[SOLVED] Calculating the input offset voltage for OpAmp

I have an OpAmp which states that its Input Offset Voltage is 2mV typical but I want to practically calculate the actual offset it will have at the room temperature. So I have connected the circuit in the following manner on the breadboard.



The Ouput voltage i got is 4.4mV, so does this means that the input offset in practice at the current temperature is (Output Offset / Gain) i.e 4.4mV/2V2 => 2mv ?

P.S : I also measure the output of 72mV when i connected the Inverting and non-Inverting inputs directly to ground. When i left the input both open(floating), the output comes out to be 50~54mV

Hi,

If you want to calculate output offset from datasheet, then you need to take care about input_bias_current, too.

Assuming input bias current is negligible:
Your circuit shows a gain for offset (and input referred noise) of (22k +10k) / 10k) = 3.2
So in your case the input refferred offset voltag is 4.4mV / 3.2 = 1.375mV.

i connected the Inverting and non-Inverting inputs directly to ground

Click to expand...

This gives no meanigful result. This method is useless.

Klaus

Your circuit shows a gain for offset (and input referred noise) of (22k +10k) / 10k) = 3.2
So in your case the input refferred offset voltag is 4.4mV / 3.2 = 1.375mV.

Click to expand...

What is input referred noise ? The gain shouldn't be Rf/Rin = 22k/10k=>2.2 in the above case ?

i connected the Inverting and non-Inverting inputs directly to ground

Click to expand...

That was just to know how the OpAmp behaves when both the inputs are applied the same PD, i.e. Zero in this case

- - - Updated - - -

Edit: KlausS you are right, By neglecting the input bias current, Vout/Vin comes out to be (10k+22k)/10k according to the above schematic.

KlausST said:

Hi,

If you want to calculate output offset from datasheet, then you need to take care about input_bias_current, too.

Assuming input bias current is negligible:
Your circuit shows a gain for offset (and input referred noise) of (22k +10k) / 10k) = 3.2
So in your case the input refferred offset voltag is 4.4mV / 3.2 = 1.375mV.


This gives no meanigful result. This method is useless.

Klaus

Click to expand...

I think they called that the output offset voltage. :wink:

Is the op amp a rail to rail input?