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
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
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.
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.