Gain Magnitudes Equal: Rejecting Common Mode Signals

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Equal gain magnitudes at the inverting and non-inverting inputs are crucial for effectively rejecting common mode signals in differential amplifiers. If the gains differ, the common mode signals will not cancel out, leading to potential clipping of the output. The discussion emphasizes the importance of understanding the resistor ratios in the voltage divider configuration of a difference amplifier, which can be derived through algebraic manipulation of the circuit equations. The concept of "virtual earth" is debated, with a preference for stating that the inputs should be at equal voltages in an ideal amplifier. Ultimately, hands-on practice with the equations is recommended to grasp these concepts fully.
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Why is it necessary for the gain magnitudes of inverting and non- inverting ends to be equal in magnitudes to be able to reject the common mode signal?

Also can someone explain the ratio obtained of voltage divider in difference amplifier.

R4/R4+R3[1+R2/R1]= R2/R1 Where,

R1 input resistance, R2 feedback resistor, R3 and R4 are voltage dividers used to attenuate the positive gain.
 
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This seems very much a homework-type question.

Vout = A+*V+ - A-*V-

For the resistor ratios, draw the circuit, establish the effect of the "virtual earth", derive the applicable equations, and manipulate the algebra.
 
because your sine waves have to be out of phase 180 degrees in order to reject the common mode noise, if the gain is different it will clip the sum. maybe someone else could elaborate on this.
 
Intuitively (and trivially): The common mode signal is an identical signal applied to both the + and - inputs, so obviously if the gains are different the inputs will not cancel each other.

But, NOxy is correct, you should grind through the equations and see why this is true. I will say I dislike the term "virtual earth". I'd rather say "just assume that the + and - inputs will be at equal voltages". That will always be true in an ideal amplifier if the output is not saturated.

There is nothing to explain about the resistor ratio equation (if what you wrote was correct). You just need to work through it for a couple of examples to understand it forever. Nothing I can explain will replace working it out yourself. It's only hard the first time.
 
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An ideal OpAmp is a differential amplifier. If you are working on that basis and then you find it will amplify A+B, the resulting output will not be what you want. It would mean, at the trivial level, that you would need to get the input bias voltage 'exact'.

I fully endorse meBig Guy's advice that:
Nothing I can explain will replace working it out yourself. It's only hard the first time.
 
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Thanks..
 
I'm late to the party...

so will just reinforce the advice to bite the bullet and do the algebra.


Solve the circuits for voltage at inverting and non-inverting inputs of the IC.
You'll get an equation for each input.
Set those voltages equal,
because that's how operational amplifiers work - they force their inputs equal,
and it is the job of the designer to surround the amplifier with a network that let's it do that.

If it cannot make them equal, it can't "operate".

So Vinverting = f1(a lot of resistors),
and Vnoninverting = f2(a lot of resistors)

Set the two functions equal one another,
The resulting equation should lead directly to your resistor ratios.

old jim
 
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