Opamp Loading question Testbench AC response

[jgk2004]

Hello all,

I am wondering about this question and can not find it anywhere.
I have a system that I am designing and have attached a simplified png to help in the question.

I am designing the first amplifier but am not 100% sure I am loading properly in my testbench. If my first amplifier is driving another amplifier do I load it like in example A or B?

I am thinking A is correct? Any advice from the experts?

Johnk

 

[erikl]

do I load it like in example A or B? ... I am thinking A is correct? Johnk

Hi John,
in terms of ac termination it doesn't really matter, however in terms of dc and ac driving requirement, example A is better (if Vcm is somewhere between VDD and GND, as I suppose), because for the output stage of the first amp it's easier to drive current to Vcm than to GND. And anyway that's what you intend in your application.
Or the other way round: you could save some current and/or area in the output stage, if you drive to Vcm instead of GND.
HTH! erikl

 

[jgk2004]

Hi Erikl,

In AC doesn't it make a difference because the DC supply on the bottom of the resistor would be open hense a resistor just hanging on the output? How do I simulate that load then in AC is it like changing the resistor to a higher value since it doesn't have to drive that DC current? I agree with you completing for the DC and DC driving requirements!

John

 

[youssef_zein]

i would choose B if i were since you r talking about amplifier, and in amplifiers you search for maximum symetrical swing to get the largest signal possible at the output, unless you are working with dc only positive supply, then you choose vcm = vdd/2
and about the resistor you were asking it will be terminated by vcm because you are using fedback so it gets the voltage from the output o the scond stage. so the positive and ngative imputs of an opamp are usually considerd to have the sam dc voltage.

 

[erikl]

... a resistor just hanging on the output? How do I simulate that load then in AC is it like changing the resistor to a higher value since it doesn't have to drive that DC current?

For AC simulation, the opAmp output "sees" the sum of ZL + Z(Vcm) as its load. If you care about Z(Vcm), you should make it more low-ohmic, e.g. by not generating it from a resistive divider, but from a diode-connected circuit, or even - if necessary - via a buffer.