The diode used in fig. below has the threshold voltage of 0.5 V and a forward resistance of 4$\Omega$ Calculate the current flow ID through and the voltage drop VD across the diode.
ID = 4.482 mA, VD = 0.52 V
Model the diode. The question gives two parameters, so use the piecewise-linear model: a conducting diode is represented by a battery equal to the threshold (cut-in) voltage Vγ in series with the forward (bulk) resistance rf.
\(V_\gamma = 0.5\ \text{V}, \qquad r_f = 4\ \Omega\)
Write KVL round the loop. The 5 V source drives the series combination of R = 1 kΩ and the diode model:
\(V_S = I_D R + V_\gamma + I_D r_f\)
Solve for the current.
\(I_D = \dfrac{V_S - V_\gamma}{R + r_f} = \dfrac{5 - 0.5}{1000 + 4} = \dfrac{4.5}{1004}\)
\(I_D = 4.482 \times 10^{-3}\ \text{A} = 4.482\ \text{mA}\)
Now the diode voltage. The drop across the device is the threshold plus the drop across its own bulk resistance:
\(V_D = V_\gamma + I_D r_f = 0.5 + (4.482\times10^{-3})(4)\)
\(V_D = 0.5 + 0.0179 = 0.5179 \approx 0.52\ \text{V}\)
Sanity checks. VD must be slightly larger than the 0.5 V threshold, never smaller — which immediately rules out the 0.45 V and 0.48 V options. And the resistor drop, \(4.482\ \text{mA}\times 1\ \text{k}\Omega = 4.482\ \text{V}\), plus 0.518 V returns the 5 V supply exactly.
Takeaway. With \(r_f \ll R\) the forward resistance barely affects the current (ignoring it gives 4.5 mA, only 0.4 % higher), but it is what makes the diode voltage rise above the threshold as the current increases. That small slope is the reciprocal of rf on the I-V characteristic.
Hence, ID = 4.482 mA and VD = 0.52 V.
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