SparkFun Forums

[8Volts]

Hello,

In my last post I had calculated Vout of an op amp according to its common/differential mode gains which was:

Vout = A_cm * V_cm + A_d * V_d
Vout = -1.1*3.5V + 1.45*3V
Vout = -3.85V + 4.35V
Vout = 0.5V

I am reading up on the CMRR of an op amp. And I Quote:

However, in real amplifiers there is always some variation and the ratio of the change to the output voltage with regards to the change in the common mode input voltage is called the Common Mode Rejection Ratio or CMRR. Opamps typically have a CMRR of around 80 dB and the higher the better.

If we calculate the CMRR of my circuit I only get:

CMRR = 20 Log (Ad/Acm)
CMRR = 10 Log (1.45/1.1)
CMRR = 2.399

Does this mean my op amp is horrible

8v

[jremington]

The choice of resistor values and their precision is a critical factor in optimizing the CMRR. After proper choice of their values, you need to use resistors with better than 1% precision and they are expensive. People usually buy differential amplifiers as a self contained unit (called an instrumentation amplifier) for just this reason. http://en.wikipedia.org/wiki/Instrumentation_amplifier

[UhClem]

It's you, not the opamp.

You should have been able to find this:

http://www.analog.com/static/imported-f ... MT-042.pdf

Which shows how it is done.

[Mee_n_Mac]

Does this mean my op amp is horrible

You had a circuit, that used an op-amp. Your circuit made no attempt to be a differential ONLY amplifier. It's neither good or bad, it is what it is. If you wanted a differential amplifier circuit, you'd have chosen the resistor values to be different from what they were. DId that circuit have any real intended usage ?

[8Volts]

hello Mee_n_Mac,

okay, I see, had I chosen all resistors of the same value
to make a purely differential amplifier the CMRR calculation
would of resulted much more appropriately. Ex: much higher value of CMRR
And if I had select the same value of resistors so to try
this, they must be precise.

Nope, that circuit was experimental only.

Thanks for your help

[8Volts]

The choice of resistor values and their precision is a critical factor in optimizing the CMRR. After proper choice of their values, you need to use resistors with better than 1% precision and they are expensive. People usually buy differential amplifiers as a self contained unit (called an instrumentation amplifier) for just this reason. http://en.wikipedia.org/wiki/Instrumentation_amplifier

yes now it makes sense to me!

Thank you