As compared to a full-wave rectifier using two diodes, the four diode bridge rectifier has the dominant advantage of :
lower peak inverse voltage requirement
Compare the two circuits on every count and only one difference is decisive.
| Centre-tapped (2 diodes) | Bridge (4 diodes) | |
|---|---|---|
| Peak inverse voltage | 2Vm | Vm |
| Ripple factor | 0.482 | 0.482 |
| Efficiency | 81.2 % | 81.2 % |
| Ripple frequency | 2f | 2f |
| Transformer | Centre tap required | None needed |
| Diodes conducting | 1 at a time | 2 in series |
Ripple factor and efficiency are identical, because both circuits produce exactly the same full-wave output waveform, so options 3 and 4 fall immediately. Current-carrying capacity is a property of the diodes chosen, not of the topology, so option 1 is no advantage either.
The PIV difference is where the bridge wins. In the centre-tapped circuit the non-conducting diode sees the voltage of both halves of the secondary — its own half plus the conducting half, which is already tied to the load:
\(PIV_{\text{centre-tap}}=2V_{m}\)
In the bridge, the two reverse-biased diodes share the reverse voltage; each blocks only
\(PIV_{\text{bridge}}=V_{m}\)
Why that matters in practice. Halving the PIV requirement halves the voltage rating — and therefore the cost — of every diode, and it is the reason bridge rectifiers dominate mains power supplies where Vm is already around 325 V for a 230 V supply. A centre-tapped design would need diodes rated beyond 650 V for the same job.
The bridge's second, related advantage is that it uses the whole secondary winding on both half-cycles, so for a given transformer it delivers roughly twice the DC output voltage of a centre-tapped circuit, and the transformer utilisation factor rises from 0.693 to 0.812. Its cost is two extra diodes and two forward drops (about 1.4 V instead of 0.7 V), which is why the centre-tapped circuit is still preferred in very-low-voltage supplies where a whole extra diode drop would matter.
Hence, the dominant advantage is the lower peak inverse voltage requirement.
Similar Questions
Consider the following rectifier circuits :
1. Half-wave rectifier without filter
2. Full-wave rectifier without filter
3. Full-wave rectifier with series inductance filter
4. Full-wave rectifier with capacitance filterThe sequence of these rectifier circuits in decreasing order of the ripple factor is
Assertion (A) : Capacitor input filter gives a high output voltage which drops as the load is increased.
Reason (R) : Capacitor input filters produce more heating in the rectifiers.
For the different values of 'L' while keeping 'RL' as constant, following graph is obtained from the given circuit.
Choose the correct ascending order from the options given below for the different values of ' L ' :
The ratio of peak to peak input ripple voltage to peak to peak ripple output voltage is known as:
Read the statements:
A. The O/P frequency of a half wave rectifier is equal to the input frequency
B. The O/P frequency of a full wave rectifier is double to the input frequency
C. The regulation of an excellent rectifier should be zero.
D. ripple factor in the full wave rectifier is 1.2.
Choose the correct statements from the options given below:
Statements regarding rectifier are :
A. RMS value of current in a half wave rectifier is \(\frac{I_m}{\sqrt{2}}\)
B. PIV of diode in a full wave bridge rectifier is Vm volts.
C. TUF and rectification efficiency for a rectifier are same parametrs.
D. Rectifier converts A.C. to rippled DC not pure D.C.
Choose the correct answer from the options given below:
In a rectifier the ripple factor can be expressed as
(a) \(\dfrac{I_{rms}}{I_{dc}}\) (b) \(\sqrt\dfrac{I_{dc}}{I_{rms}}\) (c) \(\sqrt{\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}-1}\) (d) \(\sqrt{1-\left(\dfrac{I_{rms}}{I_{dc}}\right)^{2}}\)
Which one of the above is correct ?
For a full-wave rectifier, which of the following statements are correct ?
(a) \(V_{dc}=\dfrac{2V_P}{\pi}\)
(b) \(f_{out}=2f_{in}\)
(c) \(V_{dc}=\dfrac{2\pi}{V_P}\)
(d) \(I_{dc}=\dfrac{I_m}{\sqrt{\pi}}\)
Options :
In rectifiers :
(a) The efficiency of rectification is given by \(\eta=\dfrac{P_{ac}}{P_{dc}}\)
(b) The AC component on the output side : \(V_{ac}=\sqrt{V_{rms}^{2}-V_{dc}^{2}}\)
(c) Form factor is given by : \(FF=\dfrac{V_{dc}}{V_{rms}}\)
(d) Ripple factor is given by : \(RF=\sqrt{\left(\dfrac{V_{rms}}{V_{dc}}\right)^{2}-1}\)
Of these statements :
A half wave rectifier operates from 50 Hz supply and provides a peak output voltage of 60 V across the 600 Ω resistance. The value of capacitance for the capacitor filter would limit the peak to peak ripple voltage across load is 15 V :
Important Questions from Rectifier Circuits
PIV of a non conducting diode in a bridge rectifier is
The no-load output voltage of an ideal half wave rectifier is _______.
What is the minimum PIV required for a diode used in a full-wave rectifier circuit using a center-tapped transformer with a secondary voltage of 50V-0-50 V (rms)?
One of the following is a negative resistance device
Which of the following is NOT true for a rectifier?
