[SOLVED] current SNR change after passing a ideal current mirror

Hi,

I have a question about SNR (signal-noise-ratio) change. If a current with SNR=x passes a noiseless current mirror. The current mirror scale factor is A (A can be larger or lower than 1). How's the current mirror output SNR? maintain the same or change by A. More specific question: what is the change factor of input noise?

I think the same question can be extended to a broad one: A input signal with SNR =x passes a gain A noiseless circuit. At the output, how's the noise change and how does the SNR change? Thanks.

erikl said:

By definition, the noise doesn't change when passing a noiseless amplifier. The SNR changes with A² (because SNR is a power ratio, whereas A is a voltage ratio).

Example: A=10 : SNR_new/SNR_old = 100 ≙ 10*log₁₀(100) = 20*log₁₀(10) = 20dB

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Thanks for your reply. But I am still not clear. Why the noise does not change when passing a noiseless amplifier? By what definition?

hyleeinhit said:

Why the noise does not change when passing a noiseless amplifier? By what definition?

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You are right: the incoming noise will be amplified by the same factor A , so the SNR won't be changed. SNR just won't be worsened, because the noiseless amplifier doesn't add any noise.

erikl said:

You are right: the incoming noise will be amplified by the same factor A , so the SNR won't be changed. SNR just won't be worsened, because the noiseless amplifier doesn't add any noise.

Click to expand...

I　agree with your point.