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Full Wave Rectifier is a rectifier that converts both halves of each alternating wave cycle (Alternating Current) into a pulsating DC (Direct Current) signal.
- Electric circuits that are used for the conversion of AC to DC are known as rectifiers.
- There are two types of rectifiers namely Full Wave Rectifiers and Half Wave Rectifiers.
- Full-wave rectifiers are used for a more smooth and steady supply instead of a half-wave rectifier as a significant amount of power is lost while using the half-wave rectifier.
- Full-wave rectifiers are used for the conversion of a complete cycle of AC voltage to DC voltage.
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Key Terms: Full Wave Rectifier, Full Wave Rectifier Formula, AC Voltage, Rectifiers, Centre-tapped Full Wave Rectifier, Bridge Rectifier
What is Full Wave Rectifier?
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A Full Wave Rectifier is a rectifier that converts the complete cycle of an AC signal (alternating wave) into a pulsating DC signal.
- The basic difference between a half-wave rectifier and a full-wave rectifier is that the half-wave rectifier uses only the half wave of the input AC signal whereas the full-wave rectifiers use the whole cycle of the alternating wave.
- The lower efficiency of a half-wave rectifier can be overcome by using a full-wave rectifier.
- The process through which the conversion of AC to Voltage signal takes place in a full wave rectifier is known as full wave rectification.
The video below explains this:
Full Wave Rectifier Detailed Video Explanation:
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Working Principle of Full Wave Rectifier
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The Full wave rectifier follows the given working principle–
- A very high input AC is supplied to the full wave rectifier.
- The step-down transformer in the full wave rectifier circuit converts the high-voltage AC into low-voltage.
- The anode of the center-tapped diodes is connected to the secondary winding of the transformer and to the load resistor.
- During the positive half cycle of the AC, the top half of the secondary winding or terminal 1 will become positive.
- While terminal two or the second half of the winding will become negative and the center tap will be at zero potential.
- At the time of the positive half cycle, the diode D1 is forward biased and diode D2 is reverse biased. This is because it is connected to the bottom of the secondary winding.
- Hence, D1 will let the current flow, and D2 will block the flow through it.
- In the case of a negative half cycle, the diode D1 is reverse biased and the diode D2 is forward biased.
- This is because the top half of the secondary circuit becomes negative, while the bottom half of it becomes positive.
- Therefore, in a full wave rectifier, DC voltage is obtained for both positive as well as negative half cycles.
Applications of Full Wave Rectifier
The applications of a full-wave rectifier include –
- It is used to identify the amplitude of the modulating radio signal.
- Full wave rectifiers are sued in electric welding, polarized DC voltage can be supplied through a bridge rectifier.
- The bridge rectifier circuit is used as it can convert the voltage from high AC to low DC.
- They provide the power supply to the devices that function with DC voltage such as LED and Motor.
Important Questions on Full Wave RectifiersQues. What is the main function of a full wave rectifier? (2 Marks) Ans. Full-wave rectifier fixes the negative component of the input voltage to a positive voltage. The rectifier converts it into DC (pulse current) using a diode bridge configuration. On the contrary, half-wave rectification removes only the negative voltage component using a single diode before converting to DC. Ques. What is the advantage of a full wave rectifier? (2 Marks) Ans. The key advantage of a full-wave rectifier is that it has a smaller AC ripple value for a given load and a smaller reservoir or smoothing capacitor than an equivalent half-wave rectifier. Ques. What is the difference between a half-wave and a full-wave rectifier? (2 Marks) Ans. Half wave rectifier uses only half of the AC cycle for the conversion process. When the positive half of the input is passed then the negative half is blocked and vice-versa. A full wave rectifier is an electronic circuit that converts the entire cycle of AC into Pulsating DC. |
Full Wave Rectifier Circuit
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A Full Wave Rectifier circuit can be made in two ways.
- The first way is by using two diodes and a center-tapped transformer, a center-tapped full-wave rectifier can be formed.
- For the second way, a standard transformer with four diodes is arranged as a bridge. This type is known as a bridge rectifier.
Centre-tapped Full Wave Rectifier
A center-tapped full-wave rectifier system is made up of a Centre-tapped Transformer, Two Diodes, and a Resistive Load.
- In a center-tapped transformer, an additional wire is connected to the center of the second winding for tapping. It divides the second winding into two parts.
- The voltage at this center-tapped part is zero.
- This divides the entire winding into positive (Va) and negative voltage (Vb) where the entire voltage is given by the equation, V = Va + Vb
From the above diagram, the output for both positive and negative half-cycles has been obtained.
Questions on Centre-tapped Full Wave RectifierQues. What is the difference between center-tapped and full-wave rectifiers? (2 Marks) Ans. The main difference between a bridge rectifier and a center-tapped full-wave rectifier is that: the bridge rectifier produces double the output voltage using the same secondary voltage. While a Center tapped rectifier requires a center-tapped transformer (secondary winding). Ques. What are the drawbacks of a center-tapped rectifier? (3 Marks) Ans. Disadvantages of a center-tapped full-wave rectifier:
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Full Wave Bridge Rectifier
A full wave bridge rectifier is a rectifier that uses four diodes or more than that in a bridge formation.
- A full-wave bridge rectifier system consists of Four Diodes and a Resistive Load.
- The four diodes connected in a bridge fashion are mainly done to overcome the disadvantages faced in the center-tapped full-wave rectifier.
- Here, two diodes conduct for one-half cycle and the other two diodes conduct for the remaining half cycle.
- The full-wave rectifier circuit has two diodes and a step-down transformer that are linked and centre-tapped. Across the load resistor, we get the output voltage.
- When a positive input supply is provided, the diodes D1 and D3 become forward biased, and diodes D2 and D4 become reverse biased which are in series with the load resistor.
From the above waveform diagrams, it is understood that the output across the load resistor is in the same direction for both cycles in comparison to the input where it forms a sine wave.
Full Wave Rectifier Formula
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The important wave rectifier formulas are:
Peak Inverse Voltage
Peak inverse voltage refers to the maximum voltage a diode can withstand before the breakdown in the reverse biased direction. The peak inverse voltage of a full-wave rectifier is double that of a half-wave rectifier. The PIV across D1 and D2 is:
| 2Vmax |
DC Output Voltage
The average value of the DC output voltage is represented by the following formula:
Vdc = Iac RL = \({2 \over \pi}\) Imax RL
Solved ExamplesExample 1: If the DC Output voltage is 400V with no-load attached to the power supply but decreases to 300V at full load, find the percentage voltage regulation. Solution: VNL = 400V; VFL = 300V Therefore, % Voltage Regulation = \(\frac{V_{NL}-V_{FL}}{V_{FL}} \times 100\) \(= \frac{400-300}{300} \times 100\) = 33.33% Example 2: A DC power supply is delivering 10V (normally) to a load that is altering between 0.5A and 1 A at a rate of 10 kHz. If the output impedance of the power supply is 0.01Ω at 10 kHz, find out the fluctuations in the output voltage caused by this periodic load change. Solution: For rapidly changing load levels, the normal situation in electronic systems, the output impedance of the power supply varies with the frequency of the load change. The change in frequency occurs due to the impedance of the power supply. The output impedance of the power supply is: \(Z_0 = \frac{\Delta V_L}{\Delta I_L}\) or \(\Delta V_L = Z_0 \times \Delta I_L = 0.01 \times (1-0.5)\) = 0.005V = 5 mV Thus, the output voltage will have 5mV p–p fluctuations at a rate of 10 kHz. |
RMS Value of Current
The following formula can be used for calculating the RMS value of current:
Irms = \({I_{max} \over \sqrt 2}\)
Solved ExampleCalculate the R.M.S. value of current i = 3 + 4sin(ωt + π/3). Solution: The current, i = 3 + 4sin(wt + π/3) (Given) i2 = (3 + 4sin(wt + π/3))2 i2 = 9 + 16sin2(wt + π/3) + 24sin(wt + π/3) i2 = 9 +16(1/2) + 24(0) = 17 The RMS value is 17. |
Form Factor
The form factor of the full-wave rectifier can be calculated by using the following formula:
Kf = \({RMS\ value\ of\ current \over Average\ value\ of\ Current}\)= \({I_{rms} \over I_{dc}}\)= \({I_{max}/ \sqrt 2 \over 2I_{max}\sqrt \pi}\)= \({\pi \over 2\sqrt 2} = 1.11\)
Peak Factor
The peak factor of the full wave rectifier is calculated using the formula:
Kp = \({Peak\ value\ of\ current \over RMS\ value\ of\ Current}\)= \({I_{max} \over I_{max}/ \sqrt 2} = \sqrt 2\)
Rectification Efficiency
The rectification efficiency can be obtained by the following formula:
η = \({DC\ output\ power \over AC\ Output\ Power}\)
Full-wave rectifiers have an efficiency of 81.2%.
Advantages of the Full Wave Rectifiers
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The advantages of a full wave rectifier are:
- The efficiency of the half-wave rectifier is 40.6% while for full-wave rectifiers, it is 81.2% Hence, the rectification efficiency of full-wave rectifiers is double of the half-wave rectifiers.
- The value of the ripple factor in full-wave rectifiers is 0.482 whereas in half-wave rectifiers it is 1.21. Therefore, a simple filter is required in full-wave rectifiers as the ripple factor is low.
- The output voltage, as well as the output power obtained in full-wave rectifiers, are higher as compared to that obtained by using half-wave rectifiers.
A disadvantage of the full-wave rectifier is that they need more circuit elements as compared to the half-wave rectifier which makes it a costlier choice.
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Things to Remember
- A full wave rectifier converts the complete cycle of alternating current into pulsating DC.
- The two types of full-wave rectifiers are Centre tapped and bridge rectifiers.
- The average value of the DC output voltage is represented by the following formula: Vdc = Vdc RL = 2/π Imax RL
- The following formula can be used for calculating the RMS value of current: Irms = Imax/√2
- The only disadvantage of full-wave rectifiers is that they need more circuit elements than the half-wave rectifier which makes them expensive.
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Sample Questions
Ques. What do you mean by a full wave rectifier? (2 Marks)
Ans. A full wave rectifier can be defined as a type of rectifier that converts both halves of each cycle of an alternating current (AC signal) into pulsating DC signal. In other words, full wave rectifiers are used to convert AC voltage to DC voltage and require multiple diodes to construct.
Ques. Why do we use a capacitor in a full wave rectifier circuit? (1 Mark)
Ans. In order to reduce the ripple factor, a capacitor is used in full-wave rectifiers.
Ques. What is a center-tapped full wave rectifier? (2 Marks)
Ans. A center-tapped rectifier refers to a rectifier that uses a center-tapped transformer along with two diodes to convert the complete AC signal into a DC signal.
Ques. What are the disadvantages of full wave rectifiers? (2 Marks)
Ans. When a small voltage is required to be rectified, a full wave rectifier is not applicable. As in a full wave circuit, two diodes that are connected in series, offer a double voltage drop due to internal resistances.
Ques. What are the advantages of full wave rectifiers? (3 Marks)
Ans. The advantages are:
- The rectification efficiency of the full-wave rectifiers is double as compared to the half-wave rectifiers. The rectification efficiency of the half-wave rectifiers is 40.6% while that of full-wave rectifiers is 81.2%.
- A simple filter is required in full wave rectifiers as the ripple factor is low. The value of the ripple factor in full-wave rectifiers is 0.482 whereas in half-wave rectifiers it is 1.21.
- The output voltage, as well as the output power obtained in full wave rectifiers, are higher as compared to that obtained by using half-wave rectifiers.
Ques. Where is a full wave rectifier used? (1 Mark)
Ans. The full wave rectifier is used in signal modulation and in electric welding.
Ques. State the working principle of a full wave rectifier. (3 Marks)
Ans. The full wave rectifier works on the principle where AC voltage is supplied to the input transformer. When there is a positive half cycle of the AC voltage then terminal 1 will be positive, terminal two will be negative and the center tap will be at zero potential. At the time of the positive half cycle, the diode D1 will be forward biased and diode D2 will be reverse biased. D1 will let the current flow and D2 will block the flow through it. In the case of a negative half cycle, the diode D1 is reverse biased and the diode D2 is forward biased. This is because the top half of the secondary circuit becomes negative, while the bottom half of it becomes positive. Therefore, in a full wave rectifier, DC voltage is obtained for both positive as well as negative half cycles.
Ques. In a half-wave rectifier, the circuit operates from 50 Hz main frequency. Find the fundamental frequency in the ripple. (2 Marks)
(a) 25 Hz
(b) 50 Hz
(c) 70 Hz
(d) 100 Hz
Ans. (b) 50 Hz.
Explanation: In a half-wave rectifier, the fundamental frequency in the ripple = input frequency = 50 Hz.
Ques. Draw a circuit diagram of a full wave rectifier and state how it works. (4 Marks) [All India 2017]
Ans.
Working of the full wave rectifiers: The diode Dj is forward biased during one-half cycle and current flows through the resistor, while the diode D2 is reverse biased and no current flows through it. At the time of the other half cycle, the current through the resistor flows in the same direction.
Ques. In the figure, the input waveform is converted into the output waveform by a device ‘X’. What is that device and draw a circuit diagram? 
(5 Marks) [Delhi 2017]
Ans. The device ‘X’ shown in the given figure represents the full wave rectifier.
Working of a full wave rectifier: AC voltage is supplied to the input transformer. When there is a positive half cycle of the AC voltage then terminal 1 will be positive, terminal two will be negative and the center tap will be at the zero potential.
At the time of the positive half cycle, the diode D1 will be forward biased and diode D2 will be reverse biased. D1 will let the current flow and D2 will block the flow through it.
In the case of a negative half cycle, the diode D1 is reverse biased and the diode D2 is forward biased. This is because the top half of the secondary circuit becomes negative, while the bottom half of it becomes positive. Therefore, in a full wave rectifier, DC voltage is obtained for both positive as well as negative half cycles.
Ques. Draw a circuit diagram of a full wave rectifier using a p-n junction diode. Explain its working and show the output input waveforms. (4 Marks) [Delhi 2012]
Ans. The p-n junction diode as a full wave rectifier:
A full wave rectifier comprises two diodes and a special type of transformer called a center tap transformer. The secondary of the transformer provides the desired ac voltage across A and B. During the positive half cycle of ac input, diode D1 is forward biased and conducts current, while diode D2 is reverse biased and no current flows through it. At the time of the other half cycle, the current through the resistor flows in the same direction. Thus, we get an output voltage across the load resistor RL.
Ques. Draw a neat labeled circuit diagram of a full wave rectifier using a semiconductor diode. (2 Marks)
Ans. Here is a labelled diagram of a full wave rectifier using a semiconductor diode:
Ques. A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage? (2 Marks)
Ans. The ripple voltage is (Vϒ)RMS=ϒVDC /100.0
VDC=0.636*VRMS* √2=0.636*220* √2=198V and ripple factor ϒ for full wave rectifier is 0.482.
Hence, (Vϒ)RMS=0.482*198 /100=0.954V.
Ques. A full wave rectifier uses a load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to the diode is 30V with a frequency of 50Hz. Calculate the AC power input. (2 Marks)
Ans. The AC power input PIN=IRMS2(RF+Rr).
IRMS=Im/√2=Vm/(Rf+RL)√2=30/(1500+10)*1.414=13.5mA
So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.
Ques. The transformer utilization factor of a center-tapped full wave rectifier is 0.693. Explain. (2 Marks)
Ans. The transformer utilization factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates the effectiveness of transformer usage by the rectifier. For a half-wave rectifier, it’s low and equal to 0.693.
Ques. If the peak voltage on a center-tapped full wave rectifier circuit is 5V and the diode cut-in voltage is 0.7. The peak inverse voltage on the diode is 9.3V. Explain how. (2 Marks)
Ans. PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, if the PIV rating is less than this value breakdown of the diode will occur.
For a rectifier, PIV=2Vm-Vd = 10-0.7 = 9.3V.
Ques. An ac input signal of frequency 60 Hz is rectified by a Full-wave rectifier. Write the output frequency in each case. (2 Marks)
Ans. The output frequency becomes twice the input frequency in the case of the full-wave rectifier, i.e. 120 Hz.
Ques. Draw the circuit diagram of a full-wave rectifier and explain its working. Also, give the input and output waveforms. (5 Marks) [CBSE Delhi 2019]
Ans. The circuit diagram is as below:
The two ends S1 and S2 of a center-tapped secondary of a transformer are connected to the P sections of the two diodes D1 and D2 respectively. The n-sections of the two diodes are joined together and their com¬mon junction is connected to the central tap C of the secondary winding through a load resistance RL. The input is applied across the primary and the output is obtained across the load resistance RL. The arrows show the direction of the current.
Assume that the end A of the secondary is positive during the first half cycle of the supply voltage. This makes diode D1 forward-biased and diode D2 reverse-biased. Thus diode D1 conducts and an output is obtained across the load RL. During the second half cycle of the supply voltage, the polarities of the secondary windings reverse. A becomes negative and B becomes positive with respect to the central terminal C. This makes diode D2 forward biased. Hence it conducts and an output is obtained across RL.
The input-output waveforms are:
Ques. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency? (3 Marks) [NCERT]
Ans. The output frequency of the half-wave rectifier is the same as the input frequency, while that of the full-wave rectifier is double that of the input. Therefore the frequency is 50 Hz for half-wave and 100 Hz for full-wave.
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