About the continuous-time amplifier usage problem

[elonjia]

I'm a little confused about a continuous-time amplifier in terms of signal resolution, input a 10mV waveform, and the Close Loop Gain is 1, so I think the output should be a 10mV waveform, but what's the influence of Open Loop Gain for the resolution of this amplifier? Like my amplifier can amply the minimum 10mV amp waveform when open loop gain is 1000; what's the influence for the minimum amplitude when open loop gain is 5000? Is there any difference?

 

[sidun.av]

The input resolution of the amplifier is determined by the noise, input offset, drift and a lot of other factors. Open Loop gain defines the precision of the output voltage.
Your closed-loop configuration sets the target gain and th opamp is trying to execute it as good as he can. More gain you have, more precise output would be. For example, if you set your opamp to sit in non-inverting configuration with a gain of 1, and your input is 10mV, ideally, your output should be exactly 10mV. But in reality, (I am just making up the numbers to give you an idea) it would be something like this:
Aol = 40dB, Vout = 9.5mV
Aol = 60dB, Vout = 9.8mV
Aol = 80dB, Vout = 9.9mV, and so on...

You can prove this to yourself simply by deriving the transfer function of any opamp configuration (i.e. inverting amplifier) and instead of using virtual ground (which is ideal assumption), use the following:
Vout = Aol*(V+ - V-)

Hopefully, that helps.

 

[LvW]

You only have to analyze the classical formula for the closed-loop gain Acl in case of negative feedback:

Acl=Aol/(1+Aol*beta) with open-loop gain Aol and the feedback factor beta.

Comments:
* The given expression applies for non-inverting gain (input signal connecte directly to the opamp input node). For inverting operation there is an additional factor which takes input dampung into account.
* For more exact calculation: Aol is not a constant but a frequency-dependend gain with phase shift.