A large inductive load is connected to a single
-
phase bridge semi
-
controlled rectifier circuit with a freewheeling diode. The load resistance R
=
4
Ω
,
the power supply voltage U
₂
=
2
2
0
V
,

Question

A large inductive load is connected to a single
-
phase bridge semi
-
controlled rectifier circuit with a freewheeling diode. The load resistance R
=
4
Ω
,
 ﻿the power supply voltage U
₂
=
2
2
0
V
,
 ﻿α
=
π
/
3
.
 ﻿Find: 
(
1
)
 ﻿The average output DC voltage and the average output DC current; 
(
2
)
 ﻿The effective value of the current flowing through the thyristor 
(
rectifying diode
)
; 
(
3
)
 ﻿The effective value of the current flowing through the freewheeling diode; 
(
4
)
 ﻿Draw the waveforms of u
_
d
,
 ﻿i
_
d
,
 ﻿i
_
VT
1
,
 ﻿i
_
VD
4
,
 ﻿i
_
VD
.
 ﻿Give me solution

VIPSolutions · Accepted Answer

# Solution: Single-Phase Semi-Controlled Bridge Rectifier with Inductive Load

## Given Data

- Load resistance: $ R = 4\ \Omega $
- Supply voltage (RMS): $ U_2 = 220~\mathrm{V} $
- Firing angle: $ \alpha = \frac{\pi}{3} $
- Large inductive load (Assume continuous current)
- Freewheeling diode present

---

## 1. Average Output DC Voltage and Current

### Step 1: Find Peak Supply Voltage

$$
U_m = \sqrt{2} \times U_2 = \sqrt{2} \times 220~\mathrm{V} \approx 311~\mathrm{V}
$$

### Step 2: Average Output DC Voltage ($ V_{d,avg} $)

For a single-phase semi-controlled bridge rectifier with continuous load current:

$$
V_{d,avg} = \frac{1}{\pi} \int_\alpha^\pi U_m \sin\omega t \, d(\omega t)
$$

$$
V_{d,avg} = \frac{U_m}{\pi} (1 + \cos\alpha)
$$

Plug values:

$$
V_{d,avg} = \frac{311}{\pi} \left[1 + \cos\left(\frac{\pi}{3}\right)\right]
$$

$$
\cos\left(\frac{\pi}{3}\right) = .5
$$

$$
V_{d,avg} = \frac{311}{\pi} (1 + .5) = \frac{311}{\pi} \times 1.5
$$

$$
\frac{311}{\pi} \approx 99.04
$$

$$
V_{d,avg} \approx 99.04 \times 1.5 = 148.56~\mathrm{V}
$$

### Step 3: Average Output DC Current ($ I_{d,avg} $)

$$
I_{d,avg} = \frac{V_{d,avg}}{R} = \frac{148.56}{4} \approx 37.14~\mathrm{A}
$$

---

## 2. Effective Value of Current through the Thyristor ($ I_{T, rms} $)

For a semi-controlled bridge (2 thyristors, 2 diodes, freewheeling diode), each thyristor and diode conducts for $ \pi $ radians in each cycle.

- The load current is nearly constant ($ I_d $), due to large inductance.

The RMS value of current through one thyristor:

$$
I_{T, rms} = I_d \cdot \sqrt{\frac{\text{Conduction angle}}{2\pi}}
$$

Each thyristor conducts for $ (\pi - \alpha) $ in a half cycle (since conduction starts at $ \alpha $ and ends at $ \pi $), so:

$$
I_{T, rms} = I_d \cdot \sqrt{\frac{\pi - \alpha}{2\pi}}
$$

Plug in values:

$$
\pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}
$$

$$
I_{T, rms} = 37.14 \times \sqrt{\frac{2\pi/3}{2\pi}} = 37.14 \times \sqrt{\frac{1}{3}} \approx 37.14 \times .577 = 21.44~\mathrm{A}
$$

---

## 3. Effective Value of Current through the Freewheeling Diode ($ I_{FW, rms} $)

The freewheeling diode conducts for the rest of the period, i.e., from $ \pi $ to $ 2\pi $ and from  to $ \alpha $:

$$
\text{Conduction angle} = 2\pi - (\pi - \alpha) = \pi + \alpha
$$

But for a half-cycle, it is $ \pi + \alpha $ out of $ 2\pi $:

$$
I_{FW, rms} = I_d \cdot \sqrt{\frac{\pi + \alpha}{2\pi}}
$$

Plug in values:

$$
\pi + \alpha = \pi + \frac{\pi}{3} = \frac{4\pi}{3}
$$
$$
I_{FW, rms} = 37.14 \times \sqrt{\frac{4\pi/3}{2\pi}} = 37.14 \times \sqrt{\frac{2}{3}} \approx 37.14 \times .816 = 30.31~\mathrm{A}
$$

---

## 4. Waveforms

### Notation

- $ u_d $: Output voltage across the load
- $ i_d $: Output current (constant due to large inductance)
- $ i_{VT1} $: Current through thyristor VT1
- $ i_{VD4} $: Current through diode VD4 (rectifier leg)
- $ i_{VD} $: Current through freewheeling diode

### Waveform Descriptions

#### Output Voltage ($ u_d $)

- From $ \omega t = \alpha $ to $ \pi $: follows input sine (rectified)
- From $ \omega t = \pi $ to $ \pi + \alpha $: zero (freewheeling)
- Repeats every half-cycle

#### Output Current ($ i_d $)

- Nearly constant, equal to $ I_{d,avg} $ (due to large $ L $)

#### Current Through Thyristor ($ i_{VT1} $)

- Nonzero only during its conduction: from $ \alpha $ to $ \pi $
- Square pulse, magnitude $ I_d $, duration $ \pi - \alpha $

#### Current Through Rectifier Diode ($ i_{VD4} $)

- Same as thyristor conduction

#### Current Through Freewheeling Diode ($ i_{VD} $)

- Nonzero from $ \pi $ to $ \pi + \alpha $: square pulse, magnitude $ I_d $, duration $ \pi + \alpha $

---

### Example Waveforms (ASCII, for illustration)

```plaintext
u_d:
   ____         ____         ____ 
__/    \_______/    \_______/    \___

i_d:
-------------------------------
(constant)

i_{VT1}, i_{VD4}:
   _______         _______         _______
__/       \_______/       \_______/

i_{VD} (freewheeling):
_______         _______         _______
       \_______/       \_______/        
```

---

## Summary Table

| Quantity             | Formula                                        | Value          | Unit     |
|----------------------|------------------------------------------------|----------------|----------|
| $ V_{d,avg} $      | $ \frac{U_m}{\pi}(1+\cos\alpha) $           | 148.6          | V        |
| $ I_{d,avg} $      | $ \frac{V_{d,avg}}{R} $                     | 37.1           | A        |
| $ I_{T, rms} $     | $ I_d \sqrt{\frac{\pi-\alpha}{2\pi}} $      | 21.4           | A        |
| $ I_{FW, rms} $    | $ I_d \sqrt{\frac{\pi+\alpha}{2\pi}} $      | 30.3           | A        |

---

## Reference Diagram

![Semi-controlled bridge rectifier with freewheeling diode](https://i.imgur.com/Xm9v26H.png)
*Alt text: Circuit diagram of a single-phase semi-controlled bridge rectifier

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