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Input
Filter
Regulation
Load
Experiment # 8
EE 312 & 352
Introductory Electronics Laboratory
November 1, 1999
Objective:
• Design, construct, and test a dc
power supply. The power
supply includes a transformer,
diodes, and a filter. All were
investigated in previous
experiments. This the major
design project in the course.
DC-VOLTAGE POWER SUPPLY OPERATION
Alternator
Transformer
or
Power Company
Rectifier
Filter
Regulator Load
v
t
Input: 120
VAC RMS,
60 Hz
Steps the
input
voltage up
and down
as needed
Convert the
ac to a
pulsing dc
Low-pass
type for
removing
voltage
variation
Keeps
voltage
constant
regardless of
load change
Power Company Transformer
Rectifier
Filter
v
167
118 Vrms
16.7 ms or 60Hz
t
Regulator Load
Alternator
Transformer
or
Power Company
v
167
Rectifier
Filter
Regulator Load
v
118 Vrms
25.5
18 Vrms
16.7 ms or 60Hz
t
16.7 ms or 60Hz
t
Alternator
Transformer
or
Power Company
Rectifier
Filter
v
Regulator Load
60Hz
RLoad
t
half-wave rectifier
RLoad
v
RLoad
120Hz
t
full-wave rectifier
Alternator
Transformer
or
Power Company
v
25.5
Rectifier
~ 0.6V
18 Vrms
Filter
v
24.9
Regulator Load
60 Hz
12.45 Vrms
16.7 ms or 60Hz
t
t
half-wave rectifier
t
Alternator
Transformer
or
Power Company
Input
Rectifier
Filter
Filter
Regulator Load
Regulation
Load
Input
Filter
Regulation
Load
C

v
v
e
t
RC
ripple
Vmax
t
without C
t
t
with C
t
?
Vmax
?
v
V ripple
VDC
0
t
t
t
V  V
max 
V
max 

e
t
RC
t
V  V max 
RC
V
VDC  V max 
2
VDC  V
max 
(1 
t
)
2 RC
V
max 
V
V max  e
max 
(1 
t
)
RC

t
RC
Input
Vinput
Filter
R
Regulation
Load
C
One or more filter sections are usually used to reduce the amplitude
of the ripple. The filter used in this experiment is a single-section
low-pass RC filter.
V across R < 25% of Vinput
1
1
 R
C 3
usually tolerable
Ripple Reduction Factor:
Vin
Vout , ripple
A 

Vin, ripple
Vout
1
2
1  (2fCR)
Alternator
Transformer
or
Power Company
Rectifier
Filter
Regulator Load
You are asked to design and built a
12 V dc regulated power supply.
In this experiment you use a Zener
diode that breaks down at 12 V.
IDC
ILoad
Iz
12Volts
breakdown
voltage
ID
Forward
Characteristics
VD
Reverse
Characteristics
Iz
~
I
Max
IDC
Rs
Voltage Divider
Rs
ILoad
IDC
ILoad
Vinput
RLoad
Rz
Iz
=
Rz
RLoad
Iz
ID
1
rz
VD
1
Rz
~
I
Max
Vz, DC
Rz

Vinput Rs  Rz
Vz, ripple
rz

Vinput , ripple Rs  rz
Summary
Alternator
Transformer
or
Power Company
v
t
Rectifier
Filter
Regulator Load
Procedures:
I- Design, construct, and test a
dc power supply.
II- Simulate dc power supply.
I
Components:
•
•
•
•
•
•
•
18 Volts Transformer
Variac
Bridge rectifier (use 1 diode)
Zener diode (Vbd=12Vdc, 1N4742)
12 Watt, 1000 Ohm, variable resistor
2 Resistor Substitution Boxes
50, 100, and 200 UF Fixed Capacitors
1- Design, construct, test a dc power supply
R?
R?
ILoad = 50mA
VLoad =12 V dc
C?
Peak to peak
ripple voltage
from the input
stage < 20%
Suggestion:
C?
R?
Peak to peak
ripple voltage at
the load <= 1%
Zener diode current = 21 mA
(At this current, the Zener resistance is 9 ohms)
R?
R?
ILoad = 50mA
VLoad =12 V dc
C?
C?
Peak to peak ripple
voltage from the input
stage < 20%
RLoad
VLoad
12V


 240
ILoad 50 mA
RLoad?
Peak to peak ripple
voltage at the load <=
1%
IDC
IDC
R?
ILoad = 50mA
R?
C?
Peak to peak ripple
voltage from the input
stage < 20%
C?
VLoad =12 V dc
RL  240
Iz=21mA
Peak to peak ripple
voltage at the load <=
1%
IDC  ILoad  Iz  50 mA  21mA  71mA
IDC=71mA
ILoad = 50mA
R?
R?
CI?
C?
VLoad =12 V dc
RL  240
Iz=21mA
Peak to peak ripple
voltage from the
input stage < 20%
Vmax
v
VDC
R
IDC
VDC
VMax t  IDC
CI 

0
V
VDC
24.9
0.071
CI 

 264  250
4.98 60  22.4
Peak to peak ripple
voltage at the load <=
1%
V
ripple
t
t
t
V max  e

t
RC
20% ripple
Vmax = 24.9 v
Vripple < 24.9 x 0.2=4.98 V
VDC=22.4 V
Vmin=19.92
0
t
t
t
V  V max  V max  e
t
V  V max 
RC
V max t
C

V
R

t
RC
t
 V max  V max  (1 
)
RC
IDC=71mA
ILoad = 50mA
RF?
Rs?
VLoad =12 V dc
CI=250uF
CF?
VDC=0.6V+IDC(RF+RS)+Vz
22.4=0.6+.071(RF+RS)+12
RF+RS = 138.03 ohms
RL  240
Iz=21mA
Peak to peak ripple
voltage at the load <=
1%
IDC=71mA
ILoad = 50mA
RF?
Rs?
VLoad =12 V dc
CF?
CI=250uF
Iz=21mA
6.6% ripple
20% ripple
A 
RL  240
Peak to peak ripple
voltage at the load <=
1%
1
1  ( 2fCR ) 2
1
A 
3
RF  CF  0.0075
CF  100F
RF  75
&
RF  Rs  138
Rs  63
IDC=71mA
ILoad = 50mA
RF=75
Rs=63
VLoad =12 V dc
RL  240
CI=250uF
CF=250uF
Iz=21mA
Peak to peak ripple
voltage at the load <=
1%
2% ripple
0.1% ripple
Vz , ripple
rz
18
.
1



Vinput , ripple Rs  rz 63  18
.
36
2- Simulation of Part I
Simulate the dc power supply!
Show output wave form at points 1, 2, and 3.
Comnpare these waveforms with your experiment.
1
2
3
RF=105
?
118.4 V
Rs=33
ILoad = 50mA
VLoad =12 V dc
CI=250uF
Variac
RL  240
CF=250uF
variable
Iz=21mA
18.0 V
Transformer
Report is required
for EE 312 & 352
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