DMT 121
ELECTRONIC DEVICES
CHAPTER 2
DIODE APPLICATIONS
At the end of this class, 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.

Load Line Analysis
The analysis of electronic circuits can follow one of the two paths :
1.
Actual characteristic or approximate model of the device.
2.
Approximate model will be always used in the analysis
VD= 0.7 V
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.
Point of
operation of
a circuit
Fig. 2.1 Drawing the load line and
finding the point of operation
Load Line Analysis
The intersection of load line in Fig.
2.2 can be determined by applying
Kirchhoff’s voltage in the clockwise
direction, which results in:
Fig. 2.2 Series diode configuration
 E  VD  VR  0
E  VD  IDR
ID and VD are the same for Eq. (2.1) and plotted load line in Fig. 2.2
(previous slide).
Set VD = 0 then we can get ID, where
Set ID = 0 then we get VD, where
E
ID 
R
VD  0
VD  E ID  0
Example

For the series diode configuration of Fig. 2.3a, employing the
diode characteristics of Fig. 2.3b, determine VDQ, IDQ and VR
Fig. 2.3
(a) Circuit; (b) characteristics.
Solution
E
10V
ID  V  0 
 20mA
R
0.5k
VD  E I 0  10V
D
D
From the result, plot the straight
line across ID and VD.
The resulting load line appears in
Fig. 2.4. The Q points occurred
at
VDQ  0.78 V
IDQ  18.5mA
VR=IRR=IDQR=(18.5 mA)(0.5k)
= 9.25 V
Example

For the series diode configuration of Fig. 2.13, determine VD, VR
and ID.
Solution:
VD  0.7V
VR  E  VD  8V  0.7V  7.3V
VR
7.3V
ID  IR 

 3.32mA
R 2.2k
Example

Repeat example 2.4 with the diode reversed
Solution:
Open Circuit
ID  0
E  VD  VR  0
VD   E  VR  8V  0V   8V
I d 0
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.
Half-Wave Rectifier

The diode conducts
during the positive
half cycle.

The diode does not
conducts during the
negative half cycle.
Sinusoidal Input:
Half Wave Rectification
Fig. 2.44 Half-wave
rectifier.
Fig. 2.45
Conduction region
(0  T/2).
Fig. 2.46
Nonconduction
region (T/2  T).
Average Value of Half Wave Output
Voltage
The average value of the
half-wave rectified output
voltage (also known as DC
voltage) is
Vdc  0.318Vm 
Vm

The process of removing onehalf the input signal to establish
a dc level is called half-wave
rectification
Example
What is the average value of the half-wave
rectified voltage?
Solution: Vm/π = 15.9 V
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 Vo = 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)
Example
Draw the output voltages of each rectifier for the
indicated input voltages.
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 tp.
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=Vm OR accurately
 PIV (or PRV)  Vm
• PIV = Peak inverse voltage
• PRV = Peak reverse voltage
• Vm = Peak AC voltage

Diode must capable to withstand certain
amount of repetitive reverse voltage
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.
VDC or VAVG = 2Vm/π
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
Full Wave Rectifier
Half Wave Rectifier
Example
Find average value of the full-wave rectified
voltage?
Transformer Coupling
Turns ratio, n = Nsec/Npri
V(sec) = nV(pri) (in RMS value)
Vp(sec)=√2 x V(sec)
Full-Wave Rectification
Center-Tapped Transformer
Rectifier
Requires
 Two diodes
Center-tapped transformer
VDC=0.636(Vm)
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
PIV: Full-wave Rectifier
Center-Tapped Transformer
PIV can be shown by applying
KVL for the reverse-biased
diode.
PIV across D2:


1
 V p (sec)
  V p (sec)
PIV  
 0.7V    
2
 2
 
2
PIV  V p (sec)  0.7V
3
4
V p ( out ) 
V p (sec)
V p (sec)
 0.7V
2
 2V p ( out )  1.4V



Substitute 4 to 2:
PIV=2Vp(out) + 0.7 V
Example
1. 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.
2. What minimum PIV rating must the diodes have.
Solution
1.
2. PIV = 49.3 V
Full-Wave Rectification
Bridge Rectifier
 Four diodes are required
 VDC = 0.636 Vm
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.
PIV: Full-wave Rectifier
Bridge Transformer

Vp(out)=Vp(sec) – 1.4 V

PIV=Vp(out) + 0.7 V
Example
The transformer is specified to have a 12 Vrms secondary 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
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-Wave Bridge
VDC = 0.636(Vm)
=2 Vm/π
VDC = 0.636(Vm)-2(0.7)
PIV=Vm+0.7V
VDC = 0.636(Vm)
=2 Vm/π
VDC = 0.636(Vm)-(0.7)
PIV=2Vm+0.7V
Rectifier
Center-Tapped
Transformer
Rectifier
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.
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
Power Supply Filters

Filtering is the process of smoothing the ripple from the rectifier.
Power Supply Filters and Regulators –
Capacitor-Input Filter
The capacitor
input filter
is widely
used. A
half-wave
rectifier
and the
capacitorinput filter
are shown.
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.
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.
Line regulation

 Vout 
100%
 
 Vin 
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)
Load regulation
 VNL  VFL 
100%
 
 VFL 
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.
The min. surge
Resistor values:
Rsurge 
V p (sec)  1.4V
I FSM
IFSM = forward surge current
rating specified on diode data
sheet.
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.
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]
 1 
•For the full-wave rectifier: V

V p ( rect )
r ( pp )  
 fRL C 
VDC  V AVG
Vp(rect) = unfiltered
peak

1 
V p ( rect )
 1 
 2 fRL C 
Ripple factor: indication of the effectiveness of the filter
r
Vr ( pp )
VDC
[half-wave rectifier]
Vr(pp) = peak to peak ripple voltage;
VDC = VAVG = average value of filter’s
output voltage
Example
Determine the ripple factor for the filtered bridge rectifier with a
load as indicated in the figure above.
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.
Diode Limiter/Clipper

A diode limiter is a circuit that limits (or clips) either the positive
or negative part of the input voltage.
 RL 
Vout  
Vin
 R1  RL 
Example
What would you expect to see displayed on an oscilloscope connected
across RL in the limiter shown in above figure.
Solution
 RL 
 1.0k 
Vout  
Vin  
10V  9.09V
 R1  RL 
 1.1k 
Biased Limiters (Clippers)
A positive limiter



The level to which an ac voltage is limited can be adjusted by
adding a bias voltage, VBIAS in series with the diode
The voltage at point A must equal VBIAS + 0.7 V before the diode
become forward-biased and conduct.
Once the diode begins to conduct, the voltage at point A is limited to
VBIAS + 0.7 V, so that all input voltage above this level is clipped off.
Biased Limiters (Clippers)
A negative limiter

In this case, the voltage at point A must go below –VBIAS – 0.7V to
forward-bias the diode and initiate limiting action as shown in the
above figure.
Modified Biased Limiters (Clippers)
Example
Figure above shows combining a positive limiter with a negative
limiter. Determine the output voltage waveform?
Solution
Summary Limiters (Clippers)
In this examples VD = 0
In analysis, VD = 0 or VD = 0.7 V can be used. Both are right assumption.
Summary Limiters (Clippers)
Diode Clampers



A clamper is a network constructed of a diode, a
resistor, and a capacitor that shifts a waveform to a
different dc level without changing the appearance of
the applied signal.
Sometimes known as dc restorers
Clamping networks have a capacitor connected
directly from input to output with a resistive element
in parallel with the output signal. The diode is also
parallel with the output signal but may or may not
have a series dc supply as an added elements.
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!
Diode Clampers
Positive clamper operation. (Diode pointing up – away from ground)
Diode Clampers
Negative clamper operation (Diode pointing down – toward ground)
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

Diode Clamper (Square Wave)
Diode ‘ON’ state
V – Vc = 0 ; Vc = V; Vo = 0.7 V but
ideal Vo = 0V
Diode ‘OFF’ state
-V - Vc - Vo = 0; Vc = V
Vo = -2 V
Output
Summary of Clamper Circuits
Voltage Multipliers

Voltage multiplier circuits use a combination of diodes
and capacitors to step up the output voltage of
rectifier circuits.

Voltage Doubler
Voltage Tripler
Voltage Quadrupler


Voltage Doubler

This half-wave voltage doubler’s output can be calculated by:
Vout = VC2 = 2Vm
where Vm = peak secondary voltage of the transformer
Half-Wave Voltage Doubler

Positive Half-Cycle




Negative Half-Cycle




D1 conducts
D2 is switched off
Capacitor C1 charges to Vp
D1 is switched off
D2 conducts
Capacitor C2 charges to Vp
Vout = VC2 = 2Vp
Full-Wave Voltage Doubler
Positif Half-Cycle
Negative Half-Cycle
• D1 forward-biased → C1
charges to Vp
• D1 reverse-biased
• D2 reverse-biased
• D2 forward-biased → C2 charges to
Vp
Output voltage=2Vp (across 2
capacitors in series
Voltage Tripler and Quadrupler
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
Voltage Quadrupler
Output: 4Vp across C2 and C4
The Diode Data Sheet



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.
The Diode Data Sheet
(Maximum Rating)
Rating
Symbol
Peak repetitive reverse voltage VRRM
Working peak reverse voltage
VRWM
DC blocking voltage
VR
1N4001
1N4002
1N4003
UNIT
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
A
1
Operating and storage junction Tj, Tstg
temperature range
A
30 (for 1
cycle)
-65 to
+175
oC
The Diode Data Sheet
(Maximum Rating)
Zener Diodes

The zener diode – 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
Zener Diodes





Breakdown voltage – set by controlling the doping level during
manufacture
When diode reached reverse breakdown – voltage remains constantcurrent change drastically
If zener diode is FB – operates the same as a rectifier diode
A zener diode 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
Zener Diodes







The reverse voltage (VR) is increased – the
reverse current (IR) remains extremely
small up to the “knee”of the curve
Reverse current – called the zener current,
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 zener diode maintains a constant
voltage for value of reverse current rating
from IZK to IZM
Zener Diodes
(Zener Equivalent Circuit)



Since the actual voltage is not ideally vertical, the change in
zener current produces a small change in zener voltage
By ohm’s law:
V
ZZ 
Normaly -Zz is specified at IZT
Z
I Z
Zener impedance
Zener Diodes
(Temp Coeff & Zener Power Dissipation and Derating)

As with most devices, zener diodes have given characteristics such
as temperature coefficients and power ratings that have to be
considered. The data sheet provides this information
Zener Diodes Applications

1.
2.
Zener diode can be used as
Voltage regulator for providing stable
reference voltages
Simple limiters or clippers
Zener Regulation with Varying Input
Voltage



As i/p voltage varies (within limits) – zener diode 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 zener can
operate
Resistor, R –current limiting resistor
Zener Regulation with a Variable Load

The zener diode maintains a nearly constant voltage across RL as long
as the zener current is greater than IZK and less than IZM

When the o/p terminal of the zener diode 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 zener diode continues to regulate the voltage until IZ reaches its
min value , IZK
At this point, the load current is max. , the total current through R
remains essentially constant.



Zener Limiting
Zener diode also can be used in ac applications to limit voltage swings to desired
level
(a) To limit the +ve peak of a signal voltage to the selected zener voltage
- During –ve alternation, zener arts as FB diode & limits the –ve voltage to 0.7V
(b) Zener diode is turn around
-The –ve peak is by zener action & +ve voltage is limited to +0.7V
(c) Two back-to-back zeners limit both peaks to the zener voltage ±7V
-During the +ve alternation, D2 is functioning as the zener limiter – D1 is
functioning as a FB diode.
-During the –ve alternation-the roles are reversed