14. Exercise
1) A certain 7805 regulator has a measured no-load
output voltage of 5.18V and a full load output of 5.15V.
What is the load regulation expressed as a percentage?
2) If the no-load output voltage of a regulator is 24.8V
and the full-load output is 23.9V, what is the load
regulation expressed as a percentage?
14
15. Review section
1) When a 60Hz sinusoidal voltage is applied to the
input of a half-wave rectifier, what is the output
frequency?
2) When a 60Hz sinusoidal voltage is applied to the
input of a full-wave rectifier, what is the output
frequency?
3) What cause the ripple voltage on the output of a
capacitor-input filter?
4) If the load resistance connected to a filtered power
supply is decreased, what happens to the ripple
voltage?
15
16. 5. A certain rectifier filter produce a dc o/p voltage of
75V with peak-to-peak ripple voltage of 0.5V.
Calculate the ripple factor?
6. A certain full-wave rectifier has a peak o/p voltage of
30V. A 50uF capacitor-input filter is connected to the
rectifier. Calculate the peak-to-peak ripple and the
dc o/p voltage developed across a 600ohm
resistance.
7. What is the percentage of ripple for the rectifier in
question 6?
16
24. Example
What should you expect to see displayed on an
oscilloscope connected across RL in the limiter shown
above? Sketch the waveform for 2 cycles.
Noted: +Vp= +10V; R1 = 100 ohm, RL=1kohm, diode model = 1N4001
27. Diode Limiting Circuits - Example
What is the output voltage of positive limiter shown? Sketch the
waveform
R1
10 V 1.0 k
RL 3.0 V
Vin 0 VBIAS = 0
100 k
2.3 V +
–
The diode is forward-biased when the output tries to go above +3.0 V.
This causes the output to be limited to voltages less than +3.0 V.
35. Example :
•Determine the Vo and sketch the o/p waveform
for the below network. Assume diode is ideal.
Vi RL
16
+ +
t Vi Vo
0 T/2 T
-16 V=4 V
-
-
36. Solution:
• + ve region Vi Diode state Vo
1 ON V
2 ON V
+ 3 ON V
+ 4 ON V
Vi Vo 5 OFF Vi
- 6 OFF Vi
V=4 V - 7 OFF Vi
16 OFF Vi
+ Vo
+ 16
Vi
Vo 4
- 0 T/2
t
V=4 V -
37. Vi Diode state Vo
-1 ON V
Solution (continued):
-2 ON V
• - ve region (always ON state) -3 ON V
-4 ON V
-5 ON V
+ + Vo
16 -6 ON V
Vo
Vi -7 ON V
- 4
t
T/2 T
V=4 V - 0
-16 ON V
Vo
16
The resulting o/p waveform
4
t
0 T/2 T
38. Example :
Repeat the previous example using a silicon diode with VD=0.7 V
Solution: RL
VD=0.7 V
+
+ Vo
Vi
- V=4 V
-
Vi VD V 0
Vi V VD
4 0 .7
3 . 3V
39. Solution (continued):
For i/p voltages greater than 3.3 V the diode open cct and Vo=Vi.
For i/p voltages less than 3.3 V the diode short cct and the
network result as;
RL
VD=0.7 V
+
+ Vo
Vo 4 0 .7
Vi
- V=4 V
- 3 . 3V
Vo
16
The resulting o/p waveform 3 .3
t
0 T /2 T
40. Diode Limiters (Clippers)
Summary of Series Clippers : Refer
Electronic Devices and Circuit
Theory: International Edition, 8th
Edition (Robert L. Boylestad) page 87.
40
46. Diode Clampers
Summary of clamping circuits : Refer
Electronic Devices and Circuit Theory:
International Edition, 8th Edition
(Robert L. Boylestad) page 91.
46
47. Review section
1) What component in a clamping
circuit effectively acts as a battery?
2) What is the difference between a
+ve limiter and a –ve limiter?
