Electronics > Beginners
Why is noise input-referred?
ejeffrey:
The amplifier has an input whether you feed it a signal or not!
It is just a calculation: if I had a noiseless amplifier, how much noise would have to be applied to the input to get the signal I measured.
ZeroResistance:
--- Quote from: ejeffrey on June 21, 2019, 04:12:19 pm ---The amplifier has an input whether you feed it a signal or not!
It is just a calculation: if I had a noiseless amplifier, how much noise would have to be applied to the input to get the signal I measured.
--- End quote ---
Ok, so under what conditions of the input was the signal measured? and was it measured at the output?
IDEngineer:
Short, easy to understand answer: Because "equivalent input noise" allows you to treat the amplifier as a noiseless gain block, where all noise sources are external.
A bit closer to the ideal "straight wire with gain".
You already have to consider the noise contribution from external components like resistors, so this allows you to treat the noise contribution of the gain block itself as just another external signal source.
ZeroResistance:
--- Quote from: IDEngineer on June 21, 2019, 06:21:10 pm ---Short, easy to understand answer: Because "equivalent input noise" allows you to treat the amplifier as a noiseless gain block, where all noise sources are external.
A bit closer to the ideal "straight wire with gain".
You already have to consider the noise contribution from external components like resistors, so this allows you to treat the noise contribution of the gain block itself as just another external signal source.
--- End quote ---
So if I have the following spec
Input-Referred Noise: 4 μVPP (150 Hz BW, G = 6)
Would it mean that I get a 4uVpp noise at the output of a block when the gain of that block is set to 6.
But at what input conditions?
Siwastaja:
--- Quote from: ZeroResistance on June 21, 2019, 06:26:45 pm ---So if I have the following spec
Input-Referred Noise: 4 μVPP (150 Hz BW, G = 6)
Would it mean that I get a 4uVpp noise at the output of a block when the gain of that block is set to 6.
--- End quote ---
No, you'd get 6*4µVpp. Because it's input-referred, meaning: equivalent of you inserting 4µVpp of noise to the input of an imaginary noiseless 6x amplifier.
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