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# Common mode voltage - RF Cafe Forums

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Below are all of the forum threads, including all the responses to the original posts.

Blix
Post subject: Common mode voltage Posted: Sun Jul 31, 2005 4:08 pm
I am working on an op amp project and was wondering if someone could explain common mode voltage, in the context of op amps and in general. From what I have read it is voltage that is common to both inputs of the op amp and is not desirable.

Thanks

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Guest
Post subject: Posted: Sun Jul 31, 2005 9:27 pm
Equal voltage at both inputs should result in no voltage at the output.

How well an op amp acheives this is called common mode rejection.

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Guest
Post subject: Common-mode voltagePosted: Sat Aug 06, 2005 11:32 am
Hi!

Whenever you have two terminals (let's call them A and B), neither of which is ground, you can talk about the voltages two ways:

1. You can measure each to ground, getting VA and VB
2. You can measure the voltage between the terminals, getting the differential voltage Vd = VA - VB, and then measure the average voltage of the two terminals to ground: VC = (VA + VB)/2. This last voltage is the common-mode voltage, and whether it's good or bad depends on what you're trying to do.

For example, if you're using a 5-volt-only opamp, then the inputs might both want to be at about 2.5 volts, so that they can both go up and down equal amounts. That 2.5 volts would be common-mode voltage.

In some opamp circuits, you want to look ONLY at the voltage difference between the two terminals. In that case, you don't want your opamp to respond to ANY change in the common-mode voltage AT ALL. Unfortunately, you can't build an opamp perfect in that way, so all opamps have a "common mode rejection ratio".

An example of wanting only the difference voltage would be a professional microphone with balanced output. Induced hum due to nearby AC power will generally be equal on each of the two signal wires, so it won't be in the difference at all. Any common mode rejection less than perfect would mean that you'd hear some hum in the output.

Good Luck!

Posted  11/12/2012
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