1. Common Mode Rejection Ratio

2. Introduction
The common-mode rejection ratio (CMRR) of a differential amplifier (or other device) measures the tendency of the device to reject input signals common to both input leads.
A high CMRR is important in applications where the signal of interest is represented by a small voltage fluctuation superimposed on a (possibly large) voltage offset, or when relevant information is contained in the voltage difference between two signals.

3. Cont..,
Ideally, a differential amplifier takes the voltages V + and V − on its two inputs and produces an output voltage Vo = Ad(V + − V − ), where Ad is the differential gain. However, the output of a real differential amplifier is better described as
where Acm is the common-mode gain, which is typically much smaller than the differential gain.

4. Definition
The CMRR is defined as the ratio of the powers of the differential gain over the common-mode gain, measured in positive decibels (thus using the 20 log rule):
As differential gain should exceed common-mode gain, this will be a positive number, and the higher the better.

5. The CMRR is a very important specification, as it indicates how much of the common-mode signal will appear in your measurement.
The value of the CMRR often depends on signal frequency as well, and must be specified as a function thereof.
It is often important in reducing noise on transmission lines. For example, when measuring the resistance of a thermocouple in a noisy environment, the noise from the environment appears as an offset on both input leads, making it a common-mode voltage signal.
The CMRR of the measurement instrument determines the attenuation applied to the offset or noise.

6. Measuring Common Mode Rejection Ratio
Common-mode rejection ratio can be measured in several ways. The method shown in Figure below uses four precision resistors to configure the op amp as a differential amplifier, a signal is applied to both inputs, and the change in output is measured—an amplifier with infinite CMRR would have no change in output.
The disadvantage inherent in this circuit is that the ratio match of the resistors is as important as the CMRR of the op amp.

7. A mismatch of 0.1% between resistor pairs will result in a CMR of only 66 dB—no matter how good the op amp! Since most op amps have a low frequency CMR of between 80 dB and 120 dB, it is clear that this circuit is only marginally useful for measuring CMRR (although it does an excellent job in measuring the matching of the resistors!).

8. Simple Common-Mode Rejection Ratio (CMRR) Test Circuit

9. The slightly more complex circuit shown in Figure below measures CMRR without requiring accurately matched resistors. In this circuit, the common-mode voltage is changed by switching the power supply voltages. (This is easy to implement in a test facility, and the same circuit with different supply voltage connections can be used to measure power supply rejection ratio).

10. CMRR Test Circuit Does Not Require Precision Resistors

11. The power supply values shown in the circuit are for a ±15 V DUT op amp, with a common-mode voltage range of ±10 V. Other supplies and common-mode ranges can also be accommodated by changing voltages, as appropriate. The integrating amplifier A1 should have high gain, low VOS and low IB, such as an OP97 family device.

12. Given: A 741 op amp with CMRR = 90 dB and a noise gain, AN = 1 k
Find: The common mode gain, Acm
Example 1:
Solution:
Acm = AN =
log-1 (CMRR / 20) log-1 (90 / 20) =
It is very desirable for the common-mode gain to be small.

13. Example 2: Solution: From the defining equation for CMRR: 5000
Common-Mode Rejection Ratio (CMRR)
Example 2:
A certain diff-amp has a differential voltage gain of 500 and a common-mode gain of 0.1. What is the CMRR?
Solution:
From the defining equation for CMRR:
5000
Expressed in decibels, it is
74 dB

14. Common-Mode Rejection Ratio (CMRR)
Example 3:
A certain diff-amp has Ad = 100 and a CMRR of 90 dB. Describe the output if the input is a 50 mV differential signal and a common mode noise of 1.0 V is present.
Solution:
The differential signal is amplified by 100. Therefore, the signal output is
Vout = Av(d) x Vin = 100 x 50 mV =
5.0 V
The common-mode gain can be found by
The noise is amplified by Therefore,
Vnoise = Acm x Vin = x 1.0 V =
3.2 mV

15. Quiz
1. When two identical in-phase signals are applied to the inputs of a differential amplifier, they are said to be
a. feedback signals.
b. noninverting signals.
c. differential-mode signals.
d. common-mode signals.

16. Quiz
2. Assume a differential amplifier has an input signal applied to the base of Q1 as shown. An inverted replica of this signal will appear at the
emitter terminals.
collector of Q1
collector of Q2
all of the above.
RC1
RC2 Q1 Q2 RE
-VEE

17. Quiz 3. A differential amplifier will tend to reject
noise that is in differential-mode.
noise that is in common-mode.
only high frequency noise.
all noise.

18. Quiz
4. The average of two input currents required to bias the first stage of an op-amp is called the
a. input offset current.
b. open-loop input current.
c. feedback current.
d. input bias current.

19. Quiz 5. The slew rate illustrated is a. 0.5 V/ms b. 1.0 V/ms
c. 2.0 V/ms
d. 2.4 V/ms
Vout (V) 12 10
-10
-12
10 ms

20. Quiz 6. For the circuit shown, Vf is approximately equal to a. Vin
b. Vout
c. ground.
d. none of the above. - +
Feedback network Vf
Vin Rf Ri
Vout

21. Quiz
7. For the inverting amplifier shown, the input resistance is closest to
a. zero
b. 10 kW
c. 2 MW
d. 8 GW Rf
150 kW Ri
Vin -
Vout
10 kW +

22. Quiz
8. For the inverting amplifier shown, the output resistance is closest to
a. zero
b. 10 kW
c. 150 kW
d. 8 GW Rf
150 kW Ri
Vin -
Vout
10 kW +

23. Quiz 9. The gain of the inverting amplifier shown is a. -1 b. -10
c. -15
d. -16 Rf
150 kW Ri
Vin -
Vout
10 kW +

24. Quiz 10. A voltage follower has a. current gain. b. voltage gain.
c. both of the above.
d. none of the above.

25. Quiz Answers: 1. d 2. b 3. b 6. a 7. b 4. d 8. a 5. c 9. c 10. a