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EE 334
Analog and Digital
Electronics
Chapter-2: Diode circuits
Power Supply Components
Rectifier Circuits
converts an ac voltage to a
pulsating dc voltage • A^ filter
then
eliminates ac components of the
waveform
to produce a
nearly constant dc
voltage output • Rectifier circuits are used in
virtually all
electronic devices to convert the 120-V 60-Hzac power line source to the dc voltages required for operation of electronic devices.• In rectifier circuits,
the diode state changes
with time and a given piecewise linear modelis valid only for a certain time interval
Half Wave Rectification
V^ γ^
rf
If^ r
is zero, when diode is on, f^
vo^
= v
Half Wave Rectification • Diode should be capable to withstandthe
current during forward bias
and the
peak inverse voltage
during the reverse bias
PIV = V
(max) + Vs
γ
Half-Wave Rectifier as
Battery Charger
Exercise 2.
- Determine the current and voltage in a half-waverectifier circuit shown in Fig. Assume V
= 12 V,B
the peak sinusoidal voltage is V
= 30 V and theS
resistor R = 100 K
Ω, and V
= 0.6 V. Determineγ
the peak diode current, maximum reverse-biasdiode voltage, and the fraction of the cycle overwhich the diode is conducting
Full-Wave Rectifier Center-
Tapped Transformer
v= vo^
Problem 2.
- The input signal voltage to the full-waverectifier shown in Fig. is
v^ I^
= 160 sin(
ωt).
Assume
V^ γ
= 0.7 V for each diode.
Determine the required turns ratio of thetransformer to produce a peak outputvoltage of 25 V, what would be the diode PIV
ratings.
- Repeat the problemfor centered tapedfull-wave rectifier
Rectifier Topology Comparison • Filter capacitors
are a major factor in
determining
cost, size and weight
in design of rectifiers
-^ For a
given ripple voltage
, a^
full-wave rectifier
requires
half the filter capacitance as that in a half-wave rectifier
. Reduced peak current can
reduce heat dissipation
in diodes. Benefits of
full-
wave rectification outweigh increased expensesand circuit complexity
(an extra diode and center-
tapped transformer)• The
bridge rectifier eliminates the center-tapped transformer, and the PIV rating of the diodes isreduced
.^ Cost of extra diodes is negligible
Power Supply Components
Output Ripple Voltage
-^ For all the previous rectifiercircuits: -^ Voltage waveforms
have too much
variation – Variation
ripple voltage
-^ Direct-current
supplies should
have as
little ripple as practical
-^ Answer = Filtering - Common devices is a
parallel capacitor
series inductor
Half-Wave Rectifier with Filter
¾^ A capacitor-input filter willcharge and discharge suchthat it fills in the “gaps”between each peak. ¾^ This reduces variations ofvoltage. ¾^ This voltage variation is calledripple voltage.
Power Supply Circuits^ Filters and Regulators
¾^ The advantage of a full-wave rectifier over a half-wave is quite clear. ¾^ The capacitor can more effectively reduce the ripple when the timebetween peaks is shorter.
Power Supply Circuits^ Filters and Regulators
V ripple
Full-Wave Rectifier with Filter ’^ t= time after the output has reached its peak value’^ T= discharge time T= time between two peaks of the output voltagep^ V=r^
ripple voltage
¾^ Most electronic applicationsrequire smooth dc currentto operate properly. ¾^ Filtering pulsating dc circuitsaccomplishes this. ¾^ Adding a capacitor to theoutput of a half-waverectifier filters the pulsatingdc into smooth dc.
Power Supply Applications
Filter Networks
¾^ A capacitive filter added tothe output of a full-wavebridge rectifier is shown atthe left. ¾^ One drawback of a half-waverectifier is the higher level of^ ripple voltage
after filtering.
¾^ Full-wave rectificationreduces this ripple voltage.
Power Supply ApplicationsFull-wave Rectifier with Filter
^ +
^ +
^ +
M^ r
M r M
D
M r
M maxD
M r
M Davg
V V 2
V R
V 2 V
avgi
V V 2
V R
i
V V 2
V R
i
π
π π π ) (
signal input entire over current diode Average
current diode
Maximum
condcution diode during current Average Diode Current with Filter
Design Example 2.
- A full-wave rectifier is to be designedto produce a peak output voltage of12 V, deliver 120 mA to the load,and produce an output ripple of notmore than 5%. An input line voltageof 120 V, 60 Hz is available.
Root Mean SquareV
V^2
0.707V
RMS
M
M
=^
= Mean value of sinusoidal over one
period signal is zero
Example 2.3^ Design a full-wave rectifier to meet particular specification.A full-wave rectifier is to be designed to procedure a peak output voltage of
12 V
,
deliver
120 mA
to the load, and produce an output with a ripple of not more than
5.0 %. An input line voltage of
120 V
(rms
),^ 60 Hz
is available.
P
“Full-wave rectifier design”
Problem 2.
- The output resistance of the full-waverectifier given in the Fig. is
R^
Ω. A
filter capacitor is connected in parallel with R. Assume
V^ γ
= 0.7 V. The peak output
voltage is to be 12 V and the ripplevoltage is to be no more than 0.3 V. Theinput frequency is 60. (a) Determine therequired rms of
v^ , (b) Determine thes
required filter capacitance value, (c)Determine the peak current through eachdiode.
Rectifiers and Voltage
Regulation
Power Supply Components
Zener Diode Characteristics
Basic Zener Voltage
Regulator
Variable VoltageSource (from Filter)
Variable LoadConditions
V^ PS^
R^ i
V^ L - I^ L
VZ^
R^ L I^ Z
I^ I
Breakdown Region Zener
Diode Model
In breakdown, the diode ismodeled with a
voltage
source
,^ V
, and aZ
series
resistance
,^ R
.^ Z
RZ
models the slope of the
i-v
characteristic. Zener diodes
designed to
operate in reversebreakdown and use theindicated symbol.
Zener Diode Characteristics Voltage varies from maximum
to minimum
V^ PS
Ri + -
V^ L - I^ L
VZO
R^ L
I^ Z I^ I
rZ
Zener Diode Characteristics •^ Zener Voltage Regulator Concern^ –
Output from filter can vary – Load can vary
- Important considerations: -^ Case 1:
Zener diode
must remain in
the breakdown region – Case 2: Power dissipated by zenerdiode must not exceed rated valuefor diode
Design Consideration
-^ Case 1: Remain in breakdownregion -^ Minimum
—^
minimum diode current
equal 0.1 maximum diode current – More difficult design might requireminimum =
0.2 to 0.3 of maximum
current
-^ Case 2: Power not exceed rated
–^ V
Zmax
I^ Zmax
zener dissipation rating
-^ Key– size the input resistance
,^ R
i
Sizing Series Resistance
L z i ps Z
L Z
z ps i^
I
V R
V
I
Or
I
I
V
V
R^
Case 1:
Vz^
> V
zo Vps
= min, I
= min, Iz^
= maxL
Case 2:
P ≤ z^
rated diode dissipation Vps
= max, I
= max, Iz
= minl
Sizing Series ResistanceCase 1: Case 2:
(max)I
(min)I
V (min) V R
L
Z
z
ps i
−
≤
(min)I
(max)I
V
(max)
V
R
L
Z
z
ps i
Problem 2.
- The secondary voltage in the circuit shownin Fig. is
v^ s^
= 12 sin(
ωt)
V. The Zener diode
parameters
Vz
= 8 V at
I^ z^
= 100 mA and
r^ z
. Let
V^ γ
Ri
Ω. Determine
the percent of regulation for load currentsbetween
I^ L
= 0.2 and 1 A. Find the
C^
such
that the ripple voltage is no larger than0.8 V.
In the Circuit given in Fig, the resistance
R^ = 1 k
Ω ,^ V^ L
=^ 10 V at 1 mA, and
r=^^30 Z^
Ω. Given that
Demonstration of Zener diode as a voltage regulator V changes from 11 V to 20 V, calculate the Zener current change and the output voltage change. in^ When
V = 11 V in^
Example
9
Solution:
Clipper and Clamper
wave
shaping circuits • Either:– Limit or “clip” signal portion
-^ Clippers - Shift the dc voltage level of asignal -^ Clampers
Clippers
-^ Eliminate signal portion that areabove or below specified levelApplications: •^ Limit input voltage to anelectronic circuit to preventcomponent damage
Clippers
V^ b
= 5 V,
V^ γ
= 0.7 V,
then clip at 5.7 V
^ Limiting circuits limit the positive or negative amount ofan input voltage to a specific value.
This positive limiter will limit the output to
VBIAS
0.7V
Clipper (Limiter) Circuits 양^ ( 음 )^ 의 리미터는 양
( 음 )^ 의 반주기 동안에 입력 파형을 적절한
DC 값으로 잘라내는 기능을 하는 회로
Clippers
V
= 5 V,B^
V
γ^ = 0.7 V,
then clip at 4.3 V
V is clipped when o^
v< Vi^
Clippers
V
= 5 V,B^
V^ γ
= 0.7 V,
then clip at -4.3 V
V is clipped when o^ v
-i ( VB
) γ
Clamper Circuits
¾^ Shift entire signal voltage by a dc voltage level
either
positively or negatively. ¾ Application: Restoring lost dc levels in signals from transmission
(e.g
Television)
Assume
V =^0^ γ
Clampers
Action of a Diode Clamper Circuit (a) a typical diode clamper circuit
(b) the sinusoidal input signal
(c) the capacitor voltage
(d) the output voltage
t V v^
M Input
ω sin =^
Assume
V =^0^ γ^ r =^0 f^ ) 1 (sin
sin^
−
=
−=
−=
t Vt
V V v v v^
M
M M Input Capacitor Out
ω
ω
Clamper Circuits
Ideally, the capacitor cannot discharge, remains constant v^ Capacitor Kirchhoff’s voltage law
¾^ Clamping action can be used to increase peak rectified voltage. ¾^ Once C
and C 1
charges to the peak voltage they act like two batteries in 2
series, effectively doubling the voltage output. ¾ The current capacity for voltage multipliers is low.
Voltage Multipliers
¾^ Voltage triplers and quadruplers utilize three and four diode-capacitor arrangements respectively.
Voltage Multipliers
Problem 2.
- The diode circuit shown in Fig. hasparameters
V
= 0.7 and γ
r^ f^
. (a)
Plot the
v^ o
versus
v^ I
for -
vI
If the triangular wave shown in Fig. isapplied, plot the output voltageversus time.
Diode OR Logic Circuits
Test your understanding
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