A silicon diode follows the equation I = Is(e^(V/VT) - 1) with η = 1, Is = 1.0 pA and VT = 25.85 mV…

Step 1. The voltage. $V = V_T \ln\!\left(\dfrac{I}{I_s} + 1\right) = 0.02585\ \text{V} \times \ln\!\left(\dfrac{5.0 \times 10^{-3}\ \text{A}}{1.0 \times 10^{-12}\ \text{A}}\right) = 0.02585\ \text{V} \times 22.33 = 0.577\ \text{V}$.

Step 2. The dynamic resistance. $r_d = \dfrac{V_T}{I} = \dfrac{0.02585\ \text{V}}{5.0 \times 10^{-3}\ \text{A}} = 5.17\ \Omega$.

Step 3. The current at $0.65$ V. $I = 1.0 \times 10^{-12}\ \text{A} \times \left(e^{0.65/0.02585} - 1\right) = 1.0 \times 10^{-12}\ \text{A} \times 8.31 \times 10^{10} = 0.0831\ \text{A} = 83.1$ mA.

Step 4. A tenfold rise. If $I$ becomes $10I$, the voltage rises by $V_T \ln 10 = 0.02585\ \text{V} \times 2.303 = 0.0595\ \text{V} = 59.5$ mV.