---
title: "Single-phase diode bridge with constant load current"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:diode-bridge-rl-load-numerical"
source: https://www.bctengineer.com/q/semester_7/power_electronics/diode-bridge-rl-load-numerical
site: BCT Engineering
---

# Single-phase diode bridge with constant load current

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

- Asked in: 2081 Bhadra, 8 marks
- Video: [Single Phase Full Wave Rectifier | Power Electronics | Lecture 37](https://www.youtube.com/watch?v=RFwc2EOZ5CI) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/diode-bridge-rl-load-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz source. The load consists of R = 10 Ω and a large inductance so as to render the load current constant. Determine i) Average values of output voltage and output current, ii) Average and rms values of diode currents, iii) Rms values of output and input currents, and iv) Supply pf

## Short answer

Average output voltage is 207.07 V and current is 20.71 A. Each diode carries average 10.35 A and rms 14.64 A; rms output and input currents are 20.71 A. Supply power factor is 0.9003 lagging.

## Answer

**Given:** single-phase full-bridge diode rectifier, $V_s = 230$ V, 50 Hz; $R = 10\ \Omega$ with a large inductance, so the load current is constant ($I_o$).

```text
              P (+)
        +-------+-------+
        |               |
       D1              D3
        |               |
 a o----+               +----o b
        |               |
       D4              D2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load R = 10 ohm + large L (Io) between P and N
```

$D_1, D_2$ conduct in the positive half cycle and $D_3, D_4$ in the negative half cycle, each for $180^\circ$. The input current is a square wave of $\pm I_o$, in phase with $v_s$.

$$
V_m = \sqrt{2} \times 230 = 325.27\ \text{V}
$$

#### (i) Average output voltage and current

$$
\begin{aligned}
V_o &= \frac{2V_m}{\pi} = \frac{2 \times 325.27}{\pi} = 207.07\ \text{V} \\
I_o &= \frac{V_o}{R} = \frac{207.07}{10} = 20.71\ \text{A}
\end{aligned}
$$

#### (ii) Diode currents

Each diode carries $I_o$ for half the period:
$$
\begin{aligned}
I_{D(avg)} &= \frac{I_o}{2} = 10.35\ \text{A} \\
I_{D(rms)} &= \frac{I_o}{\sqrt{2}} = \frac{20.71}{\sqrt{2}} = 14.64\ \text{A}
\end{aligned}
$$

#### (iii) RMS output and input currents

- The output current is constant: $I_{o(rms)} = I_o = 20.71$ A.
- The input current is a square wave of amplitude $I_o$: $I_s = I_o = 20.71$ A.

#### (iv) Supply power factor

$$
\begin{aligned}
I_{s1} &= \frac{2\sqrt{2}}{\pi}I_o = 0.9003 \times 20.71 = 18.64\ \text{A} \\
\text{DF} &= \cos 0^\circ = 1 \\
\text{PF} &= \frac{I_{s1}}{I_s}\cos\phi_1 = 0.9003 \times 1 = 0.9003
\end{aligned}
$$

Check: $P = V_o I_o = 207.07 \times 20.71 = 4288$ W and $V_s I_s \times \text{PF} = 230 \times 20.71 \times 0.9003 = 4288$ W.

**Answer:** $V_o = 207.07$ V, $I_o = 20.71$ A; diode $I_{avg} = 10.35$ A, $I_{rms} = 14.64$ A; output and input rms currents $= 20.71$ A; supply PF $= 0.90$.

## Check yourself

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

1. For 230 V rms with constant load current, what are Vo and Io (R = 10 Ω)?
   A. Vo = 230 V, Io = 23 A
   B. Vo = 207.07 V, Io = 20.71 A
   C. Vo = 103.5 V, Io = 10.35 A
   D. Vo = 325.27 V, Io = 32.5 A
   Answer: B. Vo = 2Vm/pi = 207.07 V and Io = Vo/R = 20.71 A. (Key point: Average output voltage)

2. Each diode carries Io for half the period. What are its average and rms currents?
   A. Iavg = Io, Irms = Io
   B. Iavg = Io/2 = 10.35 A, Irms = Io/sqrt(2) = 14.64 A
   C. Iavg = Io/pi, Irms = Io/2
   D. Iavg = Io/sqrt(2), Irms = Io/2
   Answer: B. Conduction for 180° at Io gives average Io/2 and rms Io/sqrt(2). (Key point: Diode currents)

3. What is the rms value of the input current is?
   A. 0.9003 Io
   B. Io/sqrt(2)
   C. Io = 20.71 A (square wave amplitude Io)
   D. 2Io
   Answer: C. The input is a square wave of +/-Io, so its rms equals Io. (Key point: Input current rms)

4. How is the supply power factor found, and what is it?
   A. PF = Is1/Is x cos(phi1) = 0.9003
   B. PF = 1 since diodes are resistive
   C. PF = 0.707
   D. PF = 0.9003 x cos 30° = 0.78
   Answer: A. Fundamental ratio Is1/Is = 2sqrt(2)/pi = 0.9003, displacement factor is 1 (no delay), so PF = 0.9003. (Key point: Supply power factor)

## Related in Power Electronics

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