[sdiy] Resistors in feedback loop of noninverting op amp buffer

[Magnus Danielson]

Donald Tillman skrev:
>    > From: Aaron Lanterman <lanterma at ece.gatech.edu>
>    > Date: Sat, 22 Nov 2008 02:18:50 -0500
>    > 
>    > Every once in a while, I see something that's clearly a unity gain  
>    > noninverting op  buffer, but the designer has put a resistor in the  
>    > feedback loop. There's no resistor from the negative input terminal to  
>    > ground, so it's not like they're trying to provide gain. All the  
>    > textbooks show a wire from the output to the negative input terminal.
>    > 
>    > Why do people sometimes put a resistor there? When inputting  
>    > schematics into Eagle I keep wanting to just replace them with wires,  
>    > but I don't want to do that if there's a good reason they are there.
> 
> Hey Aaron,
> 
> A fine question!
> 
> In a real world bipolar opamp, the inputs are the bases of a pair of
> diff amp transistors.  These transistors are running at some quiescent
> CE current, and therefore need some dc current in their bases, 1/beta
> of the CE current.  That's called the input bias current, and it gets
> sucked in off of whatever the voltage source is on the inputs.
> 
> That feedback resistor is to insure that any voltage drop due to the
> input bias current times the input resistance at the positive input is
> roughly equal to the voltage drop due to the input bias current at the
> negative input.  So it corrects for any offsets due to input bias
> current.
> 
> The resistor is not necessary for FET input opamps.

Don nailed what I wanted to say on this. Thanks Don!

One has to recall that it is the aggregate resistive path that is 
important here, as this is a DC bias aspect. If you have 1 kOhm 
resistive into the positive path and hooks the negative input directly 
to the output, you would think that the output would follow the input. 
It does, it is just that the input is offset by the input bias current 
pulling through the 1 kOhm resistor. If this offset is of no concern to 
you, then you are fine. If you do want to reduce bias offsets, insert a 
1 kOhm resistor between the output and the negative input, then will the 
negative input be offset about the same as the positive input and the 
output will more closely follow the intended input voltage on the 
positive pin.

Naturally, there is always inperfections, so there is a difference 
between the input currents of the positive and negative, and that is the 
input offset current, which is usually much less. Exact trimming of 
resistors may not be the best means of compensation, but it would be a 
direct way.

I find Sergio Fancos book "Design with operational amplifiers and analog 
integrated circuts" a good reference with clear concepts and math. It 
should be a great extention to the TAOE which I assume is near by most 
anyways.

Cheers,
Magnus