1. E-mail:faridahhassan@unimap.edu.my Phone : 013-4676728
DNT 125
ANALOG ELECTRONIC
Chapter 2
Diode Applications
Pn Faridah Hassan
Phone :

2. Fig. Drawing the load line and finding the point of operation
Load Line Analysis
The load line plots all possible current (ID) conditions for all voltages applied to the diode (VD) in a given circuit. E/R is the maximum ID and E is the maximum VD.
Where the load line and the characteristic curve intersect is the Q-point, which specifies a particular ID and VD for a given circuit.
Fig. Drawing the load line and finding the point of operation

3. Load Line Analysis
The intersection of load line can be determined by applying Kirchhoff’s voltage in the clockwise direction, which results in:
Set VD = 0 then we can get ID,
Set ID = 0 then we get VD,

4. Fig. (a) Circuit; (b) characteristics.
Load Line Analysis
Example 1
For the series diode configuration, employ the diode characteristics and determine VDQ, IDQ and VR
Fig. (a) Circuit; (b) characteristics.

5. Load Line Analysis Solution
From the result, plot the straight line across ID and VD.
The Q points occurred at,
VDQ  0.78 V IDQ  18.5mA
VR=IRR =IDQR
=(18.5 mA)(0.5k)
=9.25V
Fig. The resulting load line

6. Diode as Rectifier
Rectifier: An electronic circuit that converts AC to pulsating DC.
Basic function of a DC power supply is to convert an AC voltage to a smooth DC voltage.

7. Half-Wave Rectifier Question:
The diode conducts during the positive half cycle.
It does not conduct during the negative half cycle.
Question:
What is the output if the diode is reversed?
See next slide…

8. Half-Wave Rectifier
Answer:
…the output if the diode is reversed?

9. Peak Inverse Voltage, PIV
Half-Wave Rectifier
Peak Inverse Voltage, PIV
The peak inverse voltage (PIV) is equal to the peak input voltage and is the maximum voltage across the diode when it is not conducting.
PIV at tp
-Vp(in)
Notice that the PIV can be found by applying Kirchhoff’s Voltage Law. The load voltage is 0 V, so the input voltage is across the diode at tp (peak time).

10. Average Value of Output Voltage, VAVG
Half-Wave Rectifier
Average Value of Output Voltage, VAVG
The average value of the half-wave rectified output voltage (also known as DC voltage) is
The process of removing one-half the input signal to establish a dc level is called half-wave rectification

11. Half-Wave Rectifier Example
What is the average value , VAVG , of the half-wave rectified voltage below?
VAVG = Vp/π = 15.9 V

12. Effect of Barrier Potential
Half-Wave Rectifier
Effect of Barrier Potential
For practical diode:
The barrier potential of 0.7V is considered.
During positive half-cycle, the input voltage must overcome the barrier potential before the diode becomes forward-biased.
Vp(out) = Vp(in) – 0.7V

13. Half-Wave Rectifier Example Answer:
Draw the output voltages of each rectifier diode (IN4001 & IN4003)for the indicated input voltages.
Answer:

14. Half-Wave Rectifier Transformer Coupling
Used to couple ac input voltage from source to rectifier.
2 advantages:
Allow the source voltage to be stepped down.
The ac source is electrically isolated from the rectifier; preventing a shock hazard in the secondary circuit.
Turns ratio, n = Vsec/Vpri
Vp(out)= Vp(sec) – 0.7V
PIV = Vp(sec)

15. Half-Wave Rectifier Example
Determine the peak value of the output voltage if the turns ratio is 0.5. What is the PIV across the diode?
Vp(pri) = Vp(in) = 170V
Vp(sec)= nVp(pri) = 0.5(170V) = 85V
Vp(out) = Vp(sec) – 0.7V = 85V – 0.7V = 84.3V
PIV = Vp(sec) = 85V

16. Half Wave Rectification
Summary
Fig. Half-wave rectifier.
+ve cycle
Fig. Conduction region for
(0  T/2).
-ve cycle
Fig. Nonconduction region for
(T/2  T).

17. Full-Wave Rectifiers
A full-wave rectifier allows current to flow during both the positive and negative half cycles or the full 360°.
Output frequency is twice the input frequency and pulsates every half-cycle of the input.
The average value on DC voltmeter,
VAVG = 2Vp/π

18. Full-Wave Rectification
Transformer Coupling
Turns ratio, n = Nsec/Npri
V(sec) = nV(pri) (in RMS value)
Vp(sec)= √2V(sec)

19. Full-Wave Rectification
Half Wave Rectifier
Full Wave Rectifier
The rectification process can be improved by using more diodes in a full-wave rectifier circuit.
Full-wave rectification produces a greater DC output:
Half-wave: VAVG =0.318Vp =Vp/π
Full-wave: VAVG =0.636Vp =2Vp/π

20. Full-Wave Rectifiers Example
Find average value ,Vavg, of the full-wave rectified voltage?

21. Center-Tapped Full-Wave Rectifier
Full-Wave Rectifiers
Center-Tapped Full-Wave Rectifier
A center-tapped transformer is used with two diodes that conduct on alternating half-cycles.
During the positive half-cycle, the upper diode is forward-biased and the lower diode is reverse-biased.
During the negative half-cycle, the lower diode is forward-biased and the upper diode is reverse-biased.

22. Center-Tapped Full-Wave Rectifier -PIV
Full-Wave Rectifiers
Center-Tapped Full-Wave Rectifier -PIV
The PIV can be shown by applying KVL around the green loop shown for the reverse-biased diode.
Apply KVL
Notice that one-half of the peak secondary voltage will be across the reverse-biased diode.
PIV=2Vp(out) V

23. Full-Wave Rectifiers Example
Show the voltage waveforms across each half of the secondary winding and across RL when a 100V peak sine wave is applied to the primary winding.
What minimum PIV rating must the diodes have.

24. Full-Wave Rectifiers Solution 1. The transformer turns ratio n=0.5
Vp(sec) = nVp(pri) = 0.5(100V) = 50V
Therefore 25V peak across each half of the secondary with respect to ground is applied. The output load voltage has a peak value of 25V, less 0.7V drop across the diode. Waveforms are shown below:
PIV = 2Vp(out) V = 2(24.3V) + 0.7V = 49.3 V

25. Full-Wave Rectification
Bridge Full-Wave Rectifier
The Bridge Full-Wave rectifier uses four diodes connected across the entire secondary as shown.
During positive half-cycle of the input, D1 and D2 are forward-biased and conduct current. D3 and D4 are reverse-biased.
Conduction path for the positive half-cycle.
During negative half-cycle of the input, D3 and D4 are forward-biased and conduct current. D1 and D2 are reverse-biased.
Question:
Conduction path for the negative half-cycle.
Ideally, what is the PIV equal to?
PIV = Vp(out)

26. Full-Wave Rectification
Bridge Full-Wave Rectifier
Vp(out)=Vp(sec)
(a) Ideal Diodes
Vp(out)=Vp(sec) – 1.4 V
(b) Practical Diodes

27. Bridge Full-Wave Rectifier -PIV
PIV=Vp(out)
(a) Ideal Diodes
PIV=Vp(out) + 0.7V
(b) Practical Diodes

28. Full-Wave Rectification
Example
The transformer is specified to have a 12Vrms secondary voltage (Vsec ) for the standard 120V across the primary.
Determine the peak output voltage, Vp(out), for the bridge rectifier.
Assuming the practical model, what PIV rating is required for the diodes?

29. Full-Wave Rectification
Solution
Vp(sec)= Vrms
= 1.414(12V) = 17V
Vp(out) = Vp(sec)- 1.4V
=17-1.4V =15.6 V
2. PIV rating for each diode is
PIV=Vp(out) + 0.7V
=15.6V + 0.7V =16.3V

30. Power Supply Filters and Regulators
In most power supply a 60Hz ac power line voltage must be converted to a constant dc voltage
60Hz of pulsating dc output of half-wave rectifier (or 120Hz- full-wave rectifier) must be filtered to reduce the large voltage variation
Small amount of fluctuation in the filter output voltage - ripple

31. Power Supply Filters and Regulators
Filtering is the process of smoothing the ripple from the rectifier.

32. Power Supply Filters and Regulators
Capacitor-Input Filter
The capacitor input filter is widely used. A half-wave rectifier and the capacitor-input filter are shown.
(a)
(b)
(c)
Initial charging capacitor(diode forward-biased) happens when power is turn on.
Capacitor discharge through RL after peak of positive (diode reverse-biased)
Capacitor charges back to peak of input(diode forward-biased)

33. Power Supply Filters and Regulators
Capacitor-Input Filter- Ripple Voltage
Ripple Voltage: the variation in the capacitor voltage due to charging and discharging is called ripple voltage
Ripple voltage is undesirable: thus, the smaller the ripple, the better the filtering action
The advantage of a full-wave rectifier over a half-wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter. Figure (a) and (b)
Easier to filter
-shorted time between
peaks.
-smaller ripple.

34. Power Supply Filters and Regulators
Capacitor-Input Filter- Ripple Voltage
Lower ripple factor  better filter.
Can be lowered by increasing the value of filter capacitor
or increasing the load resistance.
For the full-wave rectifier:
Ripple factor: indication of the effectiveness of the filter
Vr(pp) = peak to peak ripple voltage; VDC = VAVG = average value of filter’s output voltage

35. Power Supply Filters and Regulators
Capacitor-Input Filter- Ripple Voltage
Example
Determine the ripple factor for the filtered bridge rectifier with a load as indicated in the figure below.

36. Power Supply Filters and Regulators
Capacitor-Input Filter- Ripple Voltage
Solution:

37. Power Supply Filters and Regulators
Voltage Regulators
Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements.
A voltage regulator can furnish nearly constant output with excellent ripple rejection. 3-terminal regulators require only external capacitors to complete the regulation portion of the circuit.

38. Power Supply Filters and Regulators
Percent Regulation
Regulation is measured by the performance of a voltage regulator. There are two types of regulation -line and load.
Line regulation: how much the dc output changes for a given change in regulator’s input voltage.
Load regulation: how much change occurs in the output voltage for a given range of load current values from no load (NL) to full load (FL)
Line regulation
Load regulation

39. Power Supply Filters and Regulators
Percent Regulation
Question:
Assume the dc input to a regulator changes by 1.0 V due to a change in the ac line voltage. If the output changes by 1.5 mV due to the change, what is the line regulation?
Answer:
0.15%
Question:
Assume the dc output of a regulator changes from 5.00 V to 4.96 V when the output is varies from no load to full load. What is the load regulation?
Answer:
0.8 %

40. Zener Diode
The zener diode is designed to operate in the reverse breakdown region.
Ideally, the reverse breakdown has a constant breakdown voltage. This makes it useful as a voltage reference, which is its primary application.
Symbol
Characteristic curve

41. Zener Diode
Since the actual voltage is not ideally vertical, the change in zener current produces a small change in zener voltage
By ohm’s law:
Zener impedance (resistance), Zz

42. Zener Diode
The zener impedance, ZZ, is the ratio of a change in voltage in the breakdown region to the corresponding change in current:
Example:
Practical model
What is the zener impedance if the zener diode voltage changes from 4.79 V to 4.94 V when the current changes from 5.00 mA to 10.0 mA?
Solution:
30 W

43. Zener Diode
The temperature coefficient of a zener diode can be specified as the percent change in zener voltage for each degree Celsius change in temperature:
where TC has units of %/oC.
Alternatively, it can be specified in terms of change in voltage per degree Celsius change in temperature.
where TC has units of mV/oC.

44. Zener Diode
The temperature coefficient can be positive or negative, depending on the zener voltage. Above 5.6 V, zeners generally have a positive temperature coefficient; below about 5.6 V, they have a negative temperature coefficient.
Example:
A 1N756 is an 8.2 V zener diode (8.2 V at 25o C) with a positive temperature coefficient of 5.4 mV/oC. What is the output voltage if the temperature rises to 50o C?
Solution:
VZ = 8.2 V V =
8.389 V

45. Zener Diode Application
In low current applications, a zener diode can be used as a basic regulator.
Example:
A 1N756 (8.2 V at 25o C) is used as an 8.2 V regulator in the circuit shown. What is the smallest load resistor that can be used before losing regulation? Assume an ideal zener diode model.
Solution:
The no load zener current is
This is the maximum load current in regulation. Therefore,
837 W

46. Zener Diode Application
Zeners are used in three-terminal regulators to establish a reference voltage. These circuits are capable of much larger load currents than basic zener regulators.

47. Zener Diode Application
Zeners can also be used as limiters. The back-to-back zeners in this circuit limit the output to the breakdown voltage plus one diode drop.
Question:
What are the maximum positive and negative voltages if the zener breakdown voltage is 5.6 V?
± 6.3 V

48. Diode Limiters
A diode limiter is a circuit that limits (or clips) either the positive or negative part of the input voltage, without distorting the remaining part of the applied waveform.
Output signal is "clipped” away.

49. Diode Limiters a) Positive Limiters b) Negative Limiters
Limiting of the positive alternation. Diode is forward-biased during positive alternation and reverse-biased during negative alternation.
b) Negative Limiters
Limiting of the positive alternation. Diode is forward-biased during positive alternation and reverse-biased during negative alternation.
The output voltage, Vout is determined by voltage divider formed by R1 and RL.

50. Diode Limiters Example
What would you expect to see displayed on an oscilloscope connected across RL in the limiter shown in figure below.

51. Diode Limiters Solution
Diode is forward-biased when it is negative cycle, thus the clipped output voltage will be -0.7V.
The peak output voltage across RL will be determined by voltage divider rule:

52. Diode Limiters Biased Limiters
A biased limiter is one that has a bias voltage in series with the diode, so that a specific voltage level can be selected for limiting. RL is normally >> R1 to avoid loading effects.
The output will be clipped when the input voltage overcomes the bias voltage and the forward voltage of the diode.

53. Diode Limiters Question: a) Positive Limiters b) Negative Limiters
How is the output voltage when diode is turned around?
See next slide….

54. Diode Limiters ….When diode is turned around: output voltage
a) Positive Limiters
b) Negative Limiters

55. Diode Limiters Example Solution
Figure below shows combining a positive limiter with a negative limiter. Determine the output voltage waveform?
Solution

56. Summary Diode Limiters
Note: VD = 0.7V is used considering the potential barrier of the diode. VD = 0V
if ideal diode is used.

57. Summary Diode Limiters
Note: VD = 0.7V is used considering the potential barrier of the diode. VD = 0V
if ideal diode is used.

58. Diode Clampers
A clamper (dc restorer) is a circuit that adds a dc level to an ac signal. A capacitor is in series with the load. Example:
If diode is pointing up (away from ground), the circuit is a positive clamper.
If the diode is pointing down (toward ground), the circuit is a negative clamper

59. Diode Clampers Positive Clamper Operation:
When input voltage goes negative, the diode is forward-biased, allowing the capacitor to charge (Vp(in)-0.7V). Just after the negative peak, the diode is reverse-biased.
The capacitor will discharge on high resistance of RL. So the capacitor discharges very little from the peak of one negative half-cycle to the next.

60. Diode Clampers Negative Clamper Operation:
When the diode is turned around, a negative dc voltage is added to the input voltage to produce the output as shown. Vout = -(Vp(in) - 0.7V)

61. Diode Clampers Example Solution
What is the output voltage that you would expect to observe across RL in the clamping circuit below? Assume RL is large enough to prevent significant capacitor discharge.
Solution
A negative dc value equal to the input peak less the diode drop is inserted by the clamping circuit.
V(out)= VDC = -(Vp(in) - 0.7V)
= -(24V-0.7V) = -23.3V

62. Half-Wave Voltage Doubler
Voltage Multipliers
Voltage multipliers use clamping action to increase peak rectified voltages. Voltage multiplier circuits use a combination of diodes and capacitors to step up the output voltage of rectifier circuits.
Half-Wave Voltage Doubler
a) Positive Half-Cycle
D1 conducts
D2 is switched off
Capacitor C1 charges to Vp
b) Negative Half-Cycle
D1 is switched off
D2 conducts
Capacitor C2 charges to Vp
Vout = VC2 = 2Vp

63. Full-Wave Voltage Doubler
Voltage Multipliers
Full-Wave Voltage Doubler
a) Positif Half-Cycle
D1 forward-biased
C1 charges to Vp
D2 reverse-biased
b) Negative Half-Cycle
D1 reverse-biased
D2 forward-biased
C2 charges to Vp
Vout =2Vp (across 2 capacitors in series)

64. Voltage Tripler and Quadrupler
Voltage Multipliers
Voltage Tripler and Quadrupler

65. Voltage Multipliers Voltage Tripler
Positive half-cycle: C1 charges to Vp through D1
Negative half-cycle: C2 charges to 2Vp through D2
Positive half-cycle: C3 charges to 2Vp through D3
Vout = 3Vp (across C1 and C3)

66. Vout = 4Vp (across C2 and C4)
Voltage Multipliers
Voltage Quadrupler
Vout = 4Vp (across C2 and C4)

67. Diode Datasheet
Diode data sheets include maximum ratings for current, voltage and temperature as well as other electrical parameters. Some voltage and current specifications are abbreviated as follows:
VRRM The maximum peak reverse voltage that can be applied repetitively across the diode. This is the same as the PIV rating.
VR The maximum reverse dc voltage that can be applied across the diode.
VRSM The maximum peak value of nonrepetitive reverse voltage that can be applied across the diode.
IO The maximum value of a 60 Hz rectified current.
IFSM The maximum value of a nonrepetitive (one cycle) forward surge current.

68. The Diode Data Sheet (Maximum Rating)
Symbol
1N4001
1N4002
1N4003
UNIT
Peak repetitive reverse voltage
Working peak reverse voltage
DC blocking voltage
VRRM
VRWM VR 50
100
200 V
Nonrepetitive peak reverse voltage
VRSM 60
120
240
rms reverse voltage
VR(rms) 35 70
140
Average rectified forward current (single-phase, resistive load, 60Hz, TA = 75oC Io 1 A
Nonrepetitive peak surge current (surge applied at rated load conditions)
IFSM
30 (for 1 cycle)
Operating and storage junction temperature range
Tj, Tstg
-65 to +175 oC

69. The Diode Data Sheet (Maximum Rating)

70. Quiz
1. For the circuit shown, the PIV will occur when the input waveform is at point
a. A
b. B
c. C
d. D B C
A D

71. Quiz 2. The circuit shown is a a. half-wave rectifier
b. full-wave rectifier
c. bridge rectifier
d. none of the above

72. Quiz 3. The PIV for the circuit shown is equal to a. Vp(sec)/2
b. Vp(sec)
c. 2Vp(sec)
d. none of the above

73. Quiz
4. During the positive input cycle shown, the conduction path is through diodes
a. D1 and D2
b. D3 and D4
c. D1 and D4
d. D2 and D3

74. Quiz
5. The formula to calculate the load regulation is, a. b. c. d.

75. Quiz
6. The bias voltage is set to +4.3 V. The output of the biased limiter shown will be clipped
a. above +3.6 V
b. below +3.6 V
c. above +5.0 V
d. below +5.0 V

76. Quiz
7. The bias voltage is set to +4.3 V. The output of the biased limiter shown will be clipped
a. above +3.6 V
b. below +3.6 V
c. above +5.0 V
d. below +5.0 V

77. Quiz 8. The circuit shown is a a. negative clipping circuit
b. positive clipping circuit
c. negative clamping circuit
d. positive clamping circuit

78. Quiz 9. The circuit shown is a a. full-wave rectifier
b. full-wave voltage doubler
c. positive clamping circuit
d. negative clamping circuit

79. Quiz
10. A diode abbreviation that means the same thing as the PIV is the
a. VRRM
b. VRSM
c. IO
d. IFSM