# What is a common mode signal in a layman's language and why do we care for it while studying Differential Amplifiers?

I searched on the topic 'Common Mode' on the internet but what I found is the application of common mode and without any simple explanation. Well, one needs to have a master's degree in order to understand what the internet says about common mode signals. Can anyone help me with the basics of common mode signal and its use in the differential amplifier, assuming that I am a 5year old kid and does not know ABC of electronics?

## Example to motivate the answer

Imagine a microphone at the end to two long wires. You connect these two wires to the microphone input of a amplifier. What comes out of the speakers is a lot of hum, with sometimes a faint signal in there.

What is happening is that the microphone is putting out quite small signals. These are just a few mV for ordinary human-scale sounds. However, the long wire is also capacitively coupled to the power wires in the walls, line cords on the floor, etc. That capacitive coupling is small, but you're also starting with large signals. The 60 Hz power wires may be carrying 120 V. Even if that is attenuated by 10,000 onto the wires, that's still 12 mV, which is still around a order of magnitude more than the microphone signal.

Note that both wires were picking up the 60 Hz hum. One of the wires is connected to the ground side of the microphone input of the amplifier. The hum picked up on that wire doesn't matter because it is grounded at the amplifier. The amplifier considers the voltage on the ground input 0 by definition. However, the hum on the other wire looks like real signal to the amplifier, and can't be distinguished from the microphone signals.

Now imagine you didn't have a chip on your shoulder about understanding electronics and you thought about this problem for a little bit. What if you could use the hum on the line you grounded to your advantage? Both wires are going to pick up pretty much the same hum. If you could subtract the voltage on one wire minus that on the other, then amplify the difference, you'd be able to cancel out a lot of the hum.

There are amplifiers which treat two input signals equally and perform this difference before amplification. In fact, this is common in high end microphone systems for this very reason.

One way you can do this yourself is by adding a audio transformer right in front of your existing amplifier. The magnetic field in the transformer is produced by current flowing thru the primary winding. Both wires floating up and down at the same voltage don't cause any current. The whole primary winding floats up and down in voltage with the hum, but none of that causes a magnetic field in the transformer, and therefore doesn't cause any signal coming out of the secondary.

So to now finally answer your question, the first implementation is called single ended. The second implementation with either a amplifier that takes the difference with active circuitry, or a transformer that takes the difference with basic physics, is called a differential signal.

We often talk about the differential and common mode signals on a pair of wires. The differential signal is simply the difference between the two, and the common mode signal that part of the signal that is the same between the two. You can think of the common mode signal as being the average, if that helps visualize it.

## More on handling microphone signals

In practice, just connecting two wires to a microphone and treating the result differentially isn't good enough. Remember that the power voltage is about 5 orders of magnitude greater than the microphone signals. If you attenuate it by 100,000, it's only down to about the level of the signals you want to hear. Nobody would want to listen to that. You probably need to attenuate the hum by another 20 dB to not be overwhelming, maybe another 20 dB to not be annoying, another 20 dB to be acceptable for something like telephone quality, and another 30 dB or so to be inaudible and acceptable for "HiFi" audio.

If you followed all that, the hum needs to be reduced about 90 dB below the intended signal levels. That's 4.5 orders of magnitude in voltage, in addition to the 5 we started with just to get 120 VAC down to about the same amplitude as the microphone signals. This illustrates why hum in microphone signals is something that needs to be carefully addressed. You can't have more than about 10-10 of the power line voltage get into the microphone signal.

So what do you do? Treating both wires identically, then using only the differential mode signal (ignoring the common mode signal) is certainly a important part of the solution, but not good enough by itself. This is what is meant by a balanced microphone signal, which is the standard in professional audio.

The two wires are additionally enclosed in a shield. This shield is connected to ground. It essentially gets between the noise source and the wires, and blocks the capacitive pickup of external signals.

Another trick is to twist the two signal-carrying wires around each other. That makes it more likely that whatever noise they do pick up is the same on both wires. In other words, it makes more of the noise common mode instead of differential mode. That allows the differential front end of the microphone amplifier to reject the noise.

If you are careful, use twisted pair signal wires, inside a shield, and a good differential front end on the amplifier, you can get the necessary 1010 or more attenuation of external signals.

• You are the "Guru"........... – Meem Sarkar Sep 26 '17 at 17:12
• Then there is magnetic field injection. – analogsystemsrf Sep 27 '17 at 4:47

A differential amplifier has two inputs. The differential-mode signal is the difference between the two. The common-mode signal is the shared component. Perhaps a schematic will better demonstrate:

simulate this circuit – Schematic created using CircuitLab

The common-mode can be though of as an "average" between the two. In this schematic we have a common-mode voltage of 1V, and a differential mode of 1V.

Why is this important for a differential amplifier? The key is in the name: A differential amplifier should amplify the differential. An ideal differential amplifier will ignore the common-mode voltage.

However, in practice this is not the case. As the common-mode voltage changes, due to effects such as mismatch, we might find the output changes too. This is sometimes called the "Common-mode gain". The ability to reject common-mode signals is the Common-mode-rejection-ratio, or CMRR. This can be important since it might introduce errors on differential measurements, allow noise to enter into the signal, etc.

• Do you mean +500mV into Mode1 and -500mV into Mode2? – KMC May 25 '20 at 22:56

It's true that there are bad articles on the internet, but there are some good ones, too.

So, if you have two voltages (such as the the inputs to an op amp) with voltages A and B,

1) the common mode signal is (A + B)/2.

2) the differential mode signal is (A - B) or (B - A), depending on your signal convention.

This is useful for high-gain amplifiers such as op amps, since they normally are operated with the two inputs held very close together by negative feedback. Then the differential mode signal is very small, and the common mode (as a single number) does a good job of describing the voltages at the two inputs.

In other words, if the two inputs to an op amp are 5.995 and 6.005 volts, the common mode voltage is 6 volts, and the differential mode voltage is 0.01 volts or -0.01 volts, depending on your convention. For convenience in such a situation, the positive value is likely to be the one reported, unless the specifics of polarity are important.