Half wave rectifier:
A type of circuit arrangement that only allows one halfcycle of an AC voltage waveform to pass, blocking the other halfcycle.
For square wave input, the output will be
Formula for RMS voltage:
\(I_{rms}^2 = \frac{1}{T}\mathop \smallint \limits_0^T {I^2}\left( t \right)\;dt\)
Where V_{rms }= RMS value of current
T = Time period
V(t) = Current
Calculation:
Given:
T = 2π
V(t) = 2 A
\(I_{rms}^2 = \frac{1}{{2\pi }}\mathop \smallint \limits_0^{2\pi } {I^2}\left( t \right)dt = \frac{1}{{2\pi }}\left[ {\mathop \smallint \limits_0^\pi {I^2}\left( t \right)dt + \mathop \smallint \limits_\pi ^{2\pi } 0dt} \right]\)
\(I_{rms}^2 = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi {2^2}dt\)
\(I_{rms}^2 = \frac{1}{{2\pi }}4\pi = 2\)
_{Irms} = √2 A
Peak inverse voltage of diode used in halfwave rectifier is
Peak Inverse Voltage (PIV):
The maximum voltage across a reverse bias diode is known as Peak Inverse Voltage.
PIV for different rectifiers is shown below:
Therefore, the PIV of a conducting and a nonconducting diode in a bridge rectifier is Vm i.e. Peak value of a.c. input.
CIRCUIT 
Number of Diodes 
Average DC Voltage (Vdc) 
RMS Current (Irms) 
Peak Inverse Voltage (PIV) 
HalfWave Rectifier 
1 
\(\frac{{{V_m}}}{\pi }\) 
\(\frac{{{I_{m\;}}}}{2}\) 
\({V_m}\) 
CenterTap Full Wave Rectifier 
2 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{\sqrt 2 }}\) 
\(2{V_m}\) 
BridgeType Full Wave Rectifier 
4 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{\sqrt 2 }}\) 
\({V_m}\) 
Concept:
The DC value or the average value of a waveform f(t) is given by the Area under the curve for one complete time period; i.e.,
\({f_{DC}} = \frac{1}{T}\mathop \smallint \nolimits_o^T f\left( t \right)dt\)
T = Time Period
The RMS or effective value is given by:
\({f_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {f^2}\left( t \right)dt}\)
Application:
The relation between the peak value and the average value for a halfwave rectifier is:
\({V_{avg}} = \frac{{{V_m}}}{\pi }\) for halfwave rectifier
V_{avg} = 0.318 V_{m}
= 31.8% V_{m}
For fullwave rectifier:
\(V_{avg} = \frac{V_m}{2\pi}\)
The output V_{dc} from the above circuit is:
Since Voltage supply is AC.
V_{rms} = 12 V
V_{m} = 12√2 volt
\(when\;0 < \omega t < \frac{\pi }{4}\)
V_{D} = V_{p} – V_{N } > 0 → diode in F.B. → S.C
V_{c} = V_{imax} → capacitor will charge upto max input voltage
V_{c }= 12 √2 V
When \(\omega t > \frac{\pi }{4}\)
V_{D} = V_{P} – V_{n} < 0 → diode is O.C → diode is R.B.
Since no discharge path is available for the capacitor
So, V_{c} = 12√2 volts remain constant
V_{c} = 12√2 volt = V_{DC}
Mistake Points
In the given circuit if a Resistor is present instead of a capacitor then it will work simply like a half wave rectifier (HWR)
For HWR:
V_{DC} = V_{m}/π
\( = \frac{{12\surd 2}}{\pi }\)
Input and output waveform for HWR
Identify the given waveform
Concept:
Rectifier
It is a circuit that is used to convert AC supply voltage to and Pulsating DC output voltage.
There are two types of rectifier circuits.
1) Half wave rectifier
Positive half wave rectifier
Important Points
Negative half wave rectifier
2) Fullwave rectifier
Center tape type
Bridge type
Concept:
The RMS value of a waveform x(t) is defined as:
\({x_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {x^2}\left( t \right)dt} \)
T = Time period
Application:
The given halfwave rectified square waveform with an amplitude of 2 is represented as:
The RMS value will be:
\({x_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^{T/2} {2^2}dt} \)
\({x_{rms}} = \sqrt {\frac{4}{T}\mathop \smallint \limits_0^{T/2} dt} \)
\({x_{rms}} = \sqrt {\frac{4}{T}\times\frac{T}2}\)
x_{rms} = √2 A
For sinusoidal waveforms:
CIRCUIT 
Number of Diodes 
Average DC Voltage (Vdc) 
RMS Current (Irms) 
Peak Inverse Voltage (PIV) 
HalfWave Rectifier 
1 
\(\frac{{{V_m}}}{\pi }\) 
\(\frac{{{I_{m\;}}}}{2}\) 
\({V_m}\) 
CenterTap Full Wave Rectifier 
2 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{√ 2 }}\) 
\(2{V_m}\) 
BridgeType Full Wave Rectifier 
4 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{√ 2 }}\) 
\({V_m}\) 
Concept:
For the halfwave rectifier, the output is present is only for one half of the input signal and clipped for the other half.
Positive halfwave rectifier clips the negative half of the input signal and only a positive part of the input signal is present.
Negative halfwave rectifier clips the positive half of the input signal and only negative part of the input signal is present
In a halfwave rectifier circuit, the average value of output voltage is:
\({V_{avg}} = \frac{{{V_m}}}{\pi }\)
The RMS value of the output voltage is:
\({V_{rms}} = \frac{{{V_m}}}{2}\)
Where Vm is the maximum value of supply voltage
Calculation:
Input voltage = 20 sin ωt Volt
V_{m} = 20 V
The average value of output voltage:
\({V_{avg}} = \frac{{20}}{\pi }\;V\)
The RMS value of output voltage:
\({V_{rms}} = \frac{{20}}{2} = 10\;V\)
Parameters 
Half wave 
Center Tap FWR 
Bridge FWR 
Type of Transformer 
Step Down 
Center tapped 
Step Down 
V_{DC} 
\(\frac{V_m}{\pi}\) 
\(\frac{2V_m}{\pi}\) 
\(\frac{2V_m}{\pi}\) 
I_{DC} 
\(\frac{I_m}{\pi}\) 
\(\frac{2I_m}{\pi}\) 
\(\frac{2I_m}{\pi}\) 
Vrms 
\(\frac{V_m}{2}\) 
\( \frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}\) 
\( \frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}\) 
Ripple factor 
1.21 
0.48 
0.48 
PIV 
Vm 
2Vm 
Vm 
O/P Frequency 
F 
2f 
2f 
Number of Diodes 
1 
2 
4 
Form factor 
1.57 
1.11 
1.11 
Crest factor 
2 
1.41 
1.41 
A sinusoidal current has a maximum value of 10 A. If the signal is half rectified, its rms value will be:
Half Wave Rectifier:
Consider Vm is the maximum value of Half wave rectifier,
∴ Average Value, \(I_{av}=\frac{I_m}{\pi}\)
∴ RMS Value, \(I_R=\frac{I_m}{2}\)
Conclusion:
Given Im = 10
∴ RMS Value, \(I_R=\frac{I_m}{2}=\frac{10}{2}=5 \ A\)
Given, e = E_{m} sin ωt = 400 sin 314t
Hence, E_{m} = 400 volts,
The device offers an ohmic resistance (R) of 10 Ω
Hence, the maximum value of current (I_{m}) is, \(\dfrac{E_m}{R}=\dfrac{400}{10} = 40\ A\) .... (1)
Since, the flow of current in one direction, while preventing the flow of current in the opposite direction, hence resultant waveform from actual waveform can be drawn as,
Since, the resultant waveform looks like the output waveform of the HalfWave Rectifier,
For Half Wave Rectifier,
RMS Value = \(\dfrac{A_m}{2}\) .... (2)
Average Value = \(\dfrac{A_m}{\pi}\) .... (3)
Where, A_{m} is the maximum value of an alternating quantity,
From equation (1),
A_{m} = I_{m} = 40 A
From equations (2) and, (3),
RMS Value of current = \(\dfrac{40}{2}=20\ A\)
Average Value of current = \(\dfrac{40}{\pi}=12.7\ A\)
Concept:
Ripple
‘Ripple’ is the unwanted AC component remaining when converting the AC voltage waveform into a DC waveform. Even though we try our best to remove all AC components, there is still some small amount left on the output side which pulsates the DC waveform. This undesirable AC component is called a ‘ripple’.
Ripple factor
The ripple factor is the ratio between the RMS value of the AC voltage (on the input side) and the DC voltage (on the output side) of the rectifier.
\(\Rightarrow r=\frac{\sqrt{I_{rms}^{2}I_{dc}^{2}}}{I_{dc}}\) (1)
Calculation:
Given \(I_{rms}=\frac{I_{o}}{\sqrt{2}}\) and \( I_{dc}=\frac{I_{o}}{\pi}\)
A rectifier is a device that converts an alternating current into a direct current. A pn junction can be used as a rectifier because it permits current in one direction only.
A pn junction can be used as a rectifier because it permits current in one direction only.
By equation 1,
\(\Rightarrow r=\frac{\sqrt{I_{rms}^{2}I_{dc}^{2}}}{I_{dc}}\)
\(\Rightarrow I_{rms}=\frac{\sqrt{\frac{I_{o}^{2}}{2}\frac{I_{o}^{2}}{\pi^{2}}}}{\frac{I_{o}}{\pi}}\) = 1.21
Hence, option 1 is correct.
Additional Information
CIRCUIT DIAGRAM 
Ripple factor 
HalfWave Rectifier 
1.21 
CenterTap Full Wave Rectifier 
0.48 
BridgeType Full Wave Rectifier 
0.48 
The figure shows a halfwave rectifier. The diode D is ideal. The average steadystate current (in Amperes) through the diode is approximately ____________.
This is the case of a capacitor filter.
Let,
V_{m} = peak input voltage
V_{r} = Ripple voltage
I_{DC} = D.C. average current through the load
Also, \({V_r} = \frac{{{I_{DC}}}}{{{f_0}C}}\)
Now, in the steadystate condition, the contribution of the average current by the capacitor will be zero. ∴ the average diode current will be the average load current, i.e.
\({V_{DC}} = {V_m}  \frac{{{V_r}}}{2}\)
\({V_{DC}} = {V_m}  \frac{{{I_{DC}}}}{{2{f_0}C}}\)
\({I_{DC}}{R_L} = {V_m}  \frac{{{I_{DC}}}}{{2{f_0}C}}\)
\({I_{DC}}\left[ {{R_L} + \frac{2}{{2{f_0}C}}} \right] = {V_m}\)
\({I_{DC}} = \frac{{{V_m}}}{{{R_L} + \frac{1}{{2{f_0}C}}}}\)
\({I_{DC}} = \frac{{10}}{{100 + \frac{{1000}}{{2 \times 50 \times 4}}}}\)
I_{DC} = 0.0975 A
The ripple factor for a halfwave rectifier with capacitor is given by:
Concept:
Ripple factor for Half wave rectifier with capacitor filter
\(Ripple\;factor,(r) = \frac{1}{{2\sqrt 3 fC{R_L}}}\)
Ripple factor for full wave rectifier with capacitor filter
\(Ripple\;factor,(r) = \frac{1}{{4\sqrt 3 fC{R_L}}}\)
The arrangement of a rectifier with a capacitor filter is shown below:
Values of XC and R are selected such that XC < < RL
i.e, \(\frac{1}{{{\omega _o}C}} < < {R_L}\) for Half Wave Rectifier
\(\frac{1}{{2{\omega _o}C}} < < {R_L}\) for Full Wave Rectifier
The following table shows the different Relations for FWR and HWR.

HWR 
FWR 
Ripple voltage (Vr) 
\(\frac{{{I_{DC}}}}{{2{f_o}C}}\) 
\(\frac{{{I_{DC}}}}{{{f_o}c}}\) 
Ripple factor (r) 
\(\frac{1}{{2\;\sqrt 3 {f_o}C{R_L}}}\) 
\(\frac{1}{{4\;\sqrt 3 {f_0}C{R_L}\;}}\) 
DC Output Voltage (VDC) 
\({V_m}  \frac{{{I_{DC}}}}{{2{f_o}c}}\) 
\({V_m}  \frac{{{I_{DC}}}}{{4\;{f_o}c}}\) 
Concept:
The Peak to peak ripple voltage of Half wave rectifies is given by:
\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}}\)
Where, I_{dc} = Average current
f_{0} = mains freqeucny
C = capacitor
Calculation:
Given f_{0} = 50 Hz, C = 680 μf, V_{dc} = 30 V, and R_{L} = 220 Ω, we get:
\({I_{dc}} = \frac{{{v_{dc}}}}{{{R_L}}} = \frac{{30}}{{220}}\)
I_{dc} = 0.136
\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}} = \frac{{.136}}{{50 \times 680 \times {{10}^{  6}}}}\)
\({V_r} = 4.01\;V\)
Select the option that is true regarding the following two statements labelled Assertion (A) and Reason (R).
Assertion (A): Although halfwave rectifiers without filters are theoretically possible, but inreality we use half wave rectifiers with a filter.
Reason (R): As DC equipment requires a constant waveform, we need to ‘smooth out’ this pulsating waveform for it to be of any use in the real world.
Halfwave rectifier
Important Points
A halfwave rectifier with a capacitor filter is shown
Identify the given rectifier.
Explanation:
Rectifier: A rectifier is a device that converts an alternating current into a direct current. A pn junction can be used as a rectifier because it permits current in one direction only.
There are types of rectifiers i.e. halfwave rectifier and fullwave rectifier.
Hence option (1) is the correct answer.
Calculation:
For the positive half cycle of input voltage, we get
D_{1} → off,
D_{2} → On
\({V_{01}} = \frac{{\left( {6.6\;\parallel \;2.2} \right)}}{{2.2 + \left( {6.6\;\parallel \;2.2} \right)}}{V_{imax}}\)
\(= \left( {\frac{{0.75\;}}{{1 + 0.75}}} \right) \times 10V\)
= 4.3 V
For the negative half cycle, we can see
D_{1} → On, D_{2} → Off
\({V_{02}} = \frac{{\left( {6.6\parallel 2.2} \right){V_{imax}}}}{{2.2 + \left( {6.6\parallel 2.2} \right)}} = \left( {\frac{{0.75}}{{1 + 0.75}}} \right) \times 10 = 4.3\;V\)
Concept:
Low voltage rectifiers require a stepdown transformer to reduce the strength of AC voltage.
\(\frac{V_1}{V_2}=\frac{N_1}{N_2}\)
A stepdown transformer is necessary for:
To obtain low DC voltage from the rectifier.
To protect diodes that have smaller breakdown voltage.
The RMS voltage of the halfwave rectifier is:
\(V_{rms}=\frac{V_m}{\sqrt2}\) (1)
Calculation:
Given:
V_{rms} = 240 V
\(\frac{N_1}{N_2}=\frac{8}{1}\)
V_{rms} at the output of stepdown transformer:
V_{rms} = \(\frac{240}{8}\) = 30 V
Now V_{m} can be calculated from equation (1):
V_{m} = √2× 30
V_{m} = 42.5 V
Important Points
Rectifier:
\(I_{rms}=\frac{I_m}{√2}\)
Peak inverse voltage is the maximum voltage that appears across a diode when it is under reverse bias.
Therefore, by applying KVL, the negative voltage across the diode is
V_{d} = V_{in} + V_{in} = 2V_{in}
⇒ PIV = 2V_{m}