Common Mode Rejection Ratio of op-amp

Question

CMRR of op-amp is the ratio of the differential mode gain and common mode gain. What is the difference between these two ? What is the importance of CMRR in the performace of op-amp? How does CMRR in affecting the offset voltage and output voltage ?

Answer 1

To demonstrate the difference, here is the basic form of a differential amplifier which makes up the input stage for an opamp:

Notice there are two signals input at each side. SIG and SIG_INV are a 1kHz differential input (SIG is 180° shifted in phase from SIN_INV), and SIG_COM is a 9kHz common mode input (same signal at each side referenced to ground, i.e. 0° phase difference)
These signals are both at a 10mV (20mV pk-pk) level.

Now lets have a look at the simulation:

We can see the input (referenced to ground) is the mix of both signals, but the output is only the 1kHz differential signal at roughly a gain of 100. The differential amplifier has rejected almost all of the 9kHz common mode signal.

To see exactly how much of the 9kHz signal gets through to the output, here is the simulation again with only the 9kHz signal present:

Now we can see the output is roughly 10mV pk-pk (+/-5mV), so there is a gain of 0.5. We can now calculate the CMRR as we know the differential gain is 100 and the common mode is 0.5, so 100/0.5 = 200 = 46dB.
This is not a very good ratio, but it's the most basic form of differential amplifier. An typical opamp will improve greatly on this figure by for example, using a current source instead of the common tail resistor (R3) (also other things too).
For interest's sake, I just replaced R3 with an ideal current source and this reduces the common mode output to 324uV pk-pk (for 20mV pk-pk in) so the common mode gain is 0.0162 and thus the CMRR is improved to 20 * log10 (100 / 0.0162) = ~75.8dB. A high quality opamp might reach 120dB or more.

Calculating CMRR from component values

In the differential amplifier above, we can calculate both differential gain and common mode gain pretty easily. Here are the formulas with a brief explanation:

The differential gain is:

Gdiff = Rc / (2 * (Re + re)) where Re is the emitter resistors value and re is the intrinsic emitter resistance, given by ~25mA / Ic.
So, for our circuit above, we get:

re = 25mA / 100uA = 250Ω
Gdiff = 75k / (2 * (100Ω + 250Ω)) = 107, which agrees with our simulation.

The common mode gain is given by:

Gcm = -Rc / ((2*Rtail) + Re + re) - the minus sign means the output is inverted (180° shift) Rtail is R3 in the schematic above (the differnetial pair is sometimes referred to as the "long tailed pair", so this is the "tail" resistor)
So, we get:

Gcm = -75kΩ / (2*75kΩ) + 100 Ω 250Ω) = ~-0.5, which again agrees with our simulation.

The CMRR can either be calculated using the above results, or can be calculated directly using:

20 * log10(Rtail / (Re + re)) = 20 * log10(75kΩ / (100 + 250)) = 46.6dB, which again agrees with what can can see in the simulation.

From the above formula, we can see that the ratio between the tail resistor and emitter resistor is the main factor controlling the CMRR, so using a high impedance current source improves things dramatically.

The above equations don't take everything into account (you will need to do some further reading for the more subtle effects), but get you close enough for most applications.

Answer 2

The opamp's transfer function is

\$ V_{OUT} = G \times (V_+ - V_-) \$

Where \$G\$ is the gain. So when both inputs are equal the output should be zero. For real-world opamps this is not quite so. If you apply 10 V to both inputs you'll have a small output voltage which is higher than when you apply 5 V to both inputs. A CMRR of 100 dB will attenuate this common input level by a factor 100 000, so the 10 V will be reduced to 100 µV.

The higher the CMRR the better. An ideal opamp shouldn't show anything at all of a common mode input signal.

Answer 3

Generally, the differential mode gain is the gain of difference of signals, it is often found by just taking the single ended output signal gain of a 2 input op-amp and dividing through by the input difference. Common mode gain is how much of the common input signal gets passed through to the output side divided by the differential input signal.

The extremely IMPORTANT significance to remember, is that CMRR tells how well a differential input amplifier rejects noise that is common to both input lines. Imagine that you have 60Hz noise on both lines. With a good CMRR, very little of that unwanted noise gets passed to the output. It is also a major reason why you see that differential techniques are so commonly employed in op-amps.

Answer 4

Common Mode Rejection Ratio is the ratio between differential mode voltage gain and common mode voltage gain. The greater the CMRR, the greater the ability of a DA to reject common mode signals.
There are two input signals of DA: one is common mode signal, the other is differential mode signal. When input signals of DA become same phase and same amplitude it is called common mode signal. When input signals are in same amplitude but 180 degree phase shift then it is called differential mode signal.