Question 8.13: Estimate the output noise in the circuit seen in Fig. 8.27a ......
Estimate the output noise in the circuit seen in Fig. 8.27a if the diode’s anode and cathode are swapped. Assume that the diode’s zero bias depletion capacitance CJ0 is 25 fF, its built-in potential VJ is 1 V, and its grading coefficient m is 0.5.
The diode is now reverse-biased. The small reverse leakage current that flows in the diode will result in shot noise but it will be tiny compared to the thermal noise generated by the resistor. To calculate the shot noise PSD, we use the reverse leakage current for I_{DC} in Eq. (8.49). Because the diode is reverse-biased, it can be thought of as a capacitor. The value of the diode’s depletion capacitance is calculated as
I^2_{shot}(f)=2qI_{DC}\ \text{with units of}\ A^2/Hz (8.49)
C_{j}=\frac{C_{j0}}{\left\lgroup1+\frac{V_{d}}{VJ}\right\rgroup ^{m}}=\frac{25\ fF}{\sqrt{1+\frac{1.7}{1}}}=15.2fFWe can use the circuit seen in Fig. 8.28 for the noise analysis. After examining this figure for a moment, we see that the output RMS noise is simply, from Ex. 8.6 and Eq. (8.24), kT/C noise. For this example then, V_{onoise,RMS}=521\ \mu V (verified with SPICE). The fundamental way to reduce the noise is to decrease the bandwidth of the circuit by increasing the capacitance shunting the output. Changing the resistor size has practically no effect on the RMS output noise.
V_{o n o i s e,R M S}=\sqrt{\frac{1}{2\pi R C}\cdot\frac{\pi}{2}\cdot4k T R\,}=\,\sqrt{\frac{k T}{C}} (8.24)
