The Junction Rectifier — Question 10

Question 10

A nonlinear diode detects the mean of a small ac signal

A forward-biased ideal diode has $I+I_s=I_s e^{V_D/V_T}$. Add a small sinusoidal voltage $v\cos\omega t$ to dc bias $V_Q$. Let $I_Q=I_s(e^{V_Q/V_T}-1)$ and $a=v/V_T\ll1$.

1. Expand current through second order in a and find the mean-current increase.

2. For $I_Q\gg I_s$, $V_T=.0250\,\mathrm V$, and $v=5.00\,\mathrm{mV}$, estimate the fractional dc increase and second-harmonic current amplitude relative to $I_Q$.

Question 10 - Solution

A nonlinear diode detects the mean of a small ac signal

(a) Method and calculation

$$I(t)+I_s=(I_Q+I_s)\left[1+a\cos\omega t+\tfrac12a^2\cos^2\omega t+O(a^3)\right].$$$$\boxed{\langle I\rangle-I_Q=(I_Q+I_s)a^2/4+O(a^4)}.$$

(b) Interpretation and further calculation

$$a=.200\Rightarrow\boxed{(\langle I\rangle-I_Q)/I_Q\simeq a^2/4=.0100}.$$$$\cos^2\omega t=\tfrac12(1+\cos2\omega t)\Rightarrow\boxed{I_{2\omega}/I_Q\simeq a^2/4=.0100}.$$

Physical check

Curvature of the exponential creates a positive dc shift even when the added voltage has zero mean. The result is a controlled small-amplitude approximation.