common mode rejection ratio (CMRR)

[biolycans]

Hi,

My question is related to the CMRR (common mode rejection ratio).

1) First I connected a voltage supply with an AC signal of 1V amplitude to the not inverting input of the opamp. The inverting input I connected to ground. Then I did a dc sweep to find the offset, then I included this offset to this configuration. Then I did a AC sweep between 1 Hz to 10 MHz. This plot represents the differential gain in dB versus frequency Mhz.

2) Then I connected both inputs together to a voltage supply AC 1V amplitude. I included the offset I measured before. Then I did an AC sweep between 1Hz to 10 Mhz. This plot represents the common mode gain dB versus frequency Mhz.

3) Then I chose for example 10 Hz (I did this with many frequencies), I chose for this frequency the gain (differential gain in dB and the common mode gain in dB). I did the subtraction between both. The result is the CMRR in dB.

Is this correct ?

Regards,

JC

 

[D.A.(Tony)Stewart]

Yes

CM noise and gain is also influenced by circuit balance, wiring and layout , so coax is better than twisted pair which is better than nothing.

 

[kenambo]

Hi,

In your first question you said that.. you connected inverting terminal to ground. and non-inverting terminal to AC signal..
Then where have you done the DC sweep.. and how did you measure the offset..

And.. you connected the AC source in non-inverting input .. so if you measure the gain means it is not a differential gain..
Differential gain should be measured by giving oppsite phase inputs to the inverting and non-inverting terminals.

Thanks

 

[D.A.(Tony)Stewart]

Driving (+) with (-)=0V gives a differential gain of 1 + 1million (GBW) so it is correct.
It is often done with voltage divider to lower Source resistance near zero to reduce noise. ie. 1M to 1Ohm divider with 10V gives 10uV source. Then iput offset current * R =0