[sdiy] Resistor math

[Rutger Vlek]

Thanks, yes it's indeed the summing amplifier. However, your 2 cents seem to be an answer to a different question. What I meant was that I'm using a 8:1 mux to select inputs, but I'm not selecting single inputs, but selecting different weighted sums of signals. My question is how to take the mux's ON-resistance into account when computing values for the weighting network?

Rutger.


On 21 mei 2013, at 14:12, Julien Delgoulet wrote:

> Hum …
> 
> Have you checked this pdf : 
> 
> http://www.stanford.edu/class/ee122/Parts_Info/datasheets/op_amp_circuit%20collection_AN-31.pdf
> Your circuit is the "Inverting Summing Amplifier", a classic one.
> Also if you are using an analog switch to "select" inputs, may be you could use an analog switch to select the resistor in the feedback loop of the opamp …
> 
> My 2 cents ..
> 
> Julien
> 
> 
> Le 21 mai 2013 à 11:50, Rutger Vlek <rutgervlek at gmail.com> a écrit :
> 
>> Hi all,
>> 
>> I need a little help calculating values for a network of precision resistors. The resistors are used for summing several signals with different weights (1x 2x 4x 12x, etc). Summing is done at the virtual ground of an inverting opamp. I've used 100K in the feedbackloop of the opamp, so to weight one of the inputs with 1x a resistor of 100K is normally used, and 50K for 2x and 25K for 4x. However... there is an analog switch in between the summing network and the inverting input of the opamp, with a typical ON-resistance of 40ohms. My question is, how does this affect the values that should be used for the summing network? And how to compute new values for those resistors that assure summation with the exact weights I need (and how to implement those with limited values available in a series of 0.1% tolerance precision resistors)?
>> 
>> Here's an overview of the summing network:
>> 
>> --1xresistor----|
>> --2xresistor----|
>> --4xresistor----|-----analog switch----inverting opamp---
>> --12xresistor--|
>> 
>> Any help is appreciated!
>> 
>> Rutger
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