Hi guys,
I am reading up a an article concerning CMRR at:
http://en.wikipedia.org/wiki/Common-mod … tion_ratio
There is two formulas, the first:
CMRR = 20Log(Ad/Acm)
and the 2nd:
Vout = (V+ - V-) x Gopenloop +/- Vcm/10^CMMR/20
When do we use the first one and when do we use the 2nd one?
The reason I ask is because the 2nd one seems to be used when we are testing an op amp alone (without any resistors connected)??? :?
All feedback is appreciated!
The wikipedia entry doesn’t seem to be a very helpful discussion and the example is not very useful. It deals only with an open loop op amp configuration.
Hi jremingtyon,
Thanks for reply.
“Question”
if I would figure out my Ad and Acm of a particular op amp circuit circuit, I would be able to simply use:
CMRR = 20Log(Ad/Acm)
to find the CMRR of my current circuit op amp configuration?
Yes.
Okay, so using that same CMRR that I got from:
CMRR = 20Log(Ad/Acm)
Can I simply plug it into:
Vout = (V+ - V-) x Gopenloop +/- Vcm/10^CMMR/20
to get Vout??
Personally, I don’t think so because that last formula is for open loop gain… right?
Sure why not ? But do be careful to know what V+-V- is when the circuit uses the op-amp in a feedback configuration. Especially as the open loop gain is not a well specified parameter. It's just guaranteed to exceed some number at DC. Nobody can tell you a specific number for the gain as it'll vary from op-amp to op-amp, with temperature and supply voltage, etc, etc. I think you need to understand more about feedback before my answer will make much sense to you.8Volts:
Can I simply plug it into:to get Vout??
Hi Mee_N_Mac,
Well, okay, suppose I have an op amp configuration (circuit like the old one I used) that has:
Vout = (Acm)(Vcm) + (Ad)(Vd)
0.5 Vdc = (-1.1)(3.5) + (1.45)(3)
then I do:
CMRR = 20Log(Ad/Acm)
CMRR = 20Log(1.45/1.1)
CMRR = 2.3995 dB
So your saying I can plug the CMRR back into the following equation to see the error margin CMRR would give:
Vout = (V+ - V-) x Gopenloop +/- Vcm/10^CMMR/20
Vout = (5 - 2) x 1.45 +/- (3.5/1.3181818181)
Vout = (3) x 1.45 +/- 2.655
Vout error could be:
Vout = 4.35 + 2.655 = 7.005 Vdc ??
or
Vout = 4.35 - 2.655 = 1.69 Vdc ??
confused!
Gopenloop is the gain of the operational amplifier with no feedback and is typically 1,000,000 (10^6) or higher. If you know what it is for a particular op amp, then for a given output voltage Vout, you can calculate the voltage difference at the inputs (Vin+) - (Vin-), which is typically a few microvolts. This is what Mee_n_Mac was referring to in an earlier post.
yes but my numbers came from a circuit with feedback … and
huuuummm …
okay,
I will start a new thread because I think I am not providing
all the details the way I should be!
I will do this tomorrow … in the mean time thanks so much for
your kind help!
Yes and that feedback pushes Vout so that the difference between V+ and V- (these are the voltages at the op-amp inputs) is very small. In fact the difference at those places in the circuit is proportional to the open loop gain of the op-amp, a number you won't know exactly in real-life. Note that the V+ and V- are not the inputs to the circuit, which is what you wrongly used in your post above. Also Ad of the circuit is not the open loop gain of the op-amp.8Volts:
yes but my numbers came from a circuit with feedback
ok Mee_N_Mac,
I have been crunching all sorts of numbers here and between
the verbal explanations and the math I think it’s starting
to sink in…
that last formula is for an open loop gain configuration as you
guys have been trying to tell me…using the sample at wiki
the 10v/10^ (90db/20) gives the error margin for the open
loop gain (+-) which then multiplied by Voltage differential
at the op amp inputs!!!
I don’t think I can use that same formula in its entirety
for my example though since I am using a feedback
configuration and don’t know what the open loop gain is
in that circuit!!!
So to use that last formula for an op amp in an open loop
gain configuration, where would one get the open loop
gain value … from the spec sheet?
thanks