The Junction Rectifier — Question 2

Question 2

A nonlinear load line determines the operating current

A 1.00 V source drives a $100\,\Omega$ resistor and forward diode in series. Use $I_s=1.00\,\mu\mathrm A$, $V_T=.0250\,\mathrm V$, and $I=I_s(e^{V_D/V_T}-1)$.

1. Write a scalar equation for the operating current and show that it has only one nonnegative root.

2. Find I and $V_D$ numerically and verify the source-voltage balance.

Question 2 - Solution

A nonlinear load line determines the operating current

(a) Method and calculation

$$V_s=IR+V_D\Rightarrow\boxed{f(I)=100I+.0250\ln(1+I/10^{-6})-1=0}.$$$$f(0)=-1,\quad f(\infty)>0,\quad f\prime(I)=100+\frac{.0250}{I+10^{-6}}>0.$$

(b) Interpretation and further calculation

$$\boxed{I=7.76076\,\mathrm{mA},\quad V_D=.223924\,\mathrm V}.$$$$\boxed{IR+V_D=.776076+.223924=1.00000\,\mathrm V}.$$

Physical check

The resistor limits current and makes the operating point unique for this monotone diode model. A fixed threshold voltage would discard the logarithmic current dependence.