Iterative diode analysis

Iterative diode analysis is the standard recipe for solving the transcendental equation that the Exponential diode model produces. Because $V_{DD}=i_{D}R+v_D$ together with $i_D=I_S{e}^{v_D/V_T}$ has no closed-form solution, you converge on 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 always 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, using the 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). That 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. Note how close this is to the Constant-voltage-drop model answer (0.7 V, 4.3 mA) — the iteration mostly confirms that the CVD model was already good, which is exactly why the CVD model is the default and the exponential model is reserved for cases that genuinely need the extra accuracy.

[Background from general knowledge, not the source PDF]