hi aaron,
i just had a chance to look at your equations again
aaron_do wrote on Oct 27th, 2006, 2:31am:Sorry about that...let me see. The entire circuit consists of an amplifier followed by an attenuator. The attenuator provides 50 ohm matching.
Let the entire circuit's output noise power in volts squared be Pa
Let the attenuator's output noise power in volts squared be Pb
Let the source (50 ohm) noise power at the input be Pc
For the entire circuit including the attenuator,
Noise Factor, F1 = (Pa + Pc*S21^2)/(Pc*S21^2) = (Pa/S21^2 + Pc)/Pc
This is the noise figure which i can measure. I can also measure the S21. Take the attenuator output noise Pd and refer it to the input of the circuit:
Pb/S21^2
Now we should be able to say that the NF of the amplifier minus the attenuator is
F2 = (Pa/S21^2 + Pc - Pb/S21^2)/Pc = F1 - Pb/Pc*S21^2
In other words refer all noise back to the input, and take the measured noise factor and subtract the (input refered attentuator noise)/(source noise). Is this correct?
Yes I agree... sorry for the aberation in my last post !
Noise Factor = 1+ (Circuit noise power in/ noise power in)
Circuit noise power in = Pa/S21^2 - Pb/S21^2
so yes F2=F1-Pb/(Pc*S21^2)
aaron_do wrote on Oct 27th, 2006, 2:31am:Also, the attenuator noise can be approximated as 4kT*25 while the source noise is kT*50 so the noise figure is F1-2/S21
The noise power only becomes an input noise voltage of kT*50 if the input is conjugately matched. Is your input conjugately matched ?
Any luck with working back from simulation results ?
cheers
aw