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Diode dc voltage and ac amplitude

Short answer

DC bias: ID = (10–0.7)/10k = 0.93 mA, VD ≈ 0.7 V. Dynamic resistance rd = 2×25 mV/0.93 mA ≈ 53.7 Ω. AC peak amplitude across diode = 1 V × 53.7/(10000+53.7) ≈ 5.3 mV.

Consider the circuit below for which R = 10kΩ. The power supply V+ has a dc value of 10V on which is superimposed a 60 Hz sinusoid of 1-V peak amplitude. Calculate both the dc voltage of the diode and the amplitude of the sine wave signal appearing across it. Assume the diode to have a 0.7V drop at 1 mA current and n=2. [Figure: supply V+ in series with resistor R and a diode to ground]

Treat the problem in two parts: the dc analysis gives the bias current and the diode voltage; the small-signal (ac) analysis gives the signal across the diode using its dynamic resistance.

Given

V+ = 10 V dc + 1 V peak, 60 Hz; R = 10 kΩ; diode: 0.7 V at 1 mA; η = n = 2; VT = 25 mV.

Dc analysis

The diode voltage at current I: VD = 0.7 + n VT ln(I / 1 mA) = 0.7 + 0.05 ln(I/1 mA).

Assume VD = 0.7 V first:

I = (10 - 0.7)/10 kΩ = 0.93 mA

Improve: VD = 0.7 + 0.05 ln(0.93) = 0.7 - 0.0036 = 0.6964 V, giving I = (10 - 0.6964)/10 kΩ = 0.9304 mA (no further change).

DC voltage across the diode, VD ≈ 0.696 V (≈ 0.7 V).

Ac analysis

Dynamic resistance:

rd = n VT / ID = 2 × 25 mV / 0.93 mA = 53.7 Ω

Ac equivalent: the source (1 V peak) in series with R and rd. By voltage division:

vd(peak) = 1 V × rd / (R + rd)
         = 1 × 53.7 / (10000 + 53.7)
         = 5.34 mV

Answer: dc voltage across the diode ≈ 0.7 V; ac amplitude across the diode ≈ 5.3 mV peak (10.7 mV peak-to-peak).

The small-signal approximation is valid because 5.3 mV is much smaller than n VT = 50 mV.

✓ Answer checked by Claude Opus 5.5 (high effort)

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