How to calculate noise voltage after amplification?

Question

I'm simulating the thermal noise experienced by some instrument, outputted as a voltage vs time plot. To calculate the "power" of this noise, I'm using:

P = 4kTB (k = Boltzmann's constant, T = temperature (Kelvins), and B = bandwidth (MHz))

And I then convert to voltage using:

V^2 / R = 4kTB (R = resistance (Ohms), of course)

I then produce a series of random values, distributed as a Gaussian around a mean given by this voltage, and an RMS given by:

sqrt(4kTBR)

Now I'd like to take into account an amplifier that is attached to the instrument. I know the amplifier provides 60dB of gain, and it has a Noise Figure of 5. How do I calculate the resulting change in power and/or voltage?

Answer 1

Noise figure is defined as: $$NF = 10log_{10}(\frac{SNR_i}{SNR_o})$$ Thus, $$5 = 10log_{10}(\frac{V_i^2N_o}{V_o^2N_i})$$ $$5 = -20log_{10}(\frac{V_o}{V_i})+10log_{10}(\frac{N_o}{N_i})$$ First term is the gain, so $$10log_{10}(\frac{N_o}{N_i}) = 5+60 = 65$$ $$10log_{10}(N_o) = 10log_{10}(N_i) + 65 dB$$ $$10log_{10}(N_o) = 10log_{10}(4kTRB) + 65 dB$$ Hopefully, the symbols are self explanatory.

Answer 2

Noise Figure of 3dB tells us the Amplifier contributes as much Noise Power as a 50 Ohm resistor, assuming you have a 50 ohm_reference circuit.

At 5dB, the noise power will be larger, as "sarthak" demonstrates.

The point of these 3 paragraph is ---- you will have more than double the power, IF THE SOURCE IMPEDANCE EQUALS THE AMPLIFIER REFERENCE IMPEDANCE

Thus RF Design has standardized on 50 ohms as the Reference Impedance.

You need to determine that (Reference Impedance) for your signal source and for your amplifier.