I am confused how you calculate the output impedance of an op amp. Say you have the two stage op amp here:

Usually you calculate the resistance seen a the drain of Q6 in parallel with the resistance of the drain of Q7. This would normally be something like 10k or higher since the transistor is a transconductance amplifier. But why do we say that op amps have a really low output impedance? The thevenin resistance is the exact same as the norton resistance, right?

Thanks

 

But why do we say that op amps have a really low output impedance?

Because we are normally talking about the closed-loop gain output impedance which reduces the output impedance by the loop gain.
In practice this means the closed-loop output impedance for a typical IC op amp is much less than an ohm.

 

I think my main question is more fundamental (and I'm a little embarrassed by it haha). If you have voltage source with a very low output resistance, that is a good thing. But if you convert that to a Norton's equivalent, now you have a current source with a very low output resistance, which is a bad thing. I realize the current will be way high so the circuit analysis would be the same, but you wouldn't ever create a current source that burns that much off on the low output resistance.

So are Thevenin and Norton equivalents only relevant in helping circuit analysis, but extremely inefficient if you tried to implement them in the real world?

If so, the output of the op amp is a current source and any load you connect will be in parallel with the output resistance, so why is the lower the better?

 

So are Thevenin and Norton equivalents only relevant in helping circuit analysis, but extremely inefficient if you tried to implement them in the real world?

Yes, the equivalents are basically used for circuit analysis.

If so, the output of the op amp is a current source and any load you connect will be in parallel with the output resistance, so why is the lower the better?

You want a low output resistance so that the output voltage is largely independent of a change in the load resistance.

 

You want a low output resistance so that the output voltage is largely independent of a change in the load resistance.

I understand that, but the load will always be connecting in parallel with the output resistance of an op amp, which means a higher resistance should be desired, right?

 

The output resistance is in series with the load. An ideal op amp has zero output impedance, if that was in parallel with the load it would never get any current delivered to it. That's an easy way to look at it anyways. Google equivalent circuit of an op-amp.

 

The output resistance is in series with the load. An ideal op amp has zero output impedance, if that was in parallel with the load it would never get any current delivered to it. That's an easy way to look at it anyways. Google equivalent circuit of an op-amp.

The equivalent circuit is just used for circuit analysis simplicity. I am asking about the actual design of the op amp like in the picture posted above. Q6 and Q7 in parallel both provide the actual output resistance to the op amp. Any load connected will also be added in parallel.

 

Okay let's back up a bit.
The circuit you show does have a very high open-loop output impedance.
That's different than older bipolar op amps which have a much lower open-loop output impedance.
This article discusses the difference.

So the best model for your output would be a current source in parallel with a high resistance.
But when the loop is closed the op amp looks like a voltage controlled voltage output with a low output impedance.

This low equivalent output impedance, which is desired for load insensitivity, is only obtained when a negative feedback loop is completed from the output to the negative input.
For that, the output impedance is reduced by the loop gain, as I previously mentioned.

So you need to keep in mind that there is a large difference between an op amp's open loop and closed loop equivalent output impedance.

 

I see an easy way to solve the problem. You must use Spice. Using a current source and frequency analysis, it is easy to obtain the frequency dependences of the output impedance.

 

So the best model for your output would be a current source in parallel with a high resistance. But when the loop is closed the op amp looks like a voltage controlled voltage output with a low output impedance.

This is the part that confuses me. How does it now look like a voltage source? If you connect a load to an op amp with negative feedback, you will still be connecting it in parallel with the output impedance, yes?

 

The two stage above is actually an OTA. Normally an opamp when drives resistive load will have a buffer stage following.

 

Interesting. What would the buffer stage look like, a push pull pair?

 

This is the part that confuses me. How does it now look like a voltage source? If you connect a load to an op amp with negative feedback, you will still be connecting it in parallel with the output impedance, yes?

Yes, but the apparent value of the impedance is reduced by the loop gain, as I've stated a couple times.

I believe you need to study negative feedback, as you do not seem to understand how it works.

 

I understand the negative feedback reduces the apparent value of the impedance, but that value is till being added in PARALLEL with whatever load you attach to the output. Which doesn't look like a voltage source to me.

 

but that value is till being added in PARALLEL with whatever load you attach to the output.

So convert it to the equivalent Thevenin voltage source.

 

Interesting. What would the buffer stage look like, a push pull pair?

You can read more from CMOS Analog Circuit Design by Allen. Below is an except where he discussed about the term opamp and OTA.
https://www.google.co.kr/url?sa=t&s...ghpMAg&usg=AFQjCNGC8QKMCRaVnyJE9Wtn11SGfLLw8Q

 

My question will be answered if we focus only on the open loop output impedance. From internet searches it looks like the open loop output impedance of the average op amp is on the order of 100 ohms. But if you actually do the circuit analysis on the last stage, you will get something on the order of 100k ohms (if it was near 100 ohms, there would be very little gain).

So my question is how do you get from the 100k ohms from the circuit analysis of the output stage, to the 100 ohms when you consider the output a voltage source.

 

If you specify the size of the transistors and the values of the resistor and capacitor, then I will help you calculate the output impedance. It is desirable to add models. If you do not add models, then I'll take those models that I have. I'm using LTspice

 

So my question is how do you get from the 100k ohms from the circuit analysis of the output stage, to the 100 ohms when you consider the output a voltage source.

You are comparing apples and oranges.
Your op amp does have a high output impedance.
The ones with a 100 ohms output impedance have different output circuitry which gives a low output impedance.

 

Could you show me an example of an op amp that does have a low output impedance?

For example:

Would this push pull output have low output impedance? If so, when you connect a load it will still be added in parallel with the resistances seen through the sources of T6 and T7. Adding the load in parallel with a low resistance sounds like a bad thing to me. I'm confused!! lol