to.

JFET are easier. A JFET has a multi-megohm input impedance - at DC. It has a reasonably low capacitance to source and drain, so it even keeps its impedance to fairly high frequencies. With a JFET, you can usually just consider the series/parallel combination of any biasing to be the input resistance and ignore the JFET.

MOSFETs are the same, but more so. They are an even higher gate impedance, giga- to tera-ohms, or as an alternate way to look at it, the resistance of twenty volts worth of very pure glass. But they have high gate/source capacitance so they can have a much lower impedance as frequency goes up. For both MOS and JFETs, the capacitance is nonlinear.
Fun.

And what is a practical method of measuring the input impedance?

The variable resistor way. Stick in a *big* value pot set up as a variable resistor in series with the input. Turn it to 0 ohms. Run in a voltage, set the output to some comfortable output voltage. Then turn up the resistor until the output drops by half. The pot is now equal to the input resistance.

Won't help you much at RF, but in low audio like we use, it's pretty good.