Worked example: forward current at three V_D values

A small-signal silicon diode has $I_S = 1\,\text{pA}$, ideality factor $n = 1.0$, at room temperature $T = 300\,\text{K}$. Compute $I_D$ at three forward voltages — $V_D = 0.5,\ 0.6,\ 0.7\,\text{V}$ — using the Shockley equation. Compare the results with the constant-drop model (which assumes 0 mA below 0.7 V and conducts above).

1. Given. $I_S = 10^{-12}\,\text{A}$, $n = 1$, $V_T = kT/q = 25.85\,\text{mV}$ at 300 K. The Shockley equation:
   $$I_D = I_S\left(e^{V_D / V_T} - 1\right)$$
2. Substitute V_D = 0.5 V.
   $$I_D = 10^{-12}\left(e^{0.5/0.02585} - 1\right) = 10^{-12} \cdot e^{19.34}$$
   $$e^{19.34} \approx 2.51 \times 10^{8}\quad\Rightarrow\quad I_D \approx 0.25\,\text{mA}$$
3. Substitute V_D = 0.6 V.
   $$I_D = 10^{-12} \cdot e^{23.21} \approx 10^{-12} \cdot 1.20 \times 10^{10} \approx 12\,\text{mA}$$
4. Substitute V_D = 0.7 V.
   $$I_D = 10^{-12} \cdot e^{27.08} \approx 10^{-12} \cdot 5.78 \times 10^{11} \approx 580\,\text{mA}$$
   In practice, series resistance and self-heating dominate well before this — the equation tells us the diode is well past full conduction.
5. Compare with the constant-drop model.
   • 0.5 V → constant-drop says 0 mA. Shockley says 0.25 mA. Difference matters for low-current circuits.
   • 0.6 V → constant-drop still says 0 mA. Shockley says 12 mA. Constant-drop is now wrong by orders of magnitude.
   • 0.7 V → both models agree the diode is conducting. Constant-drop just assumes V_F = 0.7 V flat.
6. Sanity-check. Each 60-mV increase in V_D multiplies I_D by ≈ e^(60/25.85) ≈ 10×. This is the famous "decade per ~60 mV" rule of thumb for silicon diodes — and it falls straight out of the Shockley exponential.