DIODE APPLICATIONS .pdf

A brief and wonderful explanation of diode applications

1. DIODE APPLICATIONS
Dr. Awadh Alqubati
Sana’a University
Faculty of Engineering
Department of Biomedical Engineering
ELECTRONICS 1
CHAPTER 2

2. At the end of this Chapter, students
should be able to:-
•Understand the concept of load-line
analysis and how it is applied to diode
networks.
•Explain the process of rectification to
establish a DC level from a sinusoidal AC
input.

3. Load Line Analysis
The analysis of electronic circuits can follow one of the two
paths :
1.Actual characteristicor approximate modelof the
device.
2.Approximate model will be always used in the analysis
VD= 0.7 V

4. 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 IDand 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. 2.1 Drawing the load line and
finding the point of operation
Point of
operation of
a circuit

5. Load Line AnalysisRIVE
VVE
DD
RD
+=
=−−+ 0
The intersection of load line in
Fig. 2.2 can be determined by
applying Kirchhoff’s voltage in the
clockwise direction, which results
in:
I
Dand V
Dare the same for Eq. (2.1) and plotted load line in Fig.
2.2 (previous slide).
Set V
D= 0 then we can get I
D, where
Set I
D= 0 then we get V
D, where 0== DVD
R
E
I 0== DIDEV
Fig. 2.2 Series diode configuration
(2.1)

6. Example (1)
•For the series diode configuration of Fig. 2.3a, employing
the diode characteristics of Fig. 2.3b, determine V
DQ, I
DQ
and V
R,and from the result, plot the straight line across
I
Dand V
D.
Fig. 2.3(a) Circuit; (b) characteristics.

7. SolutionVIDEV
mA
k
V
VD
R
E
I
D
D
100
20
5.0
10
0
===
=

===
From the result, plot the straight
line across I
Dand V
D.
The resulting load line appears in
Fig. 2.4. The Q points occurred at
V
DQ0.78 V
I
DQ18.5mA
V
R=I
RR=I
DQR=(18.5 mA)(0.5k)
= 9.25 V
Fig. 2.4. load line characteristics

8. Example (2)
•For the series diode configuration of Fig. 2.5, determine V
D,
V
Rand I
D.mA
k
V
R
V
II
VVVVEV
VV
R
RD
DR
D
32.3
2.2
3.7
3.77.08
7.0
=

===
=−=−=
=
Solution:
Fig. 2.5Circuit

9. Example (3)
•Repeat example (3) fig. 2.5 with the diode reversed0
808
0
0
=
−=+−=+−=
=−+
=
d
I
RD
RD
D
VVVVEV
VVE
I
Solution:
Open Circuit

10. 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.

11. Half-Wave Rectifier
•The diode conducts
during the positive
half cycle.
•The diode does not
conducts during the
negative half cycle.

12. Sinusoidal Input:
Half Wave Rectification
Fig. 2.44Half-wave
rectifier.
Fig. 2.46
Nonconduction
region (T/2 →T).
Fig. 2.45
Conduction region
(0 →T/2).

13. Average Value of Half Wave Output Voltage
m
mdc
V
VV ==318.0
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

14. Example
What is the average value of the
half-wave rectified voltage?
Solution: Vm/π= 15.9 V

15. Effect of Barrier Potential
(Silicon diode)
•Applied signal at least 0.7 for diode to turn on (Vk= 0.7V)
•Vi≤ 0.7 V ➔diode in open circuit and V0= 0V
•When conducting, Vk=0.7V,then Vo= Vi–Vk➔this cause
reduction in Vo, thus reduce the resulting dc voltage level.
•Now Vdc 0.318 (Vm–Vk)

16. Example
Draw the output voltages of each rectifier for the
indicated input voltages.

17. Peak Inverse Voltage (PIV)
•PIV=peak inverse
voltage and is the
maximum voltage
across the diode when
it is not
conducting/reverse
bias.
•Can be found by
applying Kirchhoff’s
voltage law. The load
voltage is 0V so the
input voltage is across
the diode at t
p.

18. Peak Inverse Voltage (PIV)
•Because the diode is only forward biased for one-half of the
AC cycle, it is also reverse biased for one-half cycle.
•It is important that the reverse breakdown voltage rating of
the diode be high enough to withstand the peak, reverse-
biasing AC voltage.
•PIV=VmOR accurately
•PIV (or PRV) Vm
•PIV = Peak inverse voltage
•PRV = Peak reverse voltage
•V
m= Peak AC voltage
Diode must capable to withstand certain
amount of repetitive reverse voltage

19. Full-Wave Rectifier
•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.
•VDCor VAVG= 2Vm/π

20. Full-Wave Rectification
•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: Vdc=0.318Vm
=Vm/π
Full-wave: Vdc=0.636Vm
=2Vm/π Half Wave Rectifier
Full Wave Rectifier

21. Example
Find average value of the full-wave rectified
voltage?

22. Transformer Coupling
Turns ratio, n = Nsec/Npri
V(sec)= nV(pri)(in RMS value)
Vp(sec)=√2 x V(sec)

23. Full-Wave Rectification
Center-Tapped Transformer
Rectifier
Requires
▪Two diodes
▪Center-tapped transformer
V
DC=0.636(Vm)

24. Full-Wave Center Tapped
•Current flow
direction during
both alternations.
The peak output is
about half of the
secondary windings
total voltage.
•Each diode is
subjected to a PIV
of the full
secondary winding
output minus one
diode voltage drop
PIV=2Vm(out)+0.7V

25. PIV: Full-wave Rectifier
Center-Tapped Transformer
•PIV can be shown by applying
KVL for the reverse-biased
diode.
•PIV across D2:







−−








−=
2
7.0
2
(sec)(sec) pp V
V
V
PIV VVPIV
p 7.0
(sec)−= VVV
V
V
V
outpp
p
outp
4.12
7.0
2
)((sec)
(sec)
)(
+=
−=
1
2
3
4
Substitute 4 to 2:
PIV=2Vp(out)+ 0.7 V

26. Example
1.Show the voltage waveforms across each half of the
secondary winding and across R
Lwhen a 100V peak sine wave
is applied to the primary winding.
2.What minimum PIV rating must the diodes have.

27. Solution
1.
2. PIV = 49.3 V

28. Full-Wave Rectification
Bridge Rectifier
▪Four diodes are required
▪VDC = 0.636 Vm

29. Full-Wave Bridge Rectifier
•The full-wave bridge
rectifier takes advantage
of the full output of the
secondary winding.
•It employs 4 diodes
arranged such that current
flows in the direction
through the load during
each half of the cycle.
During positive half-cycle of the
input, D1 and D2 are forward-
biased and conduct current. D3
and D4 are reverse-biased.
During negative half-cycle of the
input, D3 and D4 are forward-biased
and conduct current. D1 and D2 are
reverse-biased.

30. PIV: Full-wave Rectifier
Bridge Transformer
•Vp(out)=Vp(sec)–1.4 V
•PIV=Vp(out)+ 0.7 V

31. Example
The transformer is specified to have a 12 Vrmssecondary voltage
for the standard 120 V across the primary.
•Determine the peak output voltage for the bridge rectifier.
•Assuming the practical model, what PIV rating is required for the
diodes?
Solution
1. Vp(out) = 15.6 V
2. PIV = 16.3 V

32. Summary of Rectifier Circuits
Rectifier Ideal VDC Practical
(approximate) VDC
PIV
Half-Wave
Rectifier
VDC= 0.318(Vm)
= Vm/π
VDC= 0.318(Vm)-0.7 PIV=Vm
Full-WaveBridge
Rectifier
VDC= 0.636(Vm)
=2 Vm/π
VDC= 0.636(Vm)-2(0.7)PIV=Vm+0.7V
Center-Tapped
Transformer
Rectifier
VDC= 0.636(Vm)
=2 Vm/π
VDC= 0.636(Vm)-(0.7)PIV=2Vm+0.7V
Vm= peak of the AC voltage = Vp
In the center tapped transformer rectifier circuit, the peak AC
voltage is the transformer secondary voltage to the tap.

33. Power Supply Filters and Regulators
•In most power supply –60 Hz ac power line voltage →constant
dc voltage
•Pulsating dc output must be filtered to reduce the large voltage
variation
•Small amount of fluctuation in the filter o/p voltage -ripple

34. Power Supply Filters
•Filtering is the process of smoothing the ripple from the rectifier.

35. 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.

36. Power Supply Filters and 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 are require only
external capacitors to complete the regulation portion of the
circuit.

37. Power Supply Regulators
•How well the regulation is performed by a regulator is
measured by it’s regulation percentage. 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)%100










=
in
out
V
V
Line regulation%100







−
=
FL
FLNL
V
VV
Load regulation

38. Power Supply Filters and Regulators –
Capacitor-Input Filter
•Surge Current in the Capacitor-Input Filter:
•Being that the capacitor appears as a short during the initial
charging, the current through the diodes can momentarily be
quite high. To reduce risk of damaging the diodes, a surge
current limiting resistor is placed in series with the filter and
load. FSM
p
surge
I
VV
R
4.1
(sec)
−
=
I
FSM= forward surge current
rating specified on diode data
sheet.
The min. surge
Resistor values:

39. 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.

40. Capacitor Input Filter –Ripple VoltageDC
ppr
V
V
r
)(
=
Ripple factor: indication of the effectiveness of the filter
V
r(pp)= peak to peak ripple voltage;
V
DC= V
AVG= average value of filter’s
output voltage
•Lower ripple factor →better filter [can be lowered by
increasing the value of filter capacitor or increasing the load
resistance]
[half-wave rectifier]
•For the full-wave
rectifier:)(
)()(
2
1
1
1
rectp
L
AVGDC
rectp
L
ppr
V
CfR
VV
V
CfR
V








−==









V
p(rect)= unfiltered
peak

41. Example
Determine the ripple factor for the filtered bridge rectifier
with a load as indicated in the figure above.

42. Diode Limiters (Clipper)
Clippers are networks that employ diodes
to “clip” away a of an input signal
without distorting the remaining part of
the applied waveform.
Clippers used to clip-off portions of signal
voltages above or below certain levels.

43. Diode Limiter/Clipper
•A diode limiter is a circuit that limits (or clips) either the
positive or negative part of the input voltage.in
L
L
out V
RR
R
V 





+
=
1

44. Example
What would you expect to see displayed on an oscilloscope
connected across R
Lin the limiter shown in above figure.

45. SolutionVV
k
k
V
RR
R
V in
L
L
out 09.910
1.1
0.1
1
=







=





+
=

46. Biased Limiters (Clippers)
•Theleveltowhichanacvoltageislimitedcanbe
adjustedbyaddingabiasvoltage,V
BIASinserieswiththe
diode
•ThevoltageatpointAmustequalV
BIAS+0.7Vbeforethe
diodebecomeforward-biasedandconduct.
•Oncethediodebeginstoconduct,thevoltageatpointAis
limitedtoV
BIAS+0.7V,sothatallinputvoltageabovethis
levelisclippedoff.
A positive limiter

47. Biased Limiters (Clippers)
•Inthiscase,thevoltageatpointAmustgobelow–V
BIAS–0.7Vto
forward-biasthediodeandinitiatelimitingactionasshowninthe
abovefigure.
A negative limiter

48. Modified Biased Limiters (Clippers)

49. Example
Figure above shows combining a positive limiter with a
negative limiter. Determine the output voltage waveform?

50. Solution

51. Summary Limiters (Clippers)
In this examples V
D= 0
In analysis, V
D= 0 or V
D= 0.7 V can be used. Both are right
assumption.

52. Summary Limiters (Clippers)

53. Diode Clampers
•A clamper is a network constructed of a diode, a
resistor, and a capacitor that shifts a waveform to
a different dc levelwithout changing the
appearance of the applied signal.
•Sometimes known as dc restorers
•Clamping networks have a capacitor connected
directly from input to outputwith a resistive
element in parallel with the output signal. The
diode is also parallel with the output signalbut
may or may not have a series dc supply as an added
elements.

54. Clamper
•A clamper (dc restorer) is a circuit that adds a dc level to an ac
signal. A capacitor is in series with the load.
Positive clamper–the capacitor
is charged to a voltage that is one
diode drop less than the peak
voltage of the signal.
Vout= Vp(in)–0.7 V
Negative clamper
Vout= -Vp(in)+ 0.7 V
Start with forward-bias!

55. Diode Clampers
Positive clamper operation. (Diode pointing up –away from ground)

56. Diode Clampers
Negative clamper operation (Diode pointing down –toward ground)

57. Diode Clamper
•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

58. Diode Clamper (Square Wave)
Diode ‘ON’ state
Diode ‘OFF’ state
Output
V –V
c= 0 ; V
c= V; V
o= 0.7 V but
ideal V
o= 0V
-V -V
c-Vo = 0; Vc= V
V
o= -2 V

59. Summary of Clamper Circuits

60. Voltage Multipliers
▪Voltage multiplier circuits use a combination of diodes
and capacitors to step up the output voltage of
rectifier circuits.
•Voltage Doubler
•Vo0ltage Tripler
•Voltage Quadrupler

61. Voltage Doubler
•This half-wave voltage doubler’s output can be calculated by:
Vout= VC2= 2Vm
where Vm = peak secondary voltage of the transformer

62. Half-Wave Voltage Doubler
•Positive Half-Cycle
–D1 conducts
–D2 is switched off
–Capacitor C1 charges to Vp
•Negative Half-Cycle
–D1 is switched off
–D2 conducts
–Capacitor C2 charges to Vp
•Vout= VC2 = 2Vp

63. Full-Wave Voltage Doubler
PositifHalf-Cycle
•D1 forward-biased →C1
charges to Vp
•D2 reverse-biased
Negative Half-Cycle
•D1 reverse-biased
•D2 forward-biased →C2
charges to Vp
Output voltage=2Vp (across 2
capacitors in series

64. Voltage Tripler and Quadrupler

65. 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
Output: 3Vp across C1 and C3

66. Voltage Quadrupler
Output: 4Vp across C2 and C4

67. The Diode Data Sheet
•The data sheet for diodes and other devices gives detailed
information about specific characteristicssuch as the various
maximum current and voltage ratings, temperature range,
and voltage versus current curves (V-I characteristic).
•It is sometimes a very valuable piece of information, even for a
technician. There are cases when you might have to select a
replacement diode when the type of diode needed may no
longer be available.
•These are the absolute max. values under which the diode can
be operated without damage to the device.

68. The Diode Data Sheet
(Maximum Rating)
Rating Symbol1N40011N4002 1N4003UNIT
Peak repetitive reverse
voltage
Working peak reverse voltage
DC blocking voltage
V
RRM
V
RWM
V
R
50 100 200 V
Nonrepetitive peak reverse
voltage
V
RSM 60 120 240 V
rms reverse voltage V
R(rms) 35 70 140 V
Average rectified forward
current (single-phase,
resistive load, 60Hz, T
A=
75
o
C
I
o
1
A
Nonrepetitive peak surge
current (surge applied at
rated load conditions)
I
FSM
30 (for
1 cycle)
A
Operating and storage
junction temperature range
T
j, T
stg -65 to
+175
o
C

69. The Diode Data Sheet
(Maximum Rating)

70. Zener Diodes
▪The zenerdiode –silicon pn-junction device-designed for
operate in the reverse-biased region
Schematic diagram shown that this
particular zener circuit will work to
maintain 10 V across the load
Zener diode symbol

71. Zener Diodes
•Breakdown voltage –set by controlling the doping level during
manufacture
•When diode reached reverse breakdown –voltage remains
constant-current change drastically
•If zenerdiode isFB–operates the same as a rectifier diode
•A zenerdiode is much like a normal diode –but if it is placed in
the circuit in reverse bias and operates in reverse breakdown.
•Note that it’s forward characteristics are just like a normal
diode.
1.8V –200V

72. ZenerDiodes
•The reverse voltage (VR) is increased –the
reverse current (IR) remains extremely small
up to the “knee”ofthe curve
•Reverse current –called the zenercurrent,
IZ
•At the bottom of the knee-the zener
breakdown voltage (VZ) remains constant
although it increase slightly as the zener
current, IZ increase.
•IZK –min. current required to maintain
voltage regulation
•IZM –max. amount of current the diode can
handle without being damage/destroyed
•IZT –the current level at which the VZ
rating of diode is measured (specified on a
data sheet)
•The zenerdiode maintains a constant
voltage for value of reverse current rating
from IZK to IZM

73. ZenerDiodes
(ZenerEquivalent Circuit)
•Since the actual voltage is not ideally vertical, the change
in zenercurrent produces a small change in zenervoltage
•By ohm’s law:
•Normaly-Zzis specified at IZTZ
Z
Z
I
V
Z


=
Zener impedance

74. Zener Diodes
(Temp Coeff& Zener Power Dissipation and Derating)
•As with most devices, zenerdiodes have given
characteristics such as temperature coefficients
and power ratings that have to be considered. The
data sheet provides this information

76. Zener Diodes Applications
•Zener diode can be used as
1.Voltage regulator for providing stable
reference voltages
2.Simple limiters or clippers

77. ZenerRegulation with Varying Input Voltage
•As i/p voltage varies (within limits) –zenerdiode maintains a
constant o/p voltage
•But as VIN changes, IZ will change, so i/p voltage variations
are set by the min. & max. current value (IZK & IZM) which
the zenercan operate
•Resistor, R–current limiting resistor

78. Zener Regulation with a Variable Load
▪The zenerdiode maintains a nearly constant voltage across
RL as long as the zenercurrent is greater than IZK and less
than IZM
•When the o/p terminal of the zenerdiode is open (RL=∞)-
load current is zero and all of the current is through the
zener
•When a load resistor (R) is connected, current flow through
zener& load RL, IL, IZ
•The zenerdiode continues to regulate the voltage until IZ
reaches its min value , IZK
•At this point, the load current is max. , the total current
through Rremains essentially constant.

79. ZenerLimiting
Zener diode also can be used in ac applications to limit voltage swings to
desired level
(a) To limit the +vepeak of a signal voltage to the selected zenervoltage
-During –vealternation, zenerarts as FBdiode & limits the –ve
voltage to -0.7V
(b) Zener diode is turn around
-The –vepeak is by zeneraction & +vevoltage is limited to +0.7V
(c) Two back-to-back zenerslimit both peaks to the zenervoltage ±7V
-During the +vealternation, D2 is functioning as the zenerlimiter –D1
is functioning as a FBdiode.
-During the –vealternation-the roles are reversed

80. The End
Any Questions ?