[kell]

Let's say you're measuring voltage across a sense resistor on a dc line
where the voltage with respect to ground fluctuates between 12 and 15
volts. Your amp has a differential mode gain of 10, so when the sense
resistor voltage is 100 mV, the amp output is 1 V. So far so good.
But I get confused with common mode rejection ratio. Is it with
respect to the absolute common mode fluctuation, or is it with respect
to the proportional variation?
Let's say your common mode rejection ratio is 40 dB. That's 1/100.
Does that mean

the absolute error in your amp output equals 1/100 of the absolute 3
volts common mode fluctuation, i.e. 30 mV; or

the proportionate error in your amp output is 1/100 of the
proportionate fluctuation in the common mode, or 1/100 of 25%, which
would be an error of 2.5 mV when the amp is putting out 1 volt.

 

[Bob Eld]

Let's say you're measuring voltage across a sense resistor on a dc line
where the voltage with respect to ground fluctuates between 12 and 15
volts. Your amp has a differential mode gain of 10, so when the sense
resistor voltage is 100 mV, the amp output is 1 V. So far so good.
But I get confused with common mode rejection ratio. Is it with
respect to the absolute common mode fluctuation, or is it with respect
to the proportional variation?
Let's say your common mode rejection ratio is 40 dB. That's 1/100.
Does that mean

the absolute error in your amp output equals 1/100 of the absolute 3
volts common mode fluctuation, i.e. 30 mV; or

the proportionate error in your amp output is 1/100 of the
proportionate fluctuation in the common mode, or 1/100 of 25%, which
would be an error of 2.5 mV when the amp is putting out 1 volt.

If I read what you say correctly, the voltage at both ends of the sense
resistor can be 15 Volts; i.e. as much as 15 volts of common mode with
respect to ground. Is that correct? If the CMR is 40dB or one part per 100,
the offset refered to the input is as much as 15/100 or 150mV. With a gain
of 10, 1.5 volts of offset can occur on the output. It has nothing to do
with the variation except to say that a steady, known offset can be nulled
out. If you have that much common mode voltage and that poor CMR, you're in
trouble. Furthermore, 15 volts of CM is beyond what most op-amps can handle.
Bob

 

[kell]

If I read what you say correctly, the voltage at both ends of the sense
resistor can be 15 Volts; i.e. as much as 15 volts of common mode with
respect to ground. Is that correct? If the CMR is 40dB or one part per 100,
the offset refered to the input is as much as 15/100 or 150mV. With a gain
of 10, 1.5 volts of offset can occur on the output. It has nothing to do
with the variation except to say that a steady, known offset can be nulled
out. If you have that much common mode voltage and that poor CMR, you're in
trouble. Furthermore, 15 volts of CM is beyond what most op-amps can handle.
Bob

I had something like this in mind, common mode error caused by resistor
error:


12-15v
o---+----/\/\/\----+------------o
| Rs=.01 |
| |
R3 R1
| |
| +---R2---,
| | |\ |
| '--|-\ |
| | >--+---Vo
+-----------------|+/
| |/
|
R4
|
|
o----+---------------------------o

for (R2/R1)=(R4/R3) , Vo=RsIR2/R1

so for R1=R3=1K and R2=R4=10K
and I=10A, Vo=1v

Now let's say R1, R3 and R4 are exact but you have an error of 10% in
R2.

How much error does that result in at vo...

Now I get 136 mV error at vo when the input is at 15 volt

is that correct?