---
title: "Battery charger bridge 30 V: conduction period"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:battery-bridge-30v-conduction"
source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-30v-conduction
site: BCT Engineering
---

# Battery charger bridge 30 V: conduction period

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

- Asked in: 2070 Asar, 8 marks
- Video: [Single Phase Full Wave Controlled Rectifier with RLE Load | Power Electronics | Lecture 46](https://www.youtube.com/watch?v=4OqH3R5uNJ8) — 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/battery-bridge-30v-conduction
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

Fig. 2a shows a full-wave rectifier circuit used to charge a 24V battery through a 5 ohm resistor. Calculate: i) Conduction period of charging current io. ii) Average value of charging current io iii) Power supplied to the battery [Figure: single-phase diode bridge (D1, D3 top; D4, D2 bottom) fed from Vs = 30 V (rms); output Vo feeds R = 5 Ω in series with a battery Eb = 24 V; charging current io]

## Short answer

Diodes conduct only when |vs| > 24 V. Vm = 30√2 = 42.43 V, α = sin⁻¹(24/42.43) =34.45°. Conduction angle per half-cycle = 180° – 2α =111.10° (6.17 ms). Average current Io = [2Vm cosα –E(π–2α)]/(πR) = 1.492 A. Power supplied to battery =24 × 1.492 = 35.80 W.

## Answer

In a **battery-charging rectifier**, a diode conducts only when the instantaneous supply voltage exceeds the battery emf $E$. In each half cycle current flows from $\alpha$ to $\pi-\alpha$, where $V_m\sin\alpha = E$, and is limited by $R$:

$$
i_o = \frac{V_m\sin\omega t - E}{R}, \qquad \alpha < \omega t < \pi - \alpha
$$

```text
     io     R
 P o--->---/\/\/---+
                   |
                 + |
                 ===  Eb (battery)
                 - |
 N o---------------+
 (P, N = output of the diode bridge)
```

Data: $V_s = 30$ V (rms), $E_b = 24$ V, $R = 5\ \Omega$, ideal diodes. D1, D2 conduct in the positive half cycle and D3, D4 in the negative half cycle, but only while $|v_s| > 24$ V.

```text
 |vs|   ___         ___
       / | \       / | \       --- 24 V level
 -----/--+--\-----/--+--\----
 io      /\          /\
 -------/  \--------/  \-----
      a  180-a  180+a  360-a
```

#### i) Conduction period

$$
\begin{aligned}
V_m &= \sqrt2 \times 30 = 42.43\ \text{V} \\
\alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{24}{42.43} = 34.45^\circ = 0.6013\ \text{rad}
\end{aligned}
$$

$$
\begin{aligned}
\theta_c &= \pi - 2\alpha = 180^\circ - 2(34.45^\circ) = 111.10^\circ = 1.9391\ \text{rad} \\
t_c &= \frac{111.10^\circ}{360^\circ} \times 20\ \text{ms} = 6.17\ \text{ms (in each half cycle)}
\end{aligned}
$$

#### ii) Average charging current

$$
\begin{aligned}
I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\
&= \frac{1}{\pi \times 5}\left[2(42.43)(0.8246) - 24(1.9391)\right] = \frac{69.971 - 46.538}{15.7080} = 1.492\ \text{A}
\end{aligned}
$$

#### iii) Power supplied to the battery

$$
P_{batt} = E_b I_o = 24 \times 1.492 = 35.80\ \text{W}
$$

(For reference, $I_{rms} = 2.090$ A, so the loss in R is $21.84$ W and the rectifier delivers $57.65$ W in total.)

**Answer: (i) conduction 111.10° (6.17 ms) per half cycle, from 34.45° to 145.55°; (ii) $I_o = 1.492$ A; (iii) $P_{batt} = 35.80$ W.**

## Check yourself

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

1. When does a diode pair in the battery-charging bridge conduct?
   A. Throughout the whole half cycle
   B. Only when the instantaneous supply voltage |vs| exceeds the battery emf
   C. Only at the supply peak
   D. Only when |vs| is below the battery emf
   Answer: B. A diode conducts only while the supply voltage is greater than the battery voltage; current then flows from alpha to pi - alpha. (Key point: Conduction condition)

2. With Vs = 30 V rms and Eb = 24 V, what is the conduction period per half cycle?
   A. About 111.1 degrees (6.17 ms)
   B. About 34.45 degrees
   C. 180 degrees (10 ms)
   D. About 145.55 degrees
   Answer: A. alpha = asin(24/42.43) = 34.45 degrees, so conduction = 180 - 2(34.45) = 111.1 degrees = 6.17 ms. (Key point: Conduction angle)

3. What is the average charging current Io?
   A. 2.09 A
   B. 4.8 A
   C. 1.492 A
   D. 0.6 A
   Answer: C. Io = [2Vm cos(alpha) - E(pi - 2alpha)]/(pi R) = (69.97 - 46.54)/15.708 = 1.492 A. (Key point: Average charging current)

4. How is the power supplied to the battery found, and what is it?
   A. Io^2 R = 11.1 W
   B. Eb x Io = 35.8 W
   C. Vs x Io = 44.8 W
   D. Eb x Irms = 50.2 W
   Answer: B. Only the battery emf times the average current is stored in the battery: 24 x 1.492 = 35.8 W; the loss in R is separate. (Key point: Power to battery)

## Close variants set in other exams

- [Battery charger bridge 30 V: conduction, power](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-conduction-power) — Power Electronics, chapter 2, asked 2×
- [Battery charger bridge 15 V: currents, power, efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-15v-efficiency) — Power Electronics, chapter 2, asked 3×

## Related in Power Electronics

- [Battery charger bridge 15 V: limit current to 0.5 A](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-15v-extra-resistance) — Power Electronics, chapter 2
- [Battery charger bridge 50 V: power and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-50v-efficiency) — Power Electronics, chapter 2
- [Battery charging from 230 V full-wave rectifier](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-230v-200v-emf) — Power Electronics, chapter 2, asked 2×
- [Battery charger bridge: average current, efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-avg-efficiency) — Power Electronics, chapter 2
- [Half-wave diode battery charger, 10 A](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahalf-wave-battery-charger-24v) — Power Electronics, chapter 2
- [Bridge rectifier feeding R and dc source](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abridge-rectifier-dc-source-load) — Power Electronics, chapter 2

## Related in other subjects

- [Full wave bridge rectifier load power](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abridge-rectifier-power-5ohm) — Electronics Devices and Circuit, chapter 6
- [Battery charger circuit and charging methods](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Abattery-charger-two-methods) — Industrial Power Distribution and Illumination, chapter 9
- [Full wave bridge rectifier operation](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Afull-wave-bridge-rectifier) — Basic Electronics Engineering, chapter 2, asked 4×