Hello,
Can some can tell me how to convert the noise figure (in dB) to : W/√Hz or V/ √Hz .
For example an amp with 6dB noise figure.
Thanks,
Giora
Hi Glora,
Here is the expression for Noise Figure from OA-11 http://www.ti.com/litv/pdf/snoa390b:
Na: The noise power added by the amplifier (reflected to its input port) (W/Hz)
Ni: The noise power at the input due to source resistance, RS (W/Hz). Assuming 50ohm source and amplifier input terminated in 50ohm, this would be:
Ni= (0.91nV/Rthz / 2) ^2 = (0.46nV/RtHz)^2= 207e-21 (W/Hz)
G: Power Gain (square of the voltage gain)
For your example (NF= 6dB) this means that Na= 3*Ni, or SQRT(Na)= 0.79nV/RtHz
So, the amplifier will add this noise power (Na) to the inherent noise power from the source that arrives at the input (in this case assumed to be 1/2 of the noise of a 50ohm resistor). If the amplifier input is not terminated in 50ohm matching the source impedance, Ni needs to be adjusted accordingly and Na recomputed.
The output noise power (No) will then be;
No= G . (Na + Ni) (W/Hz)
Hopefully this answers your question.
Regards,
Hooman
