Half-wave diode rectifier with 20 Ω diode resistance

In a half-wave diode rectifier the diode conducts only in the positive half cycle. When the diode has a forward (internal) resistance $r_f$, the current during conduction is limited by $r_f + R_L$, and part of the voltage drops across the diode.

      D (rf = 20 ohm)
 +---->|------+
 |            |  +
(~) 230 V     RL = 1000 ohm   vL
 |            |  -
 +------------+

Data: $V_s = 230$ V rms, $V_m = \sqrt2 \times 230 = 325.27$ V, $r_f = 20\ \Omega$, $R_L = 1000\ \Omega$.

Waveforms

• $0 < \omega t < \pi$: diode ON, $i = \frac{V_m\sin\omega t}{r_f + R_L}$, $v_L = iR_L$, $v_D = ir_f$ (small).
• $\pi < \omega t < 2\pi$: diode OFF, $i = 0$, $v_L = 0$, $v_D = v_s$ (full reverse voltage, peak $-325.27$ V).

i) Peak load current

$$I_m = \frac{V_m}{r_f + R_L} = \frac{325.27}{20 + 1000} = 0.3189\ \text{A} = 318.9\ \text{mA}$$

ii) DC load current

For a half-wave rectified sine:

$$I_{dc} = \frac{1}{2\pi}\int_0^{\pi}I_m\sin\omega t\,d(\omega t) = \frac{I_m}{\pi} = \frac{0.3189}{\pi} = 0.1015\ \text{A} = 101.5\ \text{mA}$$

(RMS current $I_{rms} = I_m/2 = 0.1594$ A; dc load voltage $V_{dc} = I_{dc}R_L = 101.5$ V.)

iii) DC diode voltage

The diode voltage is $v_D = v_s - v_L$. Since the average of the sinusoidal supply over a cycle is zero,

$$\begin{aligned} V_{D,dc} &= \overline{v_s} - \overline{v_L} = 0 - I_{dc}R_L \\ &= -0.1015 \times 1000 = -101.5\ \text{V} \end{aligned}$$

Check: $V_{D,dc} = \frac{1}{2\pi}\left[\int_0^{\pi}I_m r_f\sin\omega t\,d(\omega t) + \int_{\pi}^{2\pi}V_m\sin\omega t\,d(\omega t)\right] = \frac{I_m r_f - V_m}{\pi} = \frac{6.38 - 325.27}{\pi} = -101.5$ V.

Answer: $I_m = 0.319$ A, $I_{dc} = 0.1015$ A (101.5 mA), dc diode voltage $= -101.5$ V (i.e. 101.5 V, with the cathode positive on average).