Intuitive way to understand ideal current and voltage amplifiers

[Saptarshi Sarkar]

TL;DR Summary

Need an intuitive explanation as to why input and output impedances of ideal current and voltage amplifiers are what they are.

I was reading about ideal current and voltage amplifiers and the book says that an ideal current amplifier should have 0 input impedance and infinite output impedance while an ideal voltage amplifier should have an infinite input impedance and 0 output impedance.

I am not quite sure I understand why this is. Can anyone give me an intuitive explanation as to why it is that way?

 

[berkeman]

Start with the ideal voltage amplifier: You don't want its input impedance to load the output signal from the previous stage, so the infinite input impedance keeps that from happening. Think about practical DMM and oscilloscope measuring devices. Do you see why you want them to have very high input impedances to keep from loading/affecting the signals that you are trying to measure?

And to the extent that the output impedance is very low or zero, even if there is some finite load impedance, that will not affect the value of the output voltage from that ideal voltage amp. Does that help?

The analogous things apply to an ideal current amp. Knowing the "why" for an ideal voltage amp now, can you say a bit about why you think the ideal current amp will have zero Zin and infinite/high Zout?

 

[Saptarshi Sarkar]

Start with the ideal voltage amplifier: You don't want its input impedance to load the output signal from the previous stage, so the infinite input impedance keeps that from happening.

Maybe it's because I am not from an engineering background but from a Physics background, but I don't understand what you mean by 'loading the output signal'.

What I think I understood from your explanation is that due to the infinite input impedance of the next amplifier circuit, the signal from the previous circuit would be free to enter the next circuit (which would not be the case if there was a short between the two input impedances).

As for an ideal current amplifier, having infinite input impedance would make sense as it would prevent the current from coming back in the circuit by making a U turn as current tries to follow the path of least resistance. So, whatever the load may be, the load resistance will always be less than infinity and the current will thus flow towards the load.

Is my reasoning correct? Please point out if there are any mistakes.

Thanks

 

[berkeman]

Maybe it's because I am not from an engineering background but from a Physics background, but I don't understand what you mean by 'loading the output signal'.

Are you familiar with the circuit concept of a "voltage divider"?

Have a look at this tutorial diagram to see if it makes sense how connecting a finite load to the output of a voltage divider "loads down" the voltage you are trying to sense (unless the input impedance of the load you are connecting is infinite):

https://www.electronics-tutorials.ws/wp-content/uploads/2018/05/resistor-res15.gif

 

[berkeman]

The ideal current amplifier is a bit more complicated in terms of the Zin and Zout. After we can make you comfortable with the ideal voltage amplifier concepts, we can address the "why" for the ideal current amplifier.

Have you had any basic circuits as part of your physics classes so far? Any transistor circuits? Knowing your background will help us to tune our answers to make them more understandable for you.

 

[Saptarshi Sarkar]

Have a look at this tutorial diagram to see if it makes sense how connecting a finite load to the output of a voltage divider "loads down" the voltage you are trying to sense (unless the input impedance of the load you are connecting is infinite)

Thanks! This was a great explanation. I get what loading means now and I understand how adding an infinite resistor in parallel to the output impedance of the previous amplifier can help in maintaining the signal voltage.

But, I am also a bit confused by the circuit, if I add a 0 output impedance , I am basically shorting the X and Y terminals in the voltage divider output, won't that heavily load down the output voltage? Why then should an ideal voltage amplifier have 0 output impedance?

Have you had any basic circuits as part of your physics classes so far? Any transistor circuits?

Yes, I had transistors and their CB, CE and CC configurations in our course. I also had basic Operational Amplifiers circuits like the Inverting and Non-Inverting Amplifier.

 

[berkeman]

I get what loading means now and I understand how adding an infinite resistor in parallel to the output impedance of the previous amplifier can help in maintaining the signal voltage.

Great!

But, I am also a bit confused by the circuit, if I add a 0 output impedance , I am basically shorting the X and Y terminals in the voltage divider output, won't that heavily load down the output voltage? Why then should an ideal voltage amplifier have 0 output impedance?

The output of the voltage amp does not drive the output of the previous stage, it drives the input of the next stage.

https://www.sanfoundry.com/wp-conte...ear-integrated-circuits-questions-bank-q3.png

 

[DaveE]

But, I am also a bit confused by the circuit, if I add a 0 output impedance , I am basically shorting the X and Y terminals in the voltage divider output, won't that heavily load down the output voltage? Why then should an ideal voltage amplifier have 0 output impedance?

You're confusing the simplified example of loading effects in a circuit with how amplifiers are built. That resistor circuit is very different from an amplifier. Ideal amplifiers have their outputs essentially "disconnected" from their inputs, except of course for the signal being amplified. That is why ideal amplifiers are a really useful approximation, they greatly simplify analysis.