ECT-246
T1
LINE
VOLTAGE
C1
VREG
COMMON
1. For the power supply circuit shown above, we have
40 points
Vin = 120 vac, transformer turns ration is 10:1, C1 = 100µF, Voltage
regulator is LM 7805, RL = 100 Ω
Calculate:
(a) Secondary rms voltage.
Vs(rms) = 120 /10 = 12 V
(b) Secondary peak voltage
Vs(pk) = Vs(rms) / 0.707 = 12V / 0.707 = 16.97 V
(c) Peak voltage at the output of the bridge rectifier.
VL(pk) = Vs(pk) – 1.4 = 16.97 – 1.4 = 15.57 V if voltage drop across the diodes is
considered.
(d) DC voltage across the load resistor
Idc
Vdc  Vm 
4f C
Here Vdc=?, Vm = 15.57V, Idc = Vdc/R = Vdc/100, f = 60Hz, C = 100μF
Vdc
 15.57  0.4167Vdc
Hence, Vdc  15.57 
100  4  60 100  106
15.57
 10.99 , say 11 volts.
Or, Vdc 
1.4167
(e) DC current through the load resistor.
DC current through load = Vdc/R = 11/100 = 0.11 amps.
(f) Primary peak current
Primary peak current = secondary peak current/10 = 0.607/10 = 0.0607 amps.
1
RL
(g) Secondary peak current
1
  2C 2
2
R
Im  Vm
 15.57
1
 (2 60) 2  (100 106 ) 2  0.607 amps.
2
100
(h) DC voltage at the output of bridge rectifier with no filter.
Vave = 2VL(pk) / π = 2*15.57 / π = 9.91 V
2. What is the ripple frequency for the given rectifiers:
9 points
(a) Half wave rectifier: ripple frequency = 60Hz
In the half-wave rectifier, one pulse of DC output is generated for one cycle of AC input.
The ripple frequency of a half-wave rectifier circuit powered by 60Hz AC is measured to
be 60 Hz.
(b) Full-wave rectifier: ripple frequency = 120Hz
In the full-wave rectifier, two pulses of DC output are generated for each cycle of AC
input. Since there is double the number of pulses in the full-wave rectifier's output, the
ripple frequency of a full-wave rectifier circuit, powered by the exact same 60 Hz AC
line voltage, is measured to be 120 Hz. (2*AC input frequency = 2*60Hz = 120Hz)
(c) Bridge rectifier: ripple frequency = 120Hz
The bridge rectifier also produces two pulses per cycle…..therefore, ripple frequency =
2*60Hz = 120Hz.
3. What is the effect of increasing the value for resistor and capacitor on the ripple
voltage at the output of a bridge rectifier? Express your answer in terms of increase
or decrease. Explain your answer with the help of a formula.
6 points
(a) Capacitor
Idc
Vdc
Vm
.
 Vdc  Vm 
 Vdc 
1
4f C
4 fR C
1
4 fRC
We see that 2nd term will decrease if C is increased and Vdc will increase. Vdc will
become almost equal to Vm if C becomes too large but this cannot be done as current
surge will become excessive and supply may not be capable to deliver that much current.
Vdc  Vm 
2
(b) Resistor
Idc
Vdc
Vm
.
 Vdc  Vm 
 Vdc 
1
4f C
4 fR C
1
4 fRC
Here also, 2nd term deceases with increase in R and Vdc increases.
Vdc  Vm 
4. If one diode is shorted in a bridge rectifier, what effect it would have on the output
waveform? Draw the waveform .
5 points
If one diode of the bridge rectifier is shorted, the output of the bridge rectifier is halfwave rectification and no longer full-wave rectification.
“picture is from week 1 summary in Doc sharing”
5. What is the function of a bridge rectifier?
5 points
The bridge rectifier is a full wave rectifier that alternates conduction between two diode
pairs. The purpose of a bridge rectifier is to convert AC voltage (sine wave) to (fullwave) pulsating DC voltage. Also peak inverse voltage of the diode becomes half.
6. What is the function of the filter at the output of bridge rectifier?
5 points
The function of the filter at the output of the bridge rectifier is to convert the ripple
voltage to smooth DC (it reduces ripple voltage).
7. What is the function of voltage regulator at the output of bridge rectifier and filter
in a power supply?
5 points
The function of the voltage regulator is to regulate the DC output voltage of the supply
against variations caused by changes in the load current, as well as the changes in the AC
supply voltage.
8. What is the effect of variation in input frequency in a power supply on: 9 points
(a) Output ripple voltage
Idc
Vdc
Vr 

. We see that output ripple voltage decreases with the increase in
2 f C 2 f RC
frequency.
(b) Ouput DC voltage
3
Vm
. We see that if frequency f increases, 2nd term in denominator will
1
1
4 fRC
decrease and Vdc will increase.
(c)Output DC current
Vdc
Vm
. Here also, Idc will increase with increase in frequency f.
Idc 

R

1 
R 1 

 4 fRC 
Vdc 
9. The peak voltage at the output of a bridge rectifier is 15 V. What would be the DC
voltage ?
5 points
Vave = 2*VL(pk) / π = 2*15V / π = 30 / π = 9.55V
10. What are the differences between Full-wave rectifier and bridge rectifier?
5 points

Full-wave rectifiers use two diodes where one conducts on the positive half cycle of the
AC (wave) and the other conducts on the negative half cycle.

Bridge rectifiers use four diodes, where two conduct on the positive half cycle, and the
other two conducts on the negative half cycle.

The bridge rectifier does not require the use of a center tapped transformer, whereas, the
full-rectifier does require a center tapped transformer, therefore it cannot be connected
directly to an AC line input.

The bridge rectifier provides twice the average output voltage and current and can be
directly connected to an AC line input.

Peak inverse voltage of the diodes required in case of center tap connection is double of
the peak inverse voltage of the diodes required in case of bridge rectifier. As a result,
bridge rectifier is commonly used.
4