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Diode ac amplitude with 60 Hz superimposed

Short answer

DC bias: ID = (10–0.7)/10k = 0.93 mA. Small-signal resistance rd = 25 mV/0.93 mA ≈ 26.9 Ω. The AC voltage across the diode (peak-to-peak) = 1 V × 26.9/(10000+26.9) ≈ 2.7 mV.

In the circuit given below the DC power supply Vcc = 10 V is superimposed with 60 Hz sinusoid of 1 V peak to peak amplitude. Calculate the amplitude of the sine wave signal appearing across the diode for the case R1 = 10 KΩ. Assume the constant voltage drop of 0.7 V in the diode. [Figure: supply Vcc in series with resistor R1 and a diode]

The 60 Hz signal is small, so we find the dc bias current first and then use the diode's small-signal resistance rd.

Given

Vcc = 10 V dc; ac = 1 V peak-to-peak (0.5 V peak); R1 = 10 kΩ; VD = 0.7 V (constant drop); VT = 25 mV; η = 1 is assumed, since it is not given.

Dc bias

ID = (10 - 0.7)/10 kΩ = 0.93 mA

Dynamic resistance

rd = VT / ID = 25 mV / 0.93 mA = 26.9 Ω

Ac signal across the diode (voltage divider between R1 and rd)

vd = vs × rd / (R1 + rd)
   = 1 V(p-p) × 26.9 / (10 000 + 26.9)
   = 2.68 mV  (peak-to-peak)

Answer: the ac signal across the diode is ≈ 2.7 mV peak-to-peak (≈ 1.34 mV peak). If η = 2 is taken, rd doubles to 53.8 Ω and the signal doubles to ≈ 5.4 mV peak-to-peak. Almost none of the ripple appears across the diode because rd ≪ R1; this is why a forward-biased diode can act as a nearly constant-voltage source.

✓ Answer checked by Claude Opus 5.5 (high effort)

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