Ripple factor:
The amount of AC present in the output of the signal is called as ripple.
The ripple factor indicates the number of ripples present in the DC output.
The output of the power supply is given by
\({\rm{Ripple\;factor}} = \frac{{{I_{rms}}\;of\;AC\;component}}{{{I_{DC}}\;component}}\)
It is given as:
\(\gamma = \sqrt {{{\left( {\frac{{{V_{rms}}}}{{{V_{DC}}}}} \right)}^2}  1} \)
Thus if the ripple factor is less, the power supply has less AC components and power supply output is purer (i.e more DC without much fluctuations)
Thus ripple factor is an indication of the purity of output of the power supply.
For fullwave rectifier: γ = 0.48
For halfwave rectifier: γ = 1.21
The RMS value of a halfwave rectified current is 10 Ampere. Its value for fullwave rectification would be
Half wave rectifier:
RMS Value\(= \frac{{Max.\;value}}{2}\)
Average Value =\(\frac{{Max.\;value}}{\pi }\)
Fullwave rectifier:
\(rms\;value = \frac{{maximum\;value}}{{\sqrt 2 }}\)
\(Average\;value = 2\left( {\frac{{maximum\;value}}{\pi }} \right)\)
Calculation:
The RMS value of a halfwave rectified current is 10 Ampere
RMS Value \(= \frac{{Max.\;value}}{2}\)
10 = \(= \frac{{Max.\;value}}{2}\)
Maximum value (I_{m}) = 20 A
The RMS value for fullwave rectification
\(rms\;value = \frac{{maximum\;value}}{{\sqrt 2 }}\)
\(rms\;value = \frac{{20}}{{\sqrt 2 }}\)
RMS value = 14.14 A
Rectifier:
A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction.
CenterTap Full Wave Rectifier:
It consists of two diodes as shown:
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 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{\sqrt 2 }}\) 
\({V_m}\) 
The RMS or the effective value of a waveform f(t) 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 rms value of the waveform is given as:
\(V_p=\sqrt{2} V_{rms}\)
The relation between the peak value and the average value for a half and fullwave rectifier is:
\({V_{avg}} = \frac{{{V_m}}}{\pi }\) for halfwave rectifier
\({V_{avg}} = \frac{{2{V_m}}}{\pi }\) for fullwave rectifier
Explanation:
Important:
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 
\(\frac{{2{V_m}}}{\pi }\) 
\(\frac{{{I_m}}}{{\sqrt 2 }}\) 
\({V_m}\) 
The ripple factor indicates the number of ripples present in the DC output.
The output of the power supply is given by:
Mathematically, ripple factor (r) is defined as:
\(r= \frac{{{I_{rms}}\;of\;AC\;component}}{{{I_{DC}}\;component}}\)
Thus if the ripple factor is less, the power supply has less AC components and power supply output is more pure (i.e more DC without much fluctuations)
Thus ripple factor is an indication of the purity of output of power supply
Important Points:
For Halfwave rectifier, the ripple factor is = 1.21
For Fullwave rectifier, the ripple factor is = 0.48
For Bridgewave rectifier, the ripple factor is = 0.48
Concept:
Peak Inverse Voltage:
1. It is the max voltage that appears across the diode when it is in non conducting state or in reverse biased in a rectifier.
2. For safe operation of the diode, peak inverse voltage should be less than the breakdown voltage.
PIV = Vdiode_{max}
PIV = V_{m} in case of a halfwave rectifier.
In centertapped fullwave rectifier,
1. The secondary winding is center tapped i.e. secondary winding is divided into two halves.
2. In case of a centertapped fullwave rectifier:
PIV = 2Vm (1)
3. Greater PIV is a disadvantage because devices with greater breakdown voltage will be needed which are costlier.
Calculation:
Given:
V_{m} = 300 V
From equation (1)
PIV = 2 × 300
PIV = 600 V
Hence option (3) is the correct answer.
The centre tap fullwave singlephase rectifier circuit uses 2 diodes as shown in the given figure. The rms voltage across each diode is
Concept: In full wave Rectifier two diode are present.
V_{A} and V_{B} are equal in magnitude opposite in sign
Consider diode → D_{1}
0 < α <π
D_{1} is in
F.B. V_{diode} = 0
π < α < 2π : D_{1} is in R.B.
V_{BA} + V_{diode} = 0
V_{diode} =  V_{BA} =  (V_{B} – V_{A}) = V_{m} sin α – (V_{m} sin α)
V_{diode} = 2 V_{m}sin α
Hence V_{diode} ≃ 0
0 < α < π
2V_{m}sin α
π < α < 2π
Across D_{1}, wave form will be of HWR with peak 2V_{m}
V_{peak} = 2V_{m}
For Half wave Rectifier waveform \({V_{rms}} = \frac{{{V_m}}}{2} = \frac{{{V_{peak}}}}{2}\)
\({V_{rms}} = \frac{{2{V_m}}}{2} = {V_m}\)
Solution:
Given
\({\left. {{V_{supply}}} \right_{rms}} = 280\;V\)
V_{m} = 280 √2 V
rms voltage across diode = V_{m} = 280√2 V
= 395.97 V
Concept:
Ripple factor for full wave rectifier with capacitor filter
\(Ripple\;factor = \frac{1}{{4\sqrt 3 fC{R_L}}}\)
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) 
\({V_m}  \frac{{{I_{DC}}}}{{2{f_o}c}}\) 
\({V_m}  \frac{{{I_{DC}}}}{{4\;{f_o}c}}\) 
Explanation:
A fullwave rectifier converts both the cycles of AC input to DC output whereas a halfwave rectifier converts only the positive cycle of the AC input to DC output.
Fullwave rectifiers have several advantages over halfwave rectifiers.
These advantages are mentioned below:
1) The rectification efficiency of a fullwave rectifier is double that of the halfwave rectifier because it converts both the cycles of AC to DC.
2) Ripple factor is less in fullwave rectifier so waveform is smooth.
3) The ripple frequency is also double so they are easy to filter out.
4) The fullwave rectifier produces high DC output voltage and current so the output power is higher.
5) The fullwave rectifier has a better transformer utilization factor.
In a Full Wave Rectifier circuit, two diodes are now used, one for each half of the cycle. A multiple winding transformer is used whose secondary winding is split equally into two halves with a common centretapped connection.
Hence the centertapped rectifier is more expensive than a halfwave rectifier and tends to occupy a lot of space.
So option (4) is the correct answer.
Concept:
Ripple is the fluctuating AC component present in rectified DC output.
Ripple factor: It is the ratio of the RMS value of the ac component present in the rectified output to the average value of rectified output.
Ripple factor is given by (γ)=\(\sqrt {{{\left( {\frac{{{V_{rms}}}}{{{V_{avg}}}}} \right)}^2}  1} \)
The quality of the DC waveform can be expressed in terms of the Ripple factor (or) Form factor.
For a fullwave rectifier
Vrms =\(\;\frac{{{V_m}}}{{\sqrt 2 }}\) V
Vavg = \(\frac{{2{V_m}}}{\pi }\) V
Calculation:
γ =\(\sqrt {{{\left( {\frac{{\frac{{{V_m}}}{{\sqrt 2 }}}}{{\frac{{2{V_m}}}{\pi }}}} \right)}^2}  1} \)
=\(\sqrt {{{\left( {\frac{\pi }{{2\sqrt 2 }}} \right)}^2}  1} \)
= 0.481
Parameters 
Half wave 
Center Tap FWR 
Bridge FWR 
Type of Transformer 
Step Down 
Center tapped 
Step Down 
VDC 
\(\frac{V_m}{\pi}\) 
\(\frac{2V_m}{\pi}\) 
\(\frac{2V_m}{\pi}\) 
IDC 
\(\frac{I_m}{\pi}\) 
\(\frac{2I_m}{\pi}\) 
\(\frac{2I_m}{\pi}\) 
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 
2V 
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 
Note: Ripple factor is independent of the secondary winding voltage
Concept:
Analysis:
Given f_{0} = 50 Hz.
∴ The Ripple frequency (HWR) will be 50 Hz. (Option (1) is correct)
And the Ripple frequency of a fullwave rectifier will be 100 Hz. (Option 2 is incorrect and 3 is correct)
Inductor filter:
For fullwave rectifier using inductor filter at the output is shown below:
Ripple factor for fullwave rectifier and inductor is given as:
\(r =\frac{\sqrt{2}}{3}\frac{1}{\sqrt{1+\frac{4\omega^2L^2}{R_L^2}}}\)(1)
For maximum load current R_{L} should be 0
Put R_{L} = 0 in equation (1)
r = 0
For minimum load current i.e. high value of R_{L} i.e. R_{L} should be ∞
Put RL = ∞ in equation (1)
\(r=\frac{\sqrt{2}}{3}\)
r = 0.47
Concept:
RMS value
It is the steady equivalent value of a timevarying waveform which could also develop the same amount of heat given by the original waveform for a definite period in the circuit.
It is defined as:
\({V_{RMS}} = \sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {{\left[ {V\left( t \right)} \right]}^2}dt} \)
RMS value for a fullwave is given by
\({V_{RMS}} = \frac{{{V_m}}}{{\sqrt 2 }}\)
The average value is given by
\({V_{avg}} = \frac{{2{V_m}}}{\pi }\)
V: voltage of the waveform is considered
NOTE: If the current is taken then also the same formulae applied.
Calculation:
Given that the current supplied by the centertapped transformer is 100 mA. i.e, I_{DC} = 100 mA
R_{L }= 20 Ω
The secondary resistance of the transformer is 1 Ω
The circuit is shown as
Power calculation
Power is given by
\(P = VI = \frac{{{V^2}}}{R} = {I^2}R\)
Given load current is 100 mA
P = (100 × 10^{3})^{2} × 20
P = 10^{4} × 10^{6} × 20
P = 10^{1} × 2
P = 0.2 W
Following the option elimination method, we get the answer as option 1
Calculation of the RMS voltages across each half secondary
\({I_{DC}} = \frac{{2{I_m}}}{\pi } = 100mA\)
\({I_m} = \frac{\pi }{2} \times 100mA\)
\({V_0}_{peak} = {I_m}\left( {20} \right)\)
\(= \frac{\pi }{2} \times 100mA \times 20\)
\({V_0}_{peak} = \pi \)
Applying the KVL
 V_{m1} + (0.5I_{m}) + (0.5I_{m}) + π = 0
V_{m1} = I_{m }+ π
V_{m1} = 0.157 + 3.14
V_{m1} = 3.297 V
RMS value is
\({V_{m1\left( {RMS} \right)}} = \frac{{{V_{m1}}}}{{\sqrt 2 }}\)
\({V_{m1\left( {RMS} \right)}} = \frac{{3.297}}{{\sqrt 2 }}\)
V_{m1(RMS)} = 2.336 V
From the given options it is near to option 1.