Lecture 9
ECE 331 Electronics Principles I
Grinolds FA08
Rectification
(cont)
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Filter Analysis and Rectifier Circuits in Electronics Principles I - Prof. Hugh R. Grinold, Study notes of Electrical and Electronics Engineering
An in-depth analysis of filter circuits and rectifier designs in the context of electronics principles i. Topics covered include half-wave and full-wave rectification, filter analysis, and diode power dissipation. Students will learn about the behavior of capacitors and diodes, as well as the importance of capacitor values and diode specifications in various rectifier configurations.
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Lecture 9
ECE 331 Electronics Principles IGrinolds FA
Rectification
(cont)
Filter Analysis
i
L
v
o
/R
L
i
D
= C d
v
o
/dt +
i
L
= C d(
v
i
– V
Do
)/dt +
i
L
+^ -
C
v
i
i
D
R
i L
L
i
C
v
o
= C d
v
/dt +i
i
L
ECE 331 Electronics Principles IGrinolds FA
When
v
o
v
i
v
o
= V
omax
e
-t/CR
L
V
omax
= V
I
if V
Do
<< V
I
v
i
= V
I
sin(
ω
t)
Then
V
o
~ V
I
- V
I
e
-T/CR
L
~ V
I
(T/CR
L
e
-T/CR
L
~ 1 – T/CR
L
for T<< CR
L
V
r
= V
I
(T/CR
L
Filter Analysis
The charge lost on C is
supplied during
t.
Q
supplied
= Q
lost
= CV
r
Q
supplied
t =
i
Davg
ECE 331 Electronics Principles IGrinolds FA
V
r
= I
L
/ fC
t ~ 1/
π
f
2V
r
/V
P
i
Davg
= I
L
π √
2V
P
/V
r
i
Dpeak
= I
L
π √
2V
P
/V
r
i
Davg
Half-Wave Rectifier Analysis
V
r
ms
= 12.6 (60 Hz),
R
, V
r
< 0.75V What is the
minimum for C and what are the specs for the diode?
R
C
V
rms
v
o
i
V
o
2 = V
p
= 17.8 volts
I = V
o
/R = 17.8/15 = 1.19 amps
C
> V
P
T / R V
r
ECE 331 Electronics Principles IGrinolds FA
Chap 3 -
C
> V
P
T / R V
r
0.03 F = 30,000 μF
i
Davg
= I (1 +
π√
2V
P
/V
r
) = 26.9 amps
over the time
t =
2V
r
/V
P
π
f ~ 0.8 mS
i
Dmax
i
Davg
= 53.8 amps
A constant voltage drop model
for the diode means: V
o
= V
P
– V
Do
Full Wave Rectifiers
ECE 331 Electronics Principles IGrinolds FA
ideal ideal
ideal
R
D
Full-Wave Rectifiers
ECE 331 Electronics Principles IGrinolds FA
Capacitor discharge time is half that of ahalf-wave rectifier (hwr).Thus for a given V
r
, C is one half of value
of that for a hwr.The peak diode current is ~ half of a hwr.The PIV is 2V
S
– V
Do
Reversing the diodes
negative supply.
Chap 3 -
Rectifier Topology Comparison
Parameter
Half-Wave
Full-Wave
F-W Bridge
C
1
½
½
PIV
2V
P
2V
P
V
P
ECE 331 Electronics Principles IGrinolds FA
PIV
2V
P
2V
P
V
P
i
Dpeak
I
P
I
P
/
I
P
/