Iterative diode analysis

The Exponential diode model gives a transcendental equation with no closed-form solution: $V_{DD}=i_{D}R+v_D$ together with $i_D=I_S{e}^{v_D/V_T}$. Iterate to the answer numerically by alternating between the KVL equation and the inverted Diode equation.

The recipe

1. Guess a value for $V_D$. Start with 0.7 V; the Constant-voltage-drop model answer is a good first guess.
2. Compute $I_D$ from KVL using the current guess: $I_D=(V_{DD}-V_D)/R$.
3. Refine $V_D$ from the diode equation, inverse form: $V_D=V_T\ln (I_D/I_S)$.
4. Repeat steps 2–3 with the refined $V_D$ until the value stops changing to the precision you need. Two or three passes are normally enough.

It converges fast because the logarithm in step 3 is extremely insensitive: even a large error in $I_D$ produces only a small correction to $V_D$ (a factor-of-10 error in $I_D$ moves $V_D$ by only $V_T\ln 10\approx 60$ mV). The same steepness that makes the equation annoying to solve directly is what makes the iteration converge in a few steps.

Guess $V_D$, get $I_D$ from KVL, refine $V_D$ from the diode equation, iterate to convergence.

Worked numeric pass

Take $V_{DD}=5$ V, $R=1$ k$\Omega$, and a diode with $I_S=10^{-14}$ A, $V_T=25$ mV.

Iteration 1. Guess $V_D=0.7$ V. Step 2: $I_D=(5-0.7)/1000=4.30$ mA. Step 3: $V_D=0.025\ln (4.30\times 10^{-3}/10^{-14})=0.025\ln (4.30\times 10^{11})=0.025\times 26.79=0.670$ V.

Iteration 2. Use $V_D=0.670$ V. Step 2: $I_D=(5-0.670)/1000=4.33$ mA. Step 3: $V_D=0.025\ln (4.33\times 10^{-3}/10^{-14})=0.025\times 26.80=0.670$ V.

The voltage has already stabilised at $V_D\approx 0.670$ V with $I_D\approx 4.33$ mA. This is close to the Constant-voltage-drop model answer (0.7 V, 4.3 mA): the iteration mostly confirms the CVD model was already good. That’s why CVD is the default and the exponential model gets saved for cases that need the extra accuracy.