3
\$\begingroup\$

There is an example in OrCad PSPICE which calculates noise figure for an RF amplifier.

I know the following formula for noise figure:

$$\text{NF}_\text{dB}=10\log\frac{P_{no}}{G_aP_{ni}}$$

But in this example the software uses another formula:

10*Log10(V(inoise)*V(inoise)/8.28e-19)

Where V(inoise) is the equivalent input voltage noise.

I don't understand why this formula should be true and where does 8.28e-19 come from? Is it a general formula that can be used in every simulation?

Here is the result enter image description here

and the circuit

enter image description here

\$\endgroup\$
4
\$\begingroup\$

Where does 8.28e-19 come from?

Thermal noise of a 50 ohm resistor in a 1 Hz bandwidth at 27 degC is 9.1e-10 volts

To convert this to an equivalent power it needs squaring and this produces a number of 8.28e-19.

Thermal noise calculator.

The formula also reduces to 20 log\$_{10}(\frac{V_{NOISE}}{9.1e-10}\$) i.e. it compares actual RMS noise against 1Hz-limited voltage noise from a 50 ohm resistor.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ And that 50 ohm resistor is R5 in your circuit, it is the output impedance of your source. So if your output impedance was 100 ohm, you would need a different value ! \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 14:06
  • \$\begingroup\$ @FakeMoustache Is the source required to be matched to use this formula? \$\endgroup\$ – SMA.D Oct 1 '15 at 17:37
  • 1
    \$\begingroup\$ No, a mismatch in source impedance will result in a different gain from input to output. So the noise figure will change but that will be because the output power will be lower (because of the mismatch). \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.