# Noise Calculations

by SamBam77
Tags: calculations, noise
P: 3,842
 Quote by Greg-ulate I thought thats the point, the part has a larger amount of current noise, but the voltage due to that noise is small, and the voltage is what is being amplified at the end of the day. In contrast, for a constant current source driving the input with a resistance to ground, the BJT input amplifier would be undesirable because the current noise on the input would develop a voltage on the resistance. $$e_{n\_i} = i_{n}\times R_{eq}$$ if Req is small, then the noise is small
You got me. I have no sure answer, I can only speculate. The referred input voltage and current noise is a black box model of an IDEAL op-amp with added current and voltage noise. Ideal op-amp has infinite input impedance.
In real life( not the model), the noise current actually comes out of the input of the op-amp. The current become a voltage through the external impedance. Remember the op-amp has high open loop gain, the output will change to make the input goes to zero. So if the input noise current create a noise voltage, the output of the op-amp with generate an output to neutralize the noise so to make the input back to zero.

As you can see the article I posted in post #9, equation 3.3 gave Req=R1//Rf only, the input impedance is not in it. That's is also what I studied before.

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