Due at the start of lab on Mon/Tues, Feb 12/13. You may perform your calculations right here in Python, or do the calculations on paper and show them to your TA.

$1. \hspace{5mm} I = \frac{\eta q P_{opt}}{\hbar \omega} = 0.00256633630428 \hspace{2mm} \text{A}$ $2. \hspace{5mm} V_J = \sqrt{4kTR_L B} = 4.07035406813707 \hspace{2mm} nV \cdot \sqrt{B}$ $3. \hspace{5mm} S_{V_{psn}} = \frac{V_{psn}^2}{B} = \frac{I_{psn}^2R_L^2}{B} = \frac{2qI_{dc}B R_L^2}{B} = \frac{2q^2R_L^2 \eta}{\hbar \omega} P_{opt}$ $4. \hspace{5mm} S_{V_{psn}} = S_{V_J} = \frac{V_J ^2}{B} \longrightarrow \hspace{2mm} \frac{2q^2R_L^2 \eta}{\hbar \omega}P_{opt} = 4kTR_L \hspace{2mm} \longrightarrow \hspace{2mm} P_{opt} = \frac{2kT\hbar \omega}{R_Lq^2\eta} = 20.14670777012443 \hspace{2mm}\mu W$

Possible resistor values: 1M, 10k, 1k, 100

Compared noise between resistors of different sizes, noise decreases with higher resistance.

if power stays constant, larger resistance leads to smaller voltage $P = V^2 /R$

Our resistor of choice is 1.5 M. We need to calculate the expected voltage responsivity of the circuit. = 1.5 M, and from the datasheet of the FDS100 photodiode, we get a current responsivity of: 0.65 A/W at peak frequency, but at our wavelength it is approximately 0.37 A/W. Multiplying this current responsivity by our resistor value, we get a calculated voltage responsivity of 0.555 MV/W

Okay so now we measure the optical power from the photodiode and compare it to the photodiode signal to see how accurate this prediction is. We get a power reading of 5.5 $\mu$W, and a peak-to-peak signal of 5.844(32)V. This equates to a voltage responsivity of $\frac{5.844}{5.5} = 1.063 V/\mu W$

This indicates that we maybe have a more sensitive photodiode than mentioned in the lab.

At 100KHz, the level is -28.0 dBm. We found the f3db point to be at 120 KHz with a level of -31.0 dBm. By increasing the resistance $R_L$, we increase the time constant $RC$, thereby decreasing the frequency response $1/RC$. This means that the frequency at which the system begins to become unresponsive lowers, thereby decreasing the f3db point.

Increasing the bandwidth of the rf spectrum analyzer seems to increase signal-to-noise ratio.

Okay, so now we have to measure Johnson noise as a function of resistor value. We tried using the spectrum analyzer, but didn’t get very distinguishable results. It was very inconclusive. So instead we used an oscilloscope to measure the RMS voltage across the resistor. Data located below

Note: Subtract out power!

Without our circuit connected to the oscilloscope, we get an RMS value of 430 $\mu$V

Res value: Noise RMS

```
1k: 780 $\mu$V
10k: 630 $\mu$V
100k: 660 $\mu$V
1M: 750 $\mu$V
```

But this experiment is flawed.

Data from rf spectrum analyzer:

Without: 4 nV/Hz

```
1k: 9 nV/Hz
10k: 12 nV/Hz
27k: 14 nV/Hz
33k: 15 nV/Hz
100k: 29 nV/Hz
1M: 138 nV/Hz
1.1M: 143 nV/Hz
1.5M: 168 nV/Hz
```

we made sure to subtract out the power without a resistor

a log plot is used to spread out data points, as they were taken roughly every order of magnitude

Varying power supply for the photodiode. Ambient light at exactly the right frequency. Impedance matching stuff.

coefficient for shot noise term (resistance^2) given by curve fit came out to 1.97716395e-9 $A^2/\Omega$, which improved the fit a little and from which the electron charge can be calculated from the coefficient = $2qI_{dc}B$, that is, q = $10^{-13}/I_{dc}$ C

Using a 10k $\pm$ 5% resistor

Measuring at 1.0756(5) MHz, BW = 10.0 kHz, averaging over 100 samples. With the power supplied to the photodiode and the resistor attached. In the off position, we get a noise level of 4(1) nV/Hz. In the on position with the photodiode covered, we get a noise of 14(1) nV/Hz. When illuminated, we get a noise level of 325(2) nV/Hz. The current going throught the resistor was 43.9(2) $\mu A$ DC. Definitely photon shot noise because the current changes drastically with a change in power to the light source.

so we squared the value we measured on the rf spectrum analyzer to convert to Watts per Hertz, subtracted out power of floor noise and multiplied the power by 2 because of power splitting, and finally divided by the square root of the bandwidth (10kHz) to obtain $S_{V_{psn}}$ in units of W/$\sqrt{Hz}$

$S_{V_{psn}} = 2qI_{dc} R_L^2$

from this we calculated electron charge to be: 2.4015717539863325e-19 Coulombs where the actual value is 1.6022e-19 Coulombs

Ok, so next we try a different light source. current 195(4), 39(1) nV/Hz and take more data. From the curve fit shown in the plot below, we obtained an estimated charge of 7.96383509996e-20 Coulombs, which is off by almost exactly a factor of 2.