effect of resistor tolerances on gain of op-amp

Thread Starter

karthikvjit

Joined Mar 23, 2008
1
can anybody please explain me how the gain of the op-amp gets affected with the tolerances of the resistors used in the inverting or non-inverting configurations.
 

hgmjr

Joined Jan 28, 2005
9,027
I have moved your post to its own thread so that any replies that you receive will relate to your post.

It is best if your avoid posting your questions into an active or even inactive post to avoid any confusion between replies that relate to your question with those replies relating to the original poster's question.

Welcome to the AAC forum by the way. I see that this is your first posting. Hopefully you will receive the help you seek.

Please have a look at our ebook at www.alllaboutcircuits.com where you will find some very helpful informaion on opamps as well as other electronic topics.

Good Luck,
hgmjr
 

SgtWookie

Joined Jul 17, 2007
22,230
Matching of resistors becomes quite critical when attempting to build instrumentation amplifiers for differential signals using discrete op amps. Even a small fraction of a percent can have a very serious negative impact upon the CMRR (Common Mode Rejection Ratio, or ignoring the parts of both + and - input signals that are the same) of the circuit.

Take a look at Intersil's Application Note 1298:
http://www.intersil.com/data/an/an1298.pdf
It covers a few of their instrumentation amplifiers, but also explains why the matching of the resistors in such circuits is so critical.

Fortunately, instrumentation amplifiers do exist, and they have internal resistors that have been laser-trimmed at the factory to match at very tight tolerances. You would be very hard-pressed to duplicate such performance using discrete components.
 

Audioguru

Joined Dec 20, 2007
11,248
The gain of an inverting opamp is determined by the ratio of two resistors. If the resistors have a tolerance of 5% then the gain could be perfect, +10%, -10% or anywhere in between.

The gain of a non-inverting opamp is determined by the ratio of two resistors plus 1. Then you need to calculate the error that is caused by tolerance.
 

Distort10n

Joined Dec 25, 2006
429
Fortunately, instrumentation amplifiers do exist, and they have internal resistors that have been laser-trimmed at the factory to match at very tight tolerances. You would be very hard-pressed to duplicate such performance using discrete components.
Some more information on this particular point. The absolute value of the internal resistors may vary +/-20% from a value, say 20k. But the resistors are trimmed to be matched within +/-.01% or better.

So designer may have 20k ohm resistors, but because of process tolerances the resistors may vary from 16k to 24k. However, each resistor will be 16k, 17k, 21k, etc. and matched to +/-.01%.
 
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