CHAPTER 2
Diode Applications
1
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
2
Lecture’s contents
•
•
•
•
•
•
•
•
•
2-1 The Basic DC Power Supply
2.2 Half-wave rectifier
2-3 Full-wave rectifier
2-4 Power supply filters and regulators
2-5 Diode limiting and clamping circuits
2-6 Voltage multipliers
2-7 Diode data sheet
2-8 Troubleshooting
Summary
3
2-1 The Basic DC Power Supply
Power supply is a group of circuits that convert the standard ac
voltage (120 V, 60 Hz) provided by the wall outlet to constant dc
voltage.
The voltage produced is used to power all types of electronic circuits
including:
Consumer electronics (ex: radio, television, DVD, etc.)
Computers
Industrial controllers
Most laboratory instrumentation systems and equipment
The dc voltage level required depends on the application, but most
applications require relatively low voltage.
There are two basic types of power supplies: a linear power supply
and a switching power supply. These components are described as
follows:
1. A linear power supply is one that provides a constant current
path between its input and its load.
2. A switching power supply provides an intermittent current
path between its input and its output.
4
2-1 The Basic DC Power Supply(cont.)
The basic function of a DC power supply is to convert an AC
voltage to a constant DC voltage (AC DC) Maintains a constant dc
voltage
Eliminates the fluctuations
- produce smooth dc voltage
Either half or full-wave
rectifier convert ac input voltage
to a pulsating dc voltage.
(a) Complete power supply with transformer, rectifier, filter, and regulator
(b) Half-wave rectifier
Fig. 2-1: Block diagram of a dc power supply with a load and rectifier.
5
2-1 The Basic DC Power Supply(cont.)
The basic function of a DC power supply is to convert an AC
voltage to a constant DC voltage (AC DC) Maintains a constant dc
voltage
Eliminates the fluctuations
- produce smooth dc voltage
Either half or full-wave
rectifier convert ac input voltage
to a pulsating dc voltage.
(a) Complete power supply with transformer, rectifier, filter, and regulator
(b) Half-wave rectifier
Fig. 2-1: Block diagram of a dc power supply with a load and rectifier.
6
2-1 The Basic DC Power Supply(cont.)
Transformer
A transformer is a device that changes ac electric
power at one voltage level to ac electric power at
another voltage level through the action of a
magnetic field.
Simple transformer consist of:
1. Primary winding (input winding)
2. Secondary winding (output winding)
3. Magnetic core
Φ
If the secondary has more turns than the primary,
the output voltage across the secondary will be
higher and the current will be smaller. If the
Fig.2-2: The general arrangement
secondary has fewer turns than the primary, the
of a transformer
output voltage across the secondary will be lower
and the current will be higher.
The core has a function to concentrate the magnetic
flux.
7
2-1 The Basic DC Power Supply(cont.)
There are three types of transformers: step-up, step-down, and isolation.
These components are described as follows:
1. The step-up transformer provides a secondary voltage that is greater than
the primary voltage. Ex: a step-up transformer may provides a 240 Vac
output with a 120 Vac input.
2. The step-down transformer provides a secondary voltage that is less than
the primary voltage. Ex: a step-down transformer may provides a 30 Vac
output with a 120 Vac input.
3. An isolation transformer provides an output voltage that is equal to the input
voltage. This type of transformer is used to isolate the power supply
electrically from the ac power line.
+
N
N
P
S
120
Vac
-
+
240
Vac
Step-up
(a)
-
+
N
N
P
S
120
Vac
+
+
30
Vac
120
Vac
-
-
N
P
S
+
120
Vac
-
-
Step-down
(b)
N
Isolation
(c)
Fig.2-3
8
2-1 The Basic DC Power Supply(cont.)
The turns ratio of a transformer is equal to the voltage ratio of the
component and since, the voltage ratio is the inverse of the current ratio.
By formula:
Nsec Vsec I pri
=
=
Npri Vpri Isec
(2-1)
where
NSec = the number of turns in the secondary
NPri = the number of turns in the primary
VSec = the secondary voltage
VPri = the primary voltage
ISec = the secondary current
IPri = the primary current
By the equation (2-1) can be stated that:
Step-down transformer secondary current is greater than its primary
current
(ISec > IPri).
Step-up transformer secondary current is less than its primary current
(IPri > ISec).
9
2-2 HalfHalf-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 barrier potential, VP on a half-wave rectifier
output
• Define Peak Inverse Voltage (PIV)
• Describe Transformer-couple half-wave rectifier
•
•
•
•
10
2-2 HalfHalf-Wave Rectifiers (cont.)
(The HalfHalf-Wave Rectifier)
ac source
load resistor
•A half wave rectifier(ideal) allows
conduction for only 180°or half of
a complete cycle..
•During first one cycle:
-Vin goes positive – diode FB –
conduct current
-Vin goes negative – diode RB – no
current- 0V
•The output frequency is the same
as the input (same shape).
The average value
VDC or VAVG :
VAVG =
-Measure on dc voltmeter
Vp
π
Ideal diode model
11
2-2 HalfHalf-Wave Rectifiers (cont.)
(Effect of the Barrier Potential on the HalfHalf-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.
• Example 1: Calculate the peak o/p voltage, Vp(out)?
The peak o/p voltage:
V p ( out ) = V p (in ) − 0.7V
Vp(out) =Vp(in) −0.7V
= 5V −0.7V
= 4.30V
12
2-2 HalfHalf-Wave Rectifiers (cont.)
(Effect of the Barrier Potential on the HalfHalf-Wave Rectifier Output)
Example 2:
Sketch the output V0 and determine the output level voltage for the
network in above figure.
13
2-2Half2Half-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.
PIV = Vp(in)
14
2-2 HalfHalf-Wave Rectifiers (cont.)
(Half--Wave Rectifier with Transformer(Half
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 = nVpri
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.
to couple ac input
to the rectifier
N sec
n=
N pri
If n>1, Vsec is greater than Vpri.
If n<1, Vsec is less than Vpri.
If n=1, Vsec= Vpri.
Vp(out) = Vp(sec) − 0.7V
PIV = Vp(sec)
15
Example 3:
Determine the peak value of output voltage as shown in below Figure.
16
2-3 FullFull-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.
17
2-3 FullFull-Wave Rectifiers (cont.)
(Introduction)
A full-wave rectifier allows current to flow during both the positive and
negative half cycles or the full 360º whereas 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 =
2V p
Twice output
π
63.7% of Vp
18
2-3 FullFull-Wave Rectifiers
(i - The CenterCenter-Tapped FullFull-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.
Coupled input
voltage
19
2-3 FullFull-Wave Rectifiers (cont.)
(i - The CenterCenter-Tapped FullFull-Wave Rectifier)
•+ve half-cycle input voltage (forward-bias D1 & reverse-bias D2)-the
current pass through the D1 and RL
•-ve half-cycle input voltage (reverse-bias D1 & forward-bias D2)-the
current pass 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
20
2-3 FullFull-Wave Rectifiers (cont.)
(i - The CenterCenter-Tapped FullFull-Wave Rectifier)
-Effect of the Turns Ratio on the Output Voltage-
If n=1, Vp(out)=Vp(pri) - 0.7V
2
Vp(sec)=Vp(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
minus the diode drop (barrier potential),
no matter what the turns ratio.
Vout
Vsec
=
− 0.7V
2
21
2-3 FullFull-Wave Rectifiers (cont.)
(i - The CenterCenter-Tapped FullFull-Wave Rectifier)
-Peak Inverse Voltage (PIV)V p (sec)
V p (sec)
D2 = −
Maximum anode voltage:
D1 = +
2
2
D1: forward-bias – its cathode is at the same voltage of its anode minus diode drop;
This is also the voltage on the cathode of D2.
PIV across D2 :
V
  −V

PIV =  p(sec) − 0.7V  −  p(sec) 
 2
  2 
= Vp(sec) − 0.7V
We know that
Vp(out) =
Vp(sec)
2
− 0.7V
→ Vp(sec) = 2Vp(out) +1.4V
Thus;
PIV = 2V p ( out ) + 0.7V
22
2-2 FullFull-Wave Rectifiers (cont.)
(ii - The Bridge FullFull-Wave Rectifier)
It employs four diodes arranged such that current flows in the direction
through the load during each half of the cycle.
When Vin +ve, D1 and D2 FB and conduct current. A voltage across RL
looks like +ve half of the input cycle. During this time, D3 and D4 are RB.
When Vin –ve, D3 and D4 are FB and conduct current. D1 and D2 are RB.
Used 4 diode:
2 diode in forward
2 diode in reverse
Without diode drop (ideal diode):
Vp(out) = Vp(sec)
2 diode always in series with load
resistor during +ve and –ve half cycle .
With diode drop (practical diode):
Vp(out) = Vp(sec) −1.4V
23
2-2 FullFull-Wave Rectifiers (cont.)
(ii - The Bridge FullFull-Wave Rectifier)
For ideal diode, PIV = Vp(out)
For each diode, PIV = V p ( out ) + 0.7V
PIV = V p (out )
0V (ideal diode)
PIV = V p ( out ) + 0.7V
Note that in most cases we take the diode drop into account.
24
2-3 Power Supply Filters And Regulators
(introduction)
Objective::
Objective
Explain & Analyze the operation & characteristic of
power supply filters & Regulators
Explain the purpose of a filter
Describe the capacitorcapacitor-input filter
Define ripple voltage & calculate the ripple voltage
Discuss surge current in capacitorcapacitor-input filter
Discuss voltage regulation & integrated circuit
regulator
25
2-4 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.
26
2-4 Power Supply Filters And Regulators (cont.)
(introduction)
•In most power supply – 60 Hz ac power line voltage is converted to constant
dc voltage.
•60Hz pulsating dc output must be filtered to reduce the large voltage
variation.
•Small amount of fluctuation in the filter o/p voltage - ripple
ripple
27
2-4 Power Supply Filters And Regulators (cont.)
(Capacitor--Input Filter)
(Capacitor
Capacitive Filter
Capacitive filter is simply a capacitor
connected in parallel with the load resistance
or connected from the rectifier output to
ground, as shown in Fig.2-39.
During the positive first quarter-cycle of the
input, the diode is forward-biased, allowing the
capacitor charges rapidly, as illustrated in
Fig.2-39(a).
load
capacitor
When the input begins to go negative, the
diode is reverse-biased, and the capacitor
slowly discharges through the load resistance
(Fig.2-39(b)). As the output from the rectifier
drops below the charged voltage of the
capacitor, the capacitor acts as the voltage
source for the load.
During first quarter of the next cycle, as
illustrated in part (c), the diode will again
become forward-biased when the input voltage
exceeds the capacitor voltage.
28
Ripple Voltage
Ripple voltage is the fluctuation in the capacitor voltage due to the difference
between the charge and discharge times.
The difference between the charge and discharge times is caused by two distinct
RC time constant in the circuit. One time constant is found as:
τ = RC
where R and C are the total circuit resistance and capacitance, respectively.
Since it takes five time constants for a capacitor to charge or discharge fully, this time
period (T) can be found as:
T = 5RC = 5τ
29
For example, refer to Fig. 2-40(a), the capacitor charges through the diode.
Assuming that diode has a forward resistance of 5 Ω, so the time constant for the
circuit is found as:
τ = (5 Ω)(100 µF ) = 500 µs
and the total capacitor charge time is found as:
T = (5)(500 µs ) = 2.5 ms
The discharge path for the capacitor is through the resistor as shown in Fig. 2-40(b).
For this circuit, the time constant is found as:
τ = (1 kΩ)(100 µF ) = 100 ms
and the total capacitor discharge time is found as:
T = (5)(100 ms ) = 500 ms
30
(a) Charge circuit
(b) Discharge circuit
Fig.2-40: The basic capacitive filter.
31
2-4Power Supply Filters And Regulators (cont.)
(Capacitor--Input Filter)
(Capacitor
Ripple Voltage: the variation in capacitor voltage due to the
charging and discharging times.
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.
Easier to filter
-shorted time between
peaks.
-smaller ripple.
32
2-4 Power Supply Filters And Regulators (cont.)
(Capacitor--Input Filter)
(Capacitor
Ripple factor: indication of the effectiveness of the filter
r=
Vr ( pp )
[half-wave rectifier]
VDC
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]
 1 
•For the full-wave rectifier: V

V p ( rect )
≅
r ( pp )
 fRLC 

1 
V p ( rect )
VDC = V AVG =  1 −
 2 fRLC 
Vp(rect) = unfiltered
peak.
33
2-4 Power Supply Filters And Regulators (cont.)
(Capacitor--Input Filter)
(Capacitor
Surge Current in the Capacitor-Input Filter:
When you first turn on power supply, the filter capacitor has no accumulated charge
to oppose Vp(sec). For first instant, the capacitor appears as a short circuit, thus 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
I FSM
IFSM = forward surge
current rating
specified on
diode data sheet.
34
2-4 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)
35
2-4 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
Bridge-full wave
rectifier
Gnd
filter
regulators
Fig. 2-23 : A basic +5.0V regulated power supply
36
2-4 Power Supply Filters And Regulators (cont.)
(Percent Regulations)
Percent Regulation
The regulation can be stated in a percentage in terms of input (line) regulation or
load regulation.
Line regulation specifies how much change occurs in the output voltage for a
given change in the input voltage. It is mathematically defined as a ratio of a
change in output voltage for a corresponding change in the input voltage
expressed as a percentage.
 ∆VOUT
Line regulation = 
 ∆VIN

 x100%

Load regulation specifies how much change occurs in the output voltage over a
certain range of load current values, usually from minimum current (no load, NL)
to maximum current (full load, FL). It can be mathematically determined with the
following formula:
 V −V 
Load regulation =  NL FL  x100%
 VFL 
37
2-5 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 voltagevoltage-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.
38
2-5 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.
• 4 basic clipper configuration:
– Negative series clipper
diode is in series with its load
– Positive series clipper
– Negative shunt clipper
diode is in parallel with its load
– Positive shunt clipper
39
2-5 Diode Limiting & Clamping Circuits
(Diode Limiters)
Clipper configuration
40
2-5 Diode Limiters
(series clipper/limiter)
• Negative series clipper
• Diode is forward-biased during +ve
alternation of i/p signal.
• Diode is reverse-biased when i/p
signal is –ve.
• Eliminates negatives alternation of
its i/p.
• Positive series clipper
• Diode is reverse-biased during
+ve alternation of i/p signal.
• Diode is forward-biased when i/p
signal is –ve.
• Eliminates positive alternation of
its input.
41
2-5Diode Limiters
(shunt clipper/limiter)
Positive shunt clipper
• Forward-biased diode when i/p is
+ve cycle.
• Reverse-biased diode when i/p is
in –ve cycle.
• o/p signal is limit/clip to +0.7V
during +ve cycle of i/p signal.
Negative shunt clipper
• Reverse-biased diode act as open
cct during +ve cycle.
• Forward-biased diode act as short
cct during –ve cycle.
• o/p signal is limit/clip to -0.7V
during –ve cycle of i/p signal.
42
2-5 Diode Limiting & Clamping Circuits
(Diode Limiters)
Series clipper
• When diode in –ve series
clipper is FB, load voltage is:
VL=Vin – 0.7V
• When diode is RB, doesn’t
conduct, so:
– VL = 0V
• +ve series clipper operates
the same. The only
differences are:
– O/p voltage polarities are
reversed. VL=-Vin + 0.7V
– Current direction through cct
are reversed.
Shunt clipper
• For –ve shunt, when i/p signal
+ve cycle, diode is RB (open
circuit), thus:
Vout
 RL
= 
 R1 + RL

Vin

•
During –ve cycle, diode is FB,
load voltage is equal to diode
forward voltage.
VL = -VF = -0.7V
• For +ve shunt, o/p voltage
and current direction are
reversed.
43
Question 4:
What would you expect to see displayed on an oscilloscope connected
across RL in the limiter shown below.
44
• Solution question 4
• The diode is forward biased and conducts when input
voltage goes below -0.7V. So, for –ve limiter, the peak
output voltage across RL is:
Vp(out)
 RL 
Vp(in)
= 
 R1 + RL 
 1.0kΩ 
=
10V = 9.09V
 1.1kΩ 
• The waveform is shown below:
•
45
2.5 Diode Limiting & Clamping Circuits (cont.)
(Diode Limiters)
Biased Limiters :
• Use dc biasing source, VBIAS to set limit on the circuit output voltage.
• This allow the circuit to clip input waveform at values other than diode forward voltage,
0.7V.
• In each circuit, bias voltage is in series with shunt diode. As a result, the diode conducts and
clips the i/p waveform when signal voltage equals sum of VF and VBIAS.
• 2 type of biased limiter:
positive-biased limiter
negative-biased limiter
• Positive limiter
•
The voltage at point A must equal VBIAS+0.7V before diode become FB and
conduct.
•
Once diode begin to conduct, voltage at point A is limited to VBIAS+0.7V, so all i/p
voltage above this level is clipped off.
• Negative limiter
•
Voltage at point A must go below –VBIAS - 0.7V to forward-bias the diode and
initiate limiting action.
A positive limiter
A negative limiter
46
2-5 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:
VBIAS
 R3 
VSUPPLY
= 
 R2 + R3 
• 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 will not effect VBIAS
47
48
2-5 Diode Limiting & Clamping Circuits (cont.)
(Diode Limiter Applications)
• Half-wave rectifier
– Circuit alters the shape of ac signal and change it to pulsating dc.
• Transient protection circuit
– Transient abrupt current or voltage spike in short duration.
– Many digital circuits have i/p that cannot tolerate voltage fall outside a
specified range which can cause serious damage. A clipper can be used
to protect these circuits.
• AM detector
– Eliminate –ve portion of i/p waveform, so capacitor charges and
discharges at rate of peak i/p variations. This provides a signal at load
that is a reproduction of i/p signal.
49
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
6V
Figure 5-1
50
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
51
2-5 Diode Limiting & Clamping Circuits (cont.)
(Diode Clampers)
2.7 Clampers (DC Restorers)
Clamper is a diode circuit designed to shift a waveform either above or below a given
reference voltage without distorting the waveform.
There are two types of clampers: the positive clamper and the negative clamper.
1. A positive clamper shifts its input waveform so that the negative peak of the
waveform is equal to the clamper dc reference voltage.
For example: Fig. 2-54 shows what happens when a 20 Vpp sin wave is applied to a
positive clamper with a dc reference of 0 V. The input and output waveforms have the
value of 20 Vpp. However, the clamper output waveform has the positive peak of +20 V and
the negative peak of 0 V. The positive clamper has shifted the entire waveform so that
its negative peak is equal to the circuit’s dc reference voltage.
2-5 Diode Limiting & Clamping Circuits (cont.)
(Diode Clampers)
2. A negative clamper shifts its input waveform so that the positive peak of the waveform
is equal to the clamper dc reference voltage.
For example: Fig. 2-55 shows what happens when a 20 Vpp sin wave is applied to
a negative clamper with a dc reference of 0 V. In this case, The clamper output
waveform has the positive peak of 0 V and the negative peak of –20 V. The
negative clamper has shifted the entire waveform so that its negative peak is
equal to the circuit’s dc reference voltage.
Clamper Operation
The clamper is similar to a shunt clipper; the
difference is added capacitor in the clamper,
as illustrated in Fig. 2-56.
For the circuit in Fig. 2-56(a), the diode is
forward biased and it charges the capacitor.
Thus, the charging time constant is found
as:
(a) Capacitor charge circuit
τ = RDC
(2-39)
where RD is bulk resistance of the
diode and C is capacitance of the
capacitor. The total charge time is:
TCh arg e = 5RD C
(2-40)
+
−
Reversebiased
(b) Capacitor discharge circuit
Fig.2-56: Clamper charge and discharge
When the diode is reverse biased, the capacitor starts to discharge through the resistor,
as shown in Fig. 2-56(b). Therefore, the discharge time constant is found as:
τ = RLC
(2-41)
and the total discharge time is found as:
TDischarge = 5RLC
(2-42)
The effect of the clamping action is shown in Fig. 2-56. The capacitor retains a charge
approximately equal to the input peak less the diode drop so that it acts as a battery.
Fig.2-57: Positive
clamper operation
55
If the diode is turned around, a negative dc voltage is added to the input
voltage to produce the output voltage as shown in Fig. 2-57:
Fig.2-57: Negative clamper operation
56
2-5 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.
57
2-6 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.
58
2-6 Voltage Multiplier (cont.)
(Voltage Doubler)
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.
59
2-6 Voltage Multiplier (cont.)
(Voltage Doubler)
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
Secondary voltage negative
60
2-6 Voltage Multiplier (cont.)
(Voltage Tripler & Voltage Quadrupler)
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
Voltage Quadruple
61
2-7 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.
62
2-7 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(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
A
1
A
30 (for 1
cycle)
-65 to
+175
oC
63
2-7 The Diode Data Sheet (cont.)
(Maximum Rating)
FIGURE 22-56 A selection of rectifier diodes based on maximum ratings of IO, IFSM, and IRRM.
64
2-8Troubleshooting
(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.
65
2-8 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.
66
2-8 Troubleshooting (cont.)
(Troubleshooting Techniques)
Signal tracing techniques:
Input to output
Output to input
67
2-8 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.
68
2-8 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)
69
2-8 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.
70
2-8 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
71
2-8 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.
72
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
73
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.
74
Solution 2:
The peak output voltage for the circuit is:
Vp(out) = Vp(in) – 0.7V = 5 – 0.7 = 4.30 V
75
Solution 3:
Vp(pri) = Vp(in) = 156 V
The peak secondary voltage is:
Vp(sec) = nVp(pri) = 0.5 (156 V) = 78 V
The rectified peak output voltage is:
Vp(out) = Vp(sec) – 0.7V = 78 – 0.7 = 77.3 V
where Vp(sec) is the input to the rectifier.
76
Solution 5:
1.
+6.7V
-6.7V
77
Solution 5 (cont.):
2.
 R3 
 220 
VSUPPLY = 
VBIAS = 
15V
 100 + 220 
 R2 + R3 
= 10.31V
+10.31V
-10.31V
78