The Junction Rectifier — Question 3

Question 3

Incremental resistance predicts a small signal around a dc bias

A diode obeys $I=I_s(e^{V_D/\eta V_T}-1)$, with ideality factor $\eta=2$, $V_T=.0250\,\mathrm V$, and dc current $I_Q=1.00\,\mathrm{mA}\gg I_s$. It is in series with $R=450\,\Omega$.

1. Derive and evaluate the diode incremental resistance $r_d=dV_D/dI$.

2. For a small 10.0 mV source variation, find the current variation and the fraction of source voltage appearing across the diode.

Question 3 - Solution

Incremental resistance predicts a small signal around a dc bias

(a) Method and calculation

$$\frac{dI}{dV_D}=\frac{I+I_s}{\eta V_T}\Rightarrow\boxed{r_d=\eta V_T/(I_Q+I_s)}.$$$$\boxed{r_d\simeq2(.0250)/.00100=50.0\,\Omega}.$$

(b) Interpretation and further calculation

$$\boxed{\delta I=\delta V_s/(R+r_d)=20.0\,\mu\mathrm A}.$$$$\boxed{\delta V_D/\delta V_s=r_d/(R+r_d)=.100,\quad\delta V_D=1.00\,\mathrm{mV}}.$$

Physical check

The diode variation is only .02 times eta VT, so local linearization is consistent. Static V/I and differential resistance are different quantities.