# Johnson noise with amplifier circuit

I need help with this problem. Some numerical answers would also be very helpful.

• Since this is obviously homework tell us what you've tried already and where you're stuck. – John D Oct 24 '15 at 19:07
• It's only the resistor that generates noise, look up the formula for that it is something with 4KTR and the BW (bandwidth). This is where Cinp comes in, it limits the bandwith. What is the BW of R and C ? Now multiply by the gain of the amplifier. What voltage do you get ? – Bimpelrekkie Oct 24 '15 at 20:44
• Nearly perfect what @FakeMoustache says except that the noise equivalent bandwidth of a simple 1st order filter is pi/2 * 3dB frequency. See this: onmyphd.com/?p=enbw.equivalent.noise.bandwidth&ckattempt=1 – Andy aka Oct 24 '15 at 20:47
• I took an easy shortcut there ;-) If I'd really need to know I'd just use a circuit simulator :-) – Bimpelrekkie Oct 24 '15 at 20:51

I'll augment the correct answers in the comments with the following circuit model. A resistor with noise can be modeled as a noiseless resistor in series with a voltage source representing the Johnson (thermal) noise. More on that here. So we have the following model:

simulate this circuit – Schematic created using CircuitLab

Where $V_{johnson}$ is a pure white noise source with spectral density $\sqrt{4k_BTR}$.

Now to get the rms at the voltmeter, we start with the low pass formed by the resistance $R$ and the capacitance $C$. Find the equivalent noise bandwidth which is $\frac\pi{2}\frac1{2 \pi RC}$.

Take the sqare root of the bandwidth times the spectral density to get the RMS voltage at the input to the amplifier. Then multiply by $G$ to get the RMS voltage at the voltmeter

• do you mean "at the output of the amplifier" instead? This is what the question asks – Lisa Oct 25 '15 at 15:09
• square root of the entire product of spectral density and bandwidth? – Lisa Oct 25 '15 at 15:13
• @Lisa the analysis follows the chain of the circuit. I compute the rms voltage at the input first, then multiply by G. Square root of the bandwidth only, since we computed Volts per root hertz earlier, the units need to cancel to get Volts. – Houston Fortney Oct 25 '15 at 15:17