The Junction Rectifier — Question 1

Question 1

Forward and reverse currents follow the same ideal junction law

An ideal junction obeys $I=I_s(e^{V_D/V_T}-1)$, with $I_s=2.00\,\mathrm{nA}$ and $V_T=.0250\,\mathrm V$.

1. Find the voltage required for $I=2.00\,\mathrm{mA}$.

2. Find current at $V_D=-.100\,\mathrm V$ and the percentage difference of its magnitude from $I_s$.

Question 1 - Solution

Forward and reverse currents follow the same ideal junction law

(a) Method and calculation

$$e^{V_D/V_T}=1+I/I_s\Rightarrow\boxed{V_D=V_T\ln(1+I/I_s)}.$$$$\boxed{V_D=.0250\ln(1+10^6)=.3454\,\mathrm V}.$$

(b) Interpretation and further calculation

$$\boxed{I(-.100)=2.00(e^{-4}-1)=-1.963\,\mathrm{nA}}.$$$$\boxed{(I_s-|I|)/I_s=e^{-4}=1.832\%}.$$

Physical check

The reverse-current limit applies below breakdown in the ideal model. Real leakage may include generation and surface currents absent from this law.