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I searched for information about common mode signals but I only got a vague idea about it, such as it is a noise signal generated between input signal and common ground.

However, I do not understand how it is generated in differential amplifier.

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    \$\begingroup\$ It is not generated -- it is an input. By definition, it is the average of the positive and negative inputs. \$\endgroup\$ – Scott Seidman May 6 '16 at 16:45
  • \$\begingroup\$ If you mean a differential driver... then it's 2 op-amps, one voltage follower and one inverter. driven by the same signal. \$\endgroup\$ – Spoon May 6 '16 at 20:30
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The differential amplifier amplifies the difference between the signals in the +ve terminal and the negative terminal. And no they do not generate common mode signals but rather block them

That is, Vout = Adiff(Vplus - Vminus) . Adiff is the gain with which it amplifies and usually a differntial amplifier has a differntial gain of 30-45 Db.

When both the +ve terminal and the negative terminal of the amplifier is given the same voltage then since Vout is just a scalled version of the difference between the input in the 2 terminals, it becomes 0 . For example let Vcommon be the voltage at the -ve and +ve terminal , now when you do Vout = Adiff(Vplus - Vminus)= Adiff(Vcommon - Vcommon) =0 . Since noise is common to both the -ve and +ve terminal of the diff amp it gets cancelled and thus differential amplifiers are less affected by common mode signals/noise. Though this is in theory in pratice every diff amp is slightly affected by the common mode signals/noise. They amplify the common mode signal in a very small amount and this is because of the internal mismatches in opamps internal transistor design. This is called the common mode gain of the differential amplifier. A good diff amp has a very low common mode gain. CMRR whihc is the common mode rejection ratio tells exactly this on how good a diff amp reject the common mode signal.

The image below shows the differential amp configuration and the resistors acts as a means to control the gain Adiff.

enter image description here

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A conventional op-amp is sometimes modeled as:

\$V_{OUT} = A_V (Vin^+ - Vin^-)\$ where Av is a large positive number (open-loop gain).

That's about the simplest non-ideal op-amp model.

An enhancement is to add another term to account for the fact that the op-amp does care a bit about what the two voltages actually are rather than just the difference between them.

So we get:

\$V_{OUT} = A_V ((Vin^+ - Vin^-)+ Ac(\frac{Vin^+ + Vin^-}{2}))\$

Where Ac is the input-referred common mode gain, and for a good op-amp it will be << 1.

For example, the ancient sort-of precision OP-07 has a DC common-mode rejection ratio (CMRR) of 120dB typical, so a 1V change in the common-mode voltage is equivalent to a difference of 1uV at the inputs. The open loop gain is typically 400,000 so it would represent a 400mV change at the output, in theory.

For a reasonable closed-loop gain of (say) 100, you will get only 100uV change at the output, which is small but often significant.

If you look at a differential amplifier made with an op-amp as @Bhuvanesh has shown, the CMRR will typically depend mostly on the matching of the resistors. In the example of a gain-of-100 amplifier, a 0.01% mismatch in the resistor values will completely overwhelm the op-amp rejection.

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    \$\begingroup\$ The idea that the 3-op-amp configuration has a better CMRR is simply a myth, and responsible for amps with lousy CMRR all over the world, accompanied by plenty of head scratchers wondering why. The output stage is EXACTLY the same as in @BhuvaneshN answer. What makes the CMRR really good is buying an INAMP as an IC, with the resistors precisely trimmed and placed to minimize temperature problems. \$\endgroup\$ – Scott Seidman May 6 '16 at 17:36
  • \$\begingroup\$ ... but +1 anyway! \$\endgroup\$ – Scott Seidman May 6 '16 at 17:36
  • \$\begingroup\$ @ScottSeidman Yes, you're right. Comment deleted. It does in fact reduce the sensitivity to resistor matching in the output section but only by 2:1 in the limit for very high gain, because the inputs are driven differentially. \$\endgroup\$ – Spehro Pefhany May 6 '16 at 18:17
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Common-mode means a signal that is equal=same on BOTH sides of the differential pair. A common-mode signal is UNwanted NOISE. So we use a differential amplifier input (or a transformer) to REJECT common-mode signals. The noise comes from interference getting into the wire along the path from the source. It is not "generated", it is an accidental artifact of passing through the Real World.

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