How to use a differential amplifier ?

Yup, pretty much. There's a brief discussion at

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OK, from National's AN31,:

here's a differential (difference) amplifier: (View in Courier)

R2 +--[10K]--+ R1 | | V1-----[10K]-+-|+\\ | | >----+-->Vout V2-----[10K]-+-|-/ R3 | [10K]R4 | GND

In general:

/ R1 + R2 \\ / R4 * V2 R2 * V1 \\ Vout = | --------- | | --------- - --------- | \\ R3 + R4 / \\ R1 R1 /

For R1 = R3 and R2 = R4:

R2 Vout = ---- (V2 - V1) R1

Let's say that V1 is 1V and V2 is 3V. Then the output voltage will be:

R2 10kR Vout = ---- (V2 - V1) = ------ * (3V - 1V) = 2V R1 10kR

Now let's add a little 'noise' to both of the inputs in the form of an identical DC signal added to both inputs.

Actually, let's add a whopping big signal, 5V, to each of the inputs.

Then we'll have:

R2 10kR Vout = ---- (V2 - V1) = ------ * (3V + 5V) - (1V + 5V) = 2V R1 10kR Same as before, even with a huge common mode signal in there as well. "Common mode" because it appears in common at the input ends of R1 and R3.

Amazing, huh? :-)

JF

Yes, that is mainly how they are used(in fact I do not know any other way but I'm not an expert).

You input f and -f on on the inputs and you get f - (-f) = 2f on the output(obviously when configured as an amplifier).

Now if you end up with some noise on the two inputs, say e1 and e2, then you have

f + e1 - (-f + e2) = 2f + (e1 - e2)

If you can make e1 and e2, which in general are small anyways, have a high correlation, then e1 - e2 ~= 0 and you get 2f. if e1 and e2 do not correlate, or even say that e2 = -e1 then obviously it does no good... but such things don't really happen in practice such as using twisted pair or even just straight(twisted pair makes the correlation between e1 and e2 even greater).