Download CHAPTER 2: Diode Applications (Aplikasi Diod)

CHAPTER 2
Diode Applications
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Objectives
 Explain and analyze the operation of both half and full wave
rectifiers
 Explain and analyze filters and regulators and their
characteristics
 Explain and analyze the operation of diode limiting and
clamping circuits
 Explain and analyze the operation of diode voltage multipliers
 Interpret and use a diode data sheet
 Troubleshoot simple diode circuits
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2-1 Half-Wave Rectifiers
Diode – ability to conduct current in one direction
and block current in other direction
 used in circuit called RECTIFIER (ac dc)
Objective:






Discuss the operation of half-wave rectifiers
Describe a basic dc power supply & half-wave rectifications
Determine the average value, VAVG of half-waves rectified voltage
Discuss the effect of VP on a half-wave rectifier output
Peak Inverse Voltage (PIV)
Describe Transformer-couple half-wave rectifier
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2-1 Half-Wave Rectifiers (cont.)
(The basic DC power supply)
The basic function of a DC power supply is to convert an AC
voltage to a smooth DC voltage (AC  DC)
Either half or full-wave
rectifier
Maintains a constant dc
voltage
Eliminates the fluctuations
- produce smooth dc voltage
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2-1 Half-Wave Rectifiers (cont.)
(The Half-Wave Rectifier)
ac source
•A half wave rectifier(ideal) allows
conduction for only 180° or half of
a complete cycle.
•During first one cycle:
load resistor
-Vin goes positive – diode FB –
conduct current
-Vin goes negetive – diode RB – no
current- 0V
•The output frequency is the same
as the input (same shape).
The average value
VDC or VAVG :
VAVG 
Vp

(2-1)
Ideal diode model
-Measure on dc voltmeter
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2-1 Half-Wave Rectifiers (cont.)
(Effect of the Barrier Potential on the Half-Wave Rectifier Output)


Practical Diode – barrier potential of 0.7V (Si) taken into
account.
During +ve half-cycle – Vin must overcome Vpotential for
forward bias.
V p ( out)  V p ( in)  0.7V

(2-2)
Example 1: Calculate the peak o/p voltage, Vp(out)?
The peak o/p voltage:
V p (out)  V p (in)  0.7V
 5V  0.7V
 4.30V
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2-1 Half-Wave Rectifiers (cont.)
(Effect of the Barrier Potential on the Half-Wave Rectifier Output)
Example 2:
Si
Sketch the output V0 and determine the output level voltage for the
network in above figure.
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2-1 Half-Wave Rectifiers (cont.)
[Peak Inverse Voltage (PIV)]
Peak inverse
voltage (PIV) is
the maximum
voltage across the
diode when it is in
reverse bias.
The diode must be
capable of
withstanding this
amount of voltage.
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PIV  V p (in)
8
(2-3)
2-1 Half-Wave Rectifiers (cont.)
(Half-Wave Rectifier with Transformer-Coupled Input Voltage)
Transformers are often used for voltage change and isolation.
The turns ratio, n of the primary to secondary determines the output versus the
input.
Vsec
The advantages of transformer coupling:
1) allows the source voltage to be stepped up or down
2) the ac source is electrically isolated from the rectifier, thus
prevents shock hazards in the secondary circuit.
 nV pri
to couple ac input
to the rectifier
n
N sec
N pri
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V p (out)  V p (sec)  0.7V
PIV  V p (sec)
(2-4)
Example 3:
Determine the peak value of output voltage as shown in below Figure.
5:1
120V
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info
Lab test - 17/08/2006 during lab – 1 hour
 Tutorial - 17/08/2006 after lab test


Test 1 -23/08/2006 Chapter 1,2,3
8.30pm – 9.30pm
DKG4 & DKG5
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2-2 Full-Wave Rectifiers
(Introduction)
Objective:
 Explain & Analyze the operation of Full-Wave Rectifier.
 Discuss how full wave rectifier differs from half-wave rectifier
 Determine the average value
 Describe the operation of center-tapped & bridge.
 Explain effects of the transformers turns ratio
 PIV
 Comparison between center-tapped & bridge.
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2-2 Full-Wave Rectifiers (cont.)
(Introduction)
A full-wave rectifier allows current to flow during both the positive and
negative half cycles or the full 360º but half-wave rectifier allows only
during one-half of the cycle.
The no. of +ve alternations is twice the half wave for the same time
interval
The output frequency is twice the input frequency.
The average value – the value measured on a dc voltmeter
VAVG 
2Vp

Twice output
(2-5)
67% of Vp
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2-2 Full-Wave Rectifiers
(i - The Center-Tapped Full-Wave Rectifier)
•This method of rectification employs two diodes connected to a
secondary center-tapped transformer.
•The i/p voltage is coupled through the transformer to the center-tapped
secondary.
•The peak output is only half of the transformer’s peak secondary
voltage.
Coupled input
voltage
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2-2 Full-Wave Rectifiers (cont.)
(i - The Center-Tapped Full-Wave Rectifier)
•+ve half-cycle input voltage (forward-bias D1 & reverse-bias D2)-the current
patch through the D1 and RL
•-ve half-cycle input voltage (reverse-bias D1 & forward-bias D2)-the current
patch through D2 and RL
•The output current on both portions of the input cycle – same direction
through the load.
•The o/p voltage across the load resistors – full-wave rectifiers
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2-2 Full-Wave Rectifiers (cont.)
(i - The Center-Tapped Full-Wave Rectifier)
-Effect of the Turns Ratio on the Output Voltage-
If n=1, Vp(out)=Vp(pri)/2-0.7V
 Vp(sec)=V(pri)
Vsec  2V pri
If n=2, Vsec  2V pri
 V p ( out )  V p ( pri)  0.7
• In any case, the o/p voltage is always
one-half of the total secondary voltage
less the diode drop, no matter what the
Turns ratio
Vout 
Vsec
 0.7V
2
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(2-6)
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2-2 Full-Wave Rectifiers (cont.)
(i - The Center-Tapped Full-Wave Rectifier)
-Peak Inverse Voltage (PIV)
Maximum anode voltage:
D1  
V p (sec)
2
D2  
V p (sec)
2

D1: forward-bias – cathode is at the same voltage of anode – Vpotential.

PIV across D2 :
 V p (sec)
   V p (sec) 

PIV  
 0.7V   
2
2

 

 V p (sec)  0.7V
We know that
V p (sec)
V p ( out) 
 0.7V
2
 V p (sec)  2V p ( out)  1.4V
 Thus;

PIV  2V p ( out)  0.7V
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reverse-bias
(2-7)
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2-2 Full-Wave Rectifiers (cont.)
(ii - The Bridge Full-Wave Rectifier)
 The full-wave bridge rectifier takes advantage of the full output of the
secondary winding.
 It employs four diodes arranged such that current flows in the direction
through the load during each half of the cycle.
Used 4 diode:
2 diode in forward
2 diode in reverse
2 diode always in series
(both cases).
Without diode drop (ideal diode):
V p ( out)  V p (sec)
With diode drop
(practical diode):
V p ( out)  V p (sec)  1.4V
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(2-8)
2-2 Full-Wave Rectifiers (cont.)
(ii - The Bridge Full-Wave Rectifier)
For ideal diode, PIV = Vp(out)
PIV  V p ( out )  0.7V
(2-9)
0V (ideal diode)
PIV  V p (out )
PIV  V p ( out )  0.7V
Note that in most cases we take the diode drop into account.
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Example 4:
1. For center-tapped full-wave rectifier with RL = 10k, sketch Vsec ,
Vp(out) and calculate PIV for the following problems:
(a) Transformer ratio n = 0.25; Vp(in) = 70V,
(b) Transformer ratio n = 0.5; Vp(in) = 100V
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Example 4 (cont):
2. For bridge full-wave rectifier (practical model) with RL = 10k, sketch
Vsec & Vp(out) for the following problem:
(a) Transformer specified to have a 15Vrms secondary voltage with
Vp(in) = 120V.
What PIV rating required for each diode?
Vrms 
V p (sec)
2
;
Vp(in)
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2-3 Power Supply Filters And Regulators
(introduction)
Objective:






Explain & Analyze the operation & characteristic
of power supply filters & Regulators
Explain the purpose of a filter
Describe the capacitor-input filter
Define ripple voltage & calculate the ripple
voltage
Discuss surge current in capacitor-input filter
Discuss voltage regulation & integrated circuit
regulator
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2-3 Power Supply Filters And Regulators (cont.)
(introduction)
Power Supply Filters



To reduce the fluctuations in the output voltage of half / full-wave
rectifier – produces constant-level dc voltage.
It is necessary – electronic circuits require a constant source – to
provide power & biasing for proper operation.
Filters are implemented with capacitors.
Regulators


Voltage regulation in power supply done using integrated circuit
voltage regulators.
To prevent changes in the filtered dc voltage/ to fix output dc voltage
due to variations in input voltage or load.
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2-3 Power Supply Filters And Regulators (cont.)
(introduction)
•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
ripple
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2-3 Power Supply Filters And Regulators (cont.)
(Capacitor-Input Filter)
For half-wave rectifier:
• +ve first quarter cycle –diode is FBcapacitor is charge within 0.7V of i/p
peak
load
capacitor
• I/p decreased-capacitor retains its
charge-diode become RB (cathode is
more +ve than the diode)
• Capacitor can discharge through load
resistance – at rate by the RLC time
constant (>> time constant, << capacitor
will discharge)
• Next cycle-diode is FB when i/p voltage
exceeds the Vc by 0.7V
• A capacitor-input filter will charge and
discharge such that it fills in the “gaps”
between each peak. This reduces
variations of voltage. This voltage
variation is called ripple voltage.
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2-3 Power Supply Filters And Regulators (cont.)
(Capacitor-Input Filter)
Ripple Voltage: the variation in the o/p voltage from a filter (due to the
charging and discharging)
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.
Ripple voltage
– variation in the capacitor voltage due to charging & discharging.
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Easier to filter
-shorted time between
peaks.
-smaller ripple.
2-3 Power Supply Filters And Regulators (cont.)
(Capacitor-Input Filter)
Ripple factor: indication of the effectiveness of the filter
r
Vr ( pp )
VDC
(2-10) [half-wave rectifier]
Vr(pp) = peak to peak ripple voltage; VDC = VAVG = 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]
•For the full-wave rectifier:
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 1 
Vp(rect) = unfiltered
V p ( rect )
Vr ( pp)  
(2-11)
peak.
 fRL C 

1 
V p ( rect ) (2-12)
VDC  VAVG  1 
 2 fRLC 
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2-3 Power Supply Filters And Regulators (cont.)
(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.
The min. surge
Resistor values:
Rsurge 
V p (sec)  1.4V
IFSM = forward surge
current rating
specified on
diode data sheet.
I FSM
(2-13)
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2-3 Power Supply Filters And Regulators (cont.)
(IC Regulators)
-Connected to the output of a filtered & maintains a constant output
voltage (or current) despite changes in the input, load current or
temperature.
-Combination of a large capacitor & an IC regulator – inexpensive &
produce excellent small power supply
-Popular IC regulators have 3 terminals:
(i) input terminal
(ii) output terminal
(iii) reference (or adjust) terminal
-Type number: 78xx (xx –refer to output voltage)
i.e 7805 (output voltage +5.0V); 7824 (output voltage +24V)
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2-3 Power Supply Filters And Regulators (cont.)
(IC 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.
Connected to the output
of filtered rectifier
output
Gnd
Bridge-full wave
rectifier
Adjustable regulators
Vout
filter
 R  R2 

 1.25V  1
R
1


regulators
Fig. 2-23 : A basic +5.0V regulated power supply
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2-3 Power Supply Filters And Regulators (cont.)
(IC Regulators)
Keeps a constant 1.25V between the o/p
and adjust terminals const. current in R1
Adjustable resistor
0 – 1.0kOhm
FIGURE 2-34 A basic power supply with a variable output voltage (from
1.25 V to 6.5 V).
Adjustable regulators
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 R  R2 

Vout  1.25V  1
R
1


31
min. 1.25V , max 6.5 V
2-3 Power Supply Filters And Regulators (cont.)
(Percent Regulations)
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 and load regulation percentage is simply a ratio of
change in voltage (line) or current (load) stated as a percentage.
 Vout 
Line Regulation  
 V 
100%
in 

 VNL  VFL 

Load Regulation  
100%


VFL


VNL :out put voltage with no load
VFL :output voltage with full load
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2-4 Diode Limiting & Clamping Circuits
(Introduction)
Objectives:
Analyze
the operation of diode limiting, clamping circuit,
voltage multipliers and interpret and use diode data sheet.
Determine V of biased limiter & used voltage-divider bias to set
limiting level.
Discuss voltage doublers, triplers & quadruples.
Identify V & current ratings.
Determine the electrical characteristics of a diode.
Analyze graphical data
Select an appropriate diode for a given set of specifications.
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2-4 Diode Limiting & Clamping Circuits
(Diode Limiters)
• Diode limiters/clippers – that limits/clips the portion of signal voltage above or below
certain level.
• Limiting circuits limit the positive or negative amount of an input voltage to a specific
value.
• When i/p is +ve – the diode becomes FB – limited to +0.7 V because cathode is at ground.
• When i/p << 0.7 V-diode is RB – o/p voltage likes –ve part of i/p voltage
• Turn the diode around-negetive part of i/p voltage is clipped off.
• When diode is FB-negative part of i/p voltage-diode drop -0.7V
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 RL 
Vin
Vout  
R

R
L 
 1
2-4 Diode Limiting & Clamping Circuits (cont.)
(Diode Limiters)
Biased Limiters :
• The level to which an ac voltage is limited can be adjusted by adding bias voltage
VBIAS in series with diode.
• Voltage at point A : VA = VBIAS + 0.7V (forward-biased & conduct). So, all Vin > VA is
clipped off.
• For –ve level, then VA = -VBIAS - 0.7V to forward-biased.
• Turning diode around, +ve limiter – modified to limit Vout to the portion of Vin
waveform above VBIAS – 0.7V.
• -ve limiter; below -VBIAS + 0.7V.
• by tuning the diode around- the +ve limiter can modified to limit the o/p voltage to
the portion of the i/p voltage waveform above VBIAS-0.7V
• negative limiter – limit the o/p voltage to the i/p voltage below –VBIAS+0.7V
A negative limiter
A positive limiter
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2-4 Diode Limiting & Clamping Circuits (cont.)
(Diode Limiters)
Voltage-Divider Bias:
• The bias voltage source – used to illustrate the basic operation of diode limiters can
be replace by a resistive voltage divider that derives the desired bias voltage from dc
Vsupply .
• VBIAS – set by the resistor values according to the voltage-divider formula:
 R3 
VSUPPLY
VBIAS  
R

R
3 
 2
• The desired amount of limitation can be attained by a power supply or voltage
divider. The amount clipped can be adjusted with different levels of VBIAS.
• The bias resistor << R1- the forward current through the diode – no effect
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Example 5:
1. Sketch the output voltage waveform as shown in the circuit combining
a positive limiter with negative limiter in Figure 5-1.
+15V
6V
-15V
Figure 5-1
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6V
Example 5 (cont.):
2. A student construct the circuit as shown in Figure 5-2. Describe the output
voltage waveform on oscilloscope CH2.
+15V
+20V
CH2
-20V
Figure 5-2
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2-4 Diode Limiting & Clamping Circuits (cont.)
(Diode Clampers)
- A diode clamper adds a DC level to an AC voltage. The capacitor charges to the peak of the
supply minus the diode drop. Once charged, the capacitor acts like a battery in series with the
input voltage. The AC voltage will “ride” along with the DC voltage. The polarity arrangement of
the diode determines whether the DC voltage is negative or positive.
- For negative clamper, the diode is turn around. A negative dc voltage is added to the input
voltage to produce the output.
Also known as
dc restorers.
For a good clamping action
RC time constant ~ 10 fin
Positive clamper operation
Negative clamper operation
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2-4 Diode Limiting & Clamping Circuits (cont.)
(Diode Clampers)
A Clamper Application:
A clamping circuit is often used in TV receivers as a dc
restorer.
The incoming composite video signal is normally processed
through capacitively coupled amplifiers that eliminate the dc
component, thus losing black and white reference levels and
the blanking level. Before applied to the picture tube, these
reference level must be restored.
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2-5 Voltage Multiplier
(Introduction)
• Use clamping action to increase peak rectified voltages
without necessary to increase input transformer’s voltage
rating.
• Multiplication factors: two, three or four.
• Three types of voltage multipliers:
* Voltage doubler
- Half – wave voltage doubler
- Full – wave voltage doubler
* Voltage tripler
* Voltage Quadrupler
• Voltage multipliers are used in high-voltage, low current
applications, i.e TV receivers.
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2-5 Voltage Multiplier (cont.)
(Voltage Doubler)
Half-wave voltage Doubler:
Clamping action can be used to increase peak rectified voltage. Once C1 and C2
charges to the peak voltage they act like two batteries in series, effectively
doubling the voltage output. The current capacity for voltage multipliers is low.
PIV = 2Vp
By applying Kirchhoff’s Law at (b):
VC 2  V p  VC1
~ approximately 2Vp (neglecting diode drop D2)
Half-wave voltage doubler operation. Vp is the peak secondary voltage.
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2-5 Voltage Multiplier (cont.)
(Voltage Doubler)
Full-wave voltage doubler:
Arrangement of diodes and capacitors takes advantage of both
positive and negative peaks to charge the capacitors giving it more
current capacity. forward-bias
output
charges
forward-bias
charges
Secondary voltage positive
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Secondary voltage negative
43
2-5 Voltage Multiplier (cont.)
(Voltage Tripler & Voltage Quadrupler)
• Voltage triplers and quadruplers utilize three and four diode capacitor
arrangements, respectively.
• Voltage tripler and quadrupler gives output 3Vp and 4Vp, respectively.
• Tripler output is taken across C1 and C3, thus Vout = 3Vp
• Quadrupler output is taken across C2 and C4 , thus Vout = 4Vp
• PIV for both cases: PIV = 2Vp
Voltage Triple
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Voltage Quadruple
44
2-6 The Diode Data Sheet
(Introduction)
• The data sheet for diodes and other devices gives detailed
information about specific characteristics such 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.
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2-6 The Diode Data Sheet (cont.)
(Maximum Rating)
1N4001
1N4002
1N4003
UNIT
Rating
Symbol
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
V
rms reverse voltage
VR(rms)
35
70
140
V
Average rectified forward
current (single-phase, resistive
load, 60Hz, TA = 75oC
Io
Nonrepetitive peak surge current
(surge applied at rated load
conditions)
IFSM
Operating and storage junction
temperature range
Tj, Tstg
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A
1
A
30 (for 1
cycle)
46
-65 to
+175
oC
2-6 The Diode Data Sheet (cont.)
(Maximum Rating)
FIGURE 2-56
A selection of rectifier diodes based on maximum ratings of I O, IFSM, and IRRM.
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2-7 Troubleshooting
(introduction)
Objective:



Troubleshoot diode circuits using accepted techniques.
Discuss the relationship between symptom & cause, power
check, sensory check, component replacement method and
discuss the signal tracing technique in the three variations.
Fault analysis.
Our study of these devices and how they work leads more effective
troubleshooting. Efficient troubleshooting requires us to take logical steps in
sequence. Knowing how a device, circuit, or system works when operating
properly must be known before any attempts are made to troubleshoot. The
symptoms shown by a defective device often point directly to the point of
failure. There are many different methods for troubleshooting. We will
discuss a few.
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2-7 Troubleshooting (cont.)
(Troubleshooting Techniques)
Here are some helpful troubleshooting techniques:
 Power Check: Sometimes the obvious eludes the most
proficient troubleshooters. Check for fuses blown, power
cords plugged in, and correct battery placement.
 Sensory Check: What you see or smell may lead you
directly to the failure or to a symptom of a failure.
 Component Replacement: Educated guesswork in
replacing components is sometimes effective.
 Signal Tracing: Look at the point in the circuit or
system where you first lose the signal or incorrect signal.
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2-7 Troubleshooting (cont.)
(Troubleshooting Techniques)
Signal tracing techniques:
Input to output
Output to input
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2-7 Troubleshooting (cont.)
(Fault Analysis)
Can be applied when you measure an incorrect voltage at a test point using signal
tracing and isolate the fault to a specific circuit.
Example 1:
Effect of an Open Diode in a HalfWave Rectifier:
- Zero o/p voltage
- Open diode breaks the current path from
transformer secondary winding to the
filter and load resistor – no load current.
Other faults: open transformer winding, open
fuse, or no input voltage.
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2-7 Troubleshooting (cont.)
(Fault Analysis)
Example 2:
Effect of an Open Diode in a Full-Wave Rectifier:
-The effect of either of two diodes is open diode, the o/p voltage will have large than
normal ripple voltage at 60 Hz rather than at 120 Hz.
- Another fault – open in one of the halves of the transformer secondary winding.
- Open diode give same symptom to bridge full-wave rectifier.
(See Figure 2-63)
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2-7 Troubleshooting (cont.)
(Fault Analysis)
Example 3:
Effect of a Shorted Diode in a Full-Wave Rectifier:
 Fuse should blow – cause by short circuit
 D1,D4 will probably burn open.
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2-7 Troubleshooting (cont.)
(Fault Analysis)
Example 4:
Effect of a fault filter capacitor:
 Open – o/p is full-wave rectified voltage
 Shorted – the o/p is 0V
 Leaky – increase the ripple voltage on the o/p
Example 5:
Effect of a Faulty Transformer:
 Open primary/secondary winding of a
transformer – 0V o/p
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2-7 Troubleshooting (cont.)
(The complete Troubleshooting Process)

The complete troubleshooting process:
(i) Identify the symptom(s).
(ii) Perform a power check
(iii) Perform a sensory check
(iv) Apply a signal tracing technique.
(v) Apply fault analysis
(vi) Use component replacement to fix the problem.
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Summary
 The basic function of a power supply to give us a smooth ripple
free DC voltage from an AC voltage.
Half-wave rectifiers only utilize half of the cycle to produce a DC
voltage.
Transformer Coupling allows voltage manipulation through its
windings ratio
Full-Wave rectifiers efficiently make use of the whole cycle. This
makes it easier to filter.
The full-wave bridge rectifier allows use of the full secondary
winding output whereas the center-tapped full wave uses only
half.
Filtering and Regulating the output of a rectifier helps
keep the DC voltage smooth and accurate
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Summary
Limiters are used to set the output peak(s) to a given
value.
Clampers are used to add a DC voltage to an AC voltage.
Voltage Multipliers allow a doubling, tripling, or
quadrupling of rectified DC voltage for low current
applications.
The Data Sheet gives us useful information and
characteristics of device for use in replacement or designing
circuits.
Troubleshooting requires use of common sense along
with proper troubleshooting techniques to effectively
determine the point of failure in a defective circuit or
system.
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Solution 2:
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Solution 3:
N sec 1
n
  0.2
N pri 5
V p (sec)  nV p ( pri)  0.2(120)  24V
V p ( out)  V p (sec)  0.7V  24  0.7  23.3V
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Solution 4:
+70V,+100V
-70V,-100V
+17.5V,+50V
-17.5V,-50V
+8.75V,+25V
V p (sec)  nV p ( pri)
PIV  V p (sec)  0.7V
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Solution 4:
1.
+70V,+100V
-70V,-100V
+17.5V,+50V
-17.5V,-50V
+8.75V,+25V
V p (sec)  nV p ( pri)
PIV  V p (sec)  0.7V
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Solution 4 (cont.):
2.
V p (sec)  2Vrms  2 x15  21.21V
V p ( out)  V p (sec)  0.7V  21.21  0.7  20.51V
PIV  V p ( out)  0.7V  20.51  0.7  21.21V
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Solution 5:
1.
+6.7V
-6.7V
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Solution 5 (cont.):
2.
 R3 
220 

VSUPPLY  
VBIAS  
20V
 100  220 
 R2  R3 
 13.75V
+13.75V
-18V
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