---
title: "Half-wave diode rectifier with 20 Ω diode resistance"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:half-wave-diode-internal-resistance"
source: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-diode-internal-resistance
site: BCT Engineering
---

# Half-wave diode rectifier with 20 Ω diode resistance

Power Electronics · Chapter 2: Single phase AC to DC conversion

- Asked in: 2071 Shrawan, 8 marks
- Video: [Half wave Rectifier Explained](https://www.youtube.com/watch?v=Ll0IOk_Ltfc) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-diode-internal-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

A diode whose internal resistance is 20Ω is to supply power to 1000 Ω load from a 230 V ac supply in case of half wave rectification. Calculate the following. i) Peak load current ii) dc load current iii) dc diode voltage

## Short answer

For Vm=325.27 V and total resistance 1020 Ω, peak load current is 0.319 A. DC load current is Im/π = 0.1015 A, and dc diode voltage is −101.5 V, cathode positive on average.

## Answer

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.

```text
      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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the peak load current found with diode resistance rf?
   A. Vm/RL
   B. Vm/(rf + RL) = 318.9 mA
   C. Vs/(rf + RL)
   D. Vm/rf
   Answer: B. The current is limited by the series sum rf + RL, and Vm = 325.27 V gives 0.3189 A. (Key point: Peak load current)

2. What is the dc load current for a half-wave rectified sine?
   A. Im/pi = 101.5 mA
   B. Im/2 = 159.4 mA
   C. 2Im/pi = 203 mA
   D. Im = 318.9 mA
   Answer: A. Averaging over the full cycle gives Im/pi; Im/2 is the rms value. (Key point: DC load current)

3. Why is the dc diode voltage equal to -Idc x RL?
   A. The diode drop is constant at 0.7 V
   B. The reverse voltage is zero
   C. The diode voltage is vs - vL and the average of vs over a cycle is zero
   D. Because rf is much larger than RL
   Answer: C. V_D,dc = average(vs) - average(vL) = 0 - 101.5 V, so the cathode is positive on average. (Key point: DC diode voltage)

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