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Balancing input of op amps

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nickthamma

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Why is balancing input of op amps so important? I have read that it enhances performance. In what way is the performance of an op amp affected by balancing the inputs?
 

Maybe because the transistors on the inside are inevitably slightly mismatched, or in a current shunt monitor there are resistors on the input side which it is impossible to match perfectly, as far as I have understood what I read.

I think it's because the mismatches create a minute voltage difference between the inputs which distorts the output from what it should be with an ideal device.

Hope this helps a little:

View attachment Nulling Input Offset Voltage of Operational Amplifiers sloa045 TI.pdf
 
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Not quite clear what you mean with "balancing op inputs". Please clarify.
 

2015-07-02_0033.png

For example, in an inverting amplifier configuration, adding a third resistor (balancing resistor, Rb) between the non-inverting input and ground to match the impedance looking out of the inverting input as shown above.
 

It has got nothing to do with the impedance because the impedance looking out of the inverting terminal is zero (Rf/Aopen). But in case of BJT input pair, Rb could be used for input offset cancellation.
 

AMS012 said:
It has got nothing to do with the impedance because the impedance looking out of the inverting terminal is zero (Rf/Aopen). But in case of BJT input pair, Rb could be used for input offset cancellation.
Click to expand...

I don't understand how the impedance looking out of the inverting terminal is zero. I was under the impression that it was Ri||Rf. Can you help me clear this confusion?
 

You are looking for resistance rather than "impedance". Rb has been used with classical bipolar OPs to cancel the addional offset caused by input bias current. As already mentioned "input balancing" could mean a lot of different things, I won't use the term here. Rb = Ri||Rf is the right value. In case, I would talk about source resistance balancing.

The resistor is only meaningful for OPs with |Ib| << |Ios|, that's only the case for bipolar OPs without bias current compensation.
 

The impedance looking into the inverting terminal of the closed loop opamp is zero. Any amount of current you pump into that node, its voltage will not change due to virtual ground effect. So delta V/ delta I is zero.
 

The impedance looking into the inverting terminal of the closed loop opamp is zero. Any amount of current you pump into that node, its voltage will not change due to virtual ground effect. So delta V/ delta I is zero.
Click to expand...
Obviously right, but not what the OP is asking for.
 

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