What do you mean? what problem?
Hi,
Another problem such as like we talked about, the connections on the plug board not being very good.
A mismatch in only one of the input resistors by 1% would cause a CMRR of 46db, which is less than what the data sheets says for most of the op amps we've looked at so far. That means we would not be measuring the CMRR of the op amp, we'd be measuring the CMRR caused by the resistors mostly. And that's with a perfect op amp. So we'd turn a perfect op amp into an op amp with CMRR only 46db just by changing one resistor. You can see why this is so important.
If you want to match sure the resistors are matched then get some 0.1 percent resistors or else if you have a bunch of resistors of the same value you can go through them and match them using a good Ohm Meter or a voltage divider. Find at least two sets of two that are very very close in value.
For your convenience, here is the transfer function for the single op amp set up as differential amplifier with a finite open loop gain Aol:
Vout=(Aol*(Va*R2*R4-Vb*R2*R3+Va*R2*R3-Vb*R1*R3))/((R2+R1)*(Aol*R4+R4+R3))
and here is the same with infinite open loop gain (for other theoretical purposes):
Vout=(Va*R2*R4-Vb*R2*R3+Va*R2*R3-Vb*R1*R3)/((R2+R1)*R4)
where
R1 is the input resistor to the non inverting terminal,
R4 is the input resistor to the inverting terminal,
R2 is the resistor that goes to ground,
R3 is the feedback resistor,
Aol is the open loop gain,
Va is the input voltage to the non inverting input,
Vb is the input voltage to the inverting input,
Vout is the output voltage.
Those expressions allow us to examine the effects of any mismatch condition without needed a simulator.
We could do the same for the current mirrors, but that's a little more complicated because we need the expressions for the transistors.
BTW, thanks to whomever changed the title of this thread for the spelling error. I was so sick of seeing the incorrect spelling ha ha
Thanks.