Transcript Lecture-7 - WordPress.com

Semiconductor Diode
Instructor: Farzana Rahmat Zaki
Senior Lecturer, EEE
Eastern University
Farzana R. Zaki
CSE 177/ EEE 177
1
Lecture-7
•Diode rectifier circuit
•Half wave rectifier
•Full wave rectifier
•Peak rectifier
Farzana R. Zaki
CSE 177/ EEE 177
2
Rectifier Circuits
• A rectifier is an electrical device that converts
an AC signal to DC signal.
• Rectifiers are used as components of power
supplies and as detectors of radio signals.
• Rectifiers may be made of diode components.
• There are two types of rectifier circuits: (a)
Half-wave rectifier and (b) Full-wave rectifier.
Farzana R. Zaki
3
CSE 177/ EEE 177
3
Half Wave Rectifier
• Utilizes alternate halfcycles of input sinusoid.
• Diode is modeled using
piecewise linear model.
V0 = 0,
vs< VD0
V0 = (vs – VD0)× R / (R + rD), vs ≥VD0
as, rD << R, so V0 ≈ (vs – VD0)
for vs ≥ VD0
Transfer characteristic curve
Farzana R. Zaki
4
CSE 177/ EEE 177
4
Half Wave Rectifier (contd.)
•
In selecting diodes for rectifier design, 2 parameters
must be specified :
(a) Current handling capability required of the diode,
(determined by the largest current the diode is
expected to conduct.)
(b) PIV (peak inverse voltage) that the diode must be able
to withstand without breakdown, (determined by
maximum reverse bias voltage appearing across the
diode.)
For half wave rectifier, PIV = VS
* It is usually prudent, however to select a diode that has a
reverse breakdown voltage at least 50% greater than
the expected PIV.
Farzana R. Zaki
CSE 177/ EEE 177
5
Full Wave Rectifier
• Full wave rectifier utilizes both halves of the input
sinusoid.
• To provide a uni-polar output, it converts the
negative halves of the sine wave.
• Full wave rectification can be obtained in two
ways- Full wave rectifier using a transformer with
a center-tapped secondary winding
- Full wave Bridge Rectifier
Farzana R. Zaki
6
CSE 177/ EEE 177
6
Full Wave Rectifier utilizing a transformer
with a center-tapped secondary winding
• Secondary XFR winding is
center-tapped to provide
two equal voltages (vs)
across the two halves of
secondary windings
• During +ve half cycle,
D1 -------- On
D2 --------- Off
• Current (I) will flow
through D1, R and
Back to sec. center tap
coils.
Output voltage, V0 = vs – VD
Farzana R. Zaki
7
CSE 177/ EEE 177
7
• During -ve half cycle,
D2 ---------- On
D1 ---------- Off
• Current (I) will flow
through D2 , R & will
back to sec. tap
Winding.
Output voltage, V0 = vs
– VD
Farzana R. Zaki
8
CSE 177/ EEE 177
8
PIV calculation for full-wave rectifier
(center tapped)
• During positive half cycle,
D1 is ON & D2 is OFF.
Voltage at cathode of
D2 = v0 & anode = -vs
So, reverse voltage across
D2 = v0 + vs
It will reach its maximum
value, when
v0 = VS – VD
vs = VS
So, PIV = 2VS - VD
Farzana R. Zaki
CSE 177/ EEE 177
9
The Bridge Full wave Rectifier
• The circuit is known as
bridge rectifier because
of the similarity of its
configuration to that of
Wheatstone bridge.
• No center-tapped
transformer is required
• Bridge rectifier is
inexpensive as one can
buy a bridge rectifier in
one package.
Farzana R. Zaki
CSE 177/ EEE 177
10
Operation of Bridge Rectifier
• During +ve half cycle,
D1 and D2 are ON
D3 and D4 are Off.
• Current (I) will flow
through D1,R, D2 and
back to XFR.
• O/p voltage,V0= vs2VD
Farzana R. Zaki
CSE 177/ EEE 177
11
• During -ve half cycle,
D1 and D2 are Off
D3 and D4 are On.
• Current (I) will flow
through D3,R, D4 and
back to XFR.
• O/p voltage,V0= vs2VD
Farzana R. Zaki
CSE 177/ EEE 177
12
PIV calculation for Bridge rectifier
• PIV of each diode: Let
determine reverse
voltage across D3 during
positive half cycle of
input.
• Consider loop formed by
D3, R,D2.
vD3(reverse)= v0 + vD2(forward)
so, PIV = VS – 2VD + VD
= VS – VD
Farzana R. Zaki
CSE 177/ EEE 177
13
Advantages of bridge rectifier
• ½ the PIV of the full wave
• Don’t need a center-tapped transformer
• Only need half of the turns in the secondary
winding
Farzana R. Zaki
CSE 177/ EEE 177
14
Filter capacitor
• The pulsating nature of the output voltage
produced by the rectifier circuits, makes it
unsuitable as a dc supply for electronic circuits.
• A simple way to reduce the variation of the
output voltage is to place a capacitor across the
load resistor. This capacitor is known as Filter
Capacitor.
• Filter capacitor serves to reduce substantially
the variations in the rectifier output voltage.
Farzana R. Zaki
CSE 177/ EEE 177
15
Effect of Filter Capacitor
• Operating principle: v1 be a
sinusoid input with a peak value of
VP and assume the diode to be
ideal.
• As v1 goes positive, diode
conducts & capacitor is charged so
that v0 = v1. This situation
continues until v1 reaches to VP.
• Beyond the peak, as v1 decreases,
diode goes to reverse biased and
v0 remains same (constant) as
there is no path for capacitor to
discharge.
• Thus the circuit provides a dc
voltage output equal to the peak of
input sine wave.
Farzana R. Zaki
CSE 177/ EEE 177
16
The rectifier with a filter capacitor- Peak Rectifier
• Assume, diode to be ideal. For an
sine input, the capacitor charges
to the peak of the input, Vp.
• Then, diode operates in cut off
mode and capacitor discharges
through the load R.
• The capacitor discharge will
continue for almost the entire
cycle, until v1 exceeds the
capacitor voltage.
• Then the diode turns on again,
charges the capacitor up to peak
of v1 and the process repeats
itself.
• In order to keep the output voltage
from too much decreasing during
capacitor discharge, one should
select a value for C so that time
constant, RC >> discharging
interval.
Farzana R. Zaki
CSE 177/ EEE 177
17
Expression for ripple voltage (Vr),
conduction angle, iD (avg. and max.)
• During the diode-off
• Conduction angle, ωΔt
interval,
= √(2×Vr /Vp)
v0 = VP e-t/RC
• Average diode current,
At the end of discharge
iDAV= IL{1+π√(2×Vp/Vr)}
interval,
• Maximum diode
VP – Vr = VP e-T/RC
current, iDMAX=
-t/RC = 1 – T/RC
RC>>T,
e
As
IL{1+2π√(2×Vp/Vr)}
VP – Vr = VP (1 – T/RC)
So, Vr = VpT/RC
= Vp/ (f×C×R)
Farzana R. Zaki
CSE 177/ EEE 177
18
Example-3.10
• Consider a peak rectifier fed by 60Hz sinusoid having a
peak value Vp=100V. Let the load resistance R = 10kΏ.
Find the value of the capacitance C that will result in a
peak ripple of 2V. Also calculate the fraction of the cycle
during which the diode is conducting, the average and
peak values of the diode current.
Solution:
C = Vp / (VrfR) = 83.3μF
Conduction angle, ωΔt = √(2×Vr /Vp) = 0.2 rad
IL = 100/10 = 10mA
Average diode current, iDAV= IL{1+π√(2×Vp/Vr)} = 324mA
Maximum diode current, iDMAX= IL{1+2π√(2×Vp/Vr)}=638mA
Farzana R. Zaki
CSE 177/ EEE 177
19
Uses of Peak rectifier circuit
• Used in signal processing systems where
it is required to detect the peak of an input
signal. In such a case, the circuit is
referred to as a peak detector.
• Peak detector is used in the design of a
demodulator for amplitude modulated
(AM) signal.
Farzana R. Zaki
CSE 177/ EEE 177
20