Find Voltage at Node 1 in Op-amp CMRR Circuit

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Discussion Overview

The discussion revolves around finding the voltage at node 1 in an operational amplifier (op-amp) circuit, specifically in the context of calculating the Common-Mode Rejection Ratio (CMRR). Participants explore various aspects of the op-amp's behavior, including the implications of input voltages being equal and the relationship between differential gain (Ad) and common-mode gain (Ac).

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that since no current enters the op-amp, the voltage difference (Vd) across the two nodes should be zero, implying that the voltages at nodes 1 and 2 are equal.
  • Others argue that if the voltages at nodes 1 and 2 are equal, then the differential gain (Ad) would be infinite, which raises questions about the validity of this assumption in practical scenarios.
  • A participant proposes using the voltage divider rule to express the voltage at node 1 as a function of V1 and Vo, and then setting the equations for the two nodes equal.
  • Some participants emphasize that the assumption of Vd being zero is typically used in ideal op-amp scenarios, but may not apply universally in all numerical problems.
  • Another participant notes that while the ideal op-amp is considered to have infinite gain and CMRR, real op-amps have finite values for these parameters, suggesting a need to account for non-ideal characteristics in calculations.
  • One participant clarifies that the CMRR should be calculated for the circuit as a whole, distinguishing between common-mode gain and differential gain.

Areas of Agreement / Disagreement

Participants express differing views on the implications of Vd being zero and its effect on Ad and CMRR. There is no consensus on whether the assumption of equal voltages at the nodes is valid in all contexts, and multiple competing views remain regarding the treatment of ideal versus non-ideal op-amps.

Contextual Notes

Participants highlight limitations in applying the assumption of Vd being zero, particularly in practical scenarios where real op-amp characteristics must be considered. The discussion includes unresolved mathematical steps and varying interpretations of the op-amp's behavior.

ranju
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In the given op-amp circuit we have to find CMRR..for the we will need to find voltages at the 2 nodes (1 & 2) , we can fibd voltage at node 2 by voltage divider..how to find voltage at node 1 ..??
I also have a doubt ..that since no current enters the the op=amp the voltage difference Vd across the 2 nodes will be zero , so wouls'nt the voltage at the 2 nodes will be equal??
 

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ranju said:
I also have a doubt ..that since no current enters the the op=amp the voltage difference Vd across the 2 nodes will be zero , so wouls'nt the voltage at the 2 nodes will be equal??

no help yet ?

Isn't that the basic premise of an operational amplifier? It's surrounded by a feedback network that let's it hold its inputs equal ?
What happens if you use voltage divider rule to solve for voltage at node 1 as function of V1and Vo , then set your two node equations equal...?
 
But if the voltages at node 1 & 2 are equal ..then what about Ad ( gain with different input signals) ..it will be infinity..!..since the 2 input voltages at node 1 &2 will be equal Vd will be zero then..!
 
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Ad should not appear in the method i suggested. That method will tell you for any V1 and V2 what is Vo.

Just try it.
 
before solving i out.. my main concern is..Vd ..now see.. Vd = Vnode2 -Vnode1 & Vd is zero as we know ..so both voltages are equal ..again I am saying Vd is zero that means Ad is infinity... & CMRR=Ad/Ac.. then CMRR will be infinity..!
 
ranju said:
... Vd is zero that means Ad is infinity... & CMRR=Ad/Ac.. then CMRR will be infinity..!

For an ideal op-amp, Correct.

An ideal op-amp is usually considered to have the following properties:


http://en.wikipedia.org/wiki/Operational_amplifier#Ideal_op-amps
 
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But we use this assumption (Vd=0 ) while solving for expressions of output voltage in inverting or non-inverting amplifiers..so this is applicable there only..right.>?? For numericals in which nothing as such specified..we'll find the 2 input voltages individually and solve it out..!
 
ranju said:
again I am saying Vd is zero that means Ad is infinity...

In reality Ad is something like 10^5 to 10^8
so when you say difference at inputs is zero, well, 0.000001 does round off to zero.

What is important is to get your mind working these circuits as if the gain actually were infinite and the opamp able to balance its inputs exactly,.
Then they become intuitive.
Then , as an astute student of operational amplifiers, you back up and apply corrections for the non-idealness of the amplifier.

Sometimes it's Ad you have to correct for, sometimes its the input current, sometimes it's the zero shift between inputs...

So work it as if the amp were ideal, then back up and plug in your real gain of say 10^6 and see how much the result changes.
Then plug in a couple microvolts of input offset. And then a few nanoamps of input current.
That'll give you a feel for the effects of opamp's non-idealness.
I used to do that with Basic on my TI99 because it had a great math pack for basic, TI made their home computer report as many digits as did their pocket calculators, Taught me a lot.

old jim
 
  • #10
Hey just a quick note. I think you are supposed to find the CMRR for the circuit as a whole, not the op-amp.

This means finding the common mode gain (setting V1=V2) and differential gain (keeping them independent).
 

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