1
\$\begingroup\$

I don't understand the physical significance of Common Mode Voltage.

It is arbitrarily defined as the average of two voltages and is expected to be rejected by a differential amplifier. Why? What is its physical significance?

EDIT my question was specific to differential amplifiers. Both inputs to the differential amplifier need not be equal and opposite. What would be the meaning of common mode voltage in such case?

\$\endgroup\$
7
\$\begingroup\$

For common mode signals you get two wires: one with a signal and one with the inverse of that signal. If you add them you'll get zero, if you'd take the difference you'll get the signal.

What's the cool thing about this. If the two wires (the pair of wires) picks up some noise. Than the noise will be added on both wires. If you add the two signals you only keep the noise.. But if you take the difference you'll get the signal without the noise.

Signal is S (without noise)
Cable a => S
Cable b => -S

The Difference is S - (-S) = 2*S

Signal is S (with noise)
Cable a => S + noise
Cable b => -S + noise

The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S

So the noise will be cancelled.

example image: enter image description here

When Common mode signals are used they are mostly transferred with twisted pair cables like this: enter image description here

\$\endgroup\$
  • 3
    \$\begingroup\$ The two wires are kept very close to each other and usually twisted together to make sure, as far as possible, they pick up exactly the same noise. \$\endgroup\$ – Brian Drummond Aug 29 '15 at 15:09
  • \$\begingroup\$ I added a picture of a twisted pair. Forgot to mention them, thnx. \$\endgroup\$ – Bruce Aug 29 '15 at 15:13
  • \$\begingroup\$ Thank you! But my question was specific to differential amplifiers. Both inputs to the differential amplifier need not be equal and opposite. What would be the meaning of common mode voltage in such case? \$\endgroup\$ – Aditya Patil Aug 29 '15 at 15:18
  • 1
    \$\begingroup\$ My answer is specific about differential amplifiers: "The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S" This is specific on the input of the differential amplifier \$\endgroup\$ – Bruce Aug 29 '15 at 15:24
  • 2
    \$\begingroup\$ Also there is something called CMRR common mode rejection ratio this is a very high value that represents the attenuation on the input of common signals like noise. \$\endgroup\$ – Bruce Aug 29 '15 at 15:25
3
\$\begingroup\$

Because you specifically ask for common-mode signals in differential amplifiers:

It is the purpose of a diff. amplifier to amplify ONLY the difference between two input signals with diff.gain Ad. However, this is not completely and fully possible (not ideal) because no circuit behaves ideally. Hence, the common-mode gain Acm is - more or less - a "quality figure" of a diff. amplifier. Low common mode gain gives a high quality figure - in particular we define the "common mode rejection ratio" CMRR=Ad/Acm (in dB).

In this context, it is important to know that each diff. signal (V1-V2) at the input - even if one input is zero - contains also a common-mode signal Vcm because both signals V1 and V2 can be split into two parts:

V1=Vcm+Vdd and V2=Vcm-Vdd

V1-V2=2Vdd

Vdd=(V1-V2)/2 and Vcm=(V1+V2)/2

Now - the diff amplifier should have a large diff gain Ad and a low common-mode gain Acm. For the classical diff. amplifier (long-tailed pair with two transistors) a very large CMMR (low gain Acm) can be achieved using a BJT current source in the common emitter leg (large fedback for common-mode signals Vcm).

EDIT/UPDATE: Explanation - the voltage component Vdd is called "push-pull signal" because it has different signs within V1 and V2. That means: The common-mode voltage Vcm can be seen as a arithmetic mean between V1 and V2, and Vdd is the component that must be added (subtracted) for gettuing V1 or V2, respectively.

\$\endgroup\$
1
\$\begingroup\$

The common mode voltage reaching the input of a differential amplifier is (as mentioned) the unneeded part of the input referenced to some specified circuit ground (common).

The reason it is an issue and specified as a maximum is usually due to limitations of the amplifier input circuits voltage range.

For instance if you are expecting a a differential signal of maximum 700mV peak to peak it is within the working limits of many (perhaps most) differential amplifiers if the common mode voltage limits are not exceeded. Often the type of power connection to the amplifier will determine the common mode limits. A circuit with just a single +3.3V supply may have a common mode range of just +1V to +2V while a more professional (or more old school) design might have +15V and -15V supplies and offer a common mode voltage range of -12V to +12V referenced to the circuit ground.

There are sometimes ways to increase the common mode range (with input attenuators and level shifters and such) but usually with some sacrifice.

If the input is capacitor or transformer coupled the common mode voltage is usually much less of a problem but this is many times not possible if the accuracy of a differential amplifier circuit is to be maintained for small differential voltages.

\$\endgroup\$
0
\$\begingroup\$

Important for common mode signals is that often we do not want them. The signal is the differential signal while any disturbance enters the signal as a common mode signal. So ideally a differential amplifier ignores the common mode signal

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.