Chapter (2)
Signal Processing Circuits
Full-wave rectifier analysis
(1)
The bridge rectifier circuit is shown in this this Fig.
• V on T ¾¾ is an ac(- ve and +ve) terminals
• At V ad i
¾¾ abcd
• At V da i ¾¾ dbca
(2)
• i, through R
L
¾¾ in the same direction

Chapter (1)
Signal Processing Circuits
The dc component is twice as large as in half-wave rectifier
:
I dc
=
π
2 V m
R
L
=
2
π
I m
The disadvantageous of the bridge circuit is:
• Four diodes are needed
• Their power-dissipating voltage drops are always in series with the load.
• This will reduce the power delivered to the lead

Chapter (1)
Signal Processing Circuits
The circuit shown in this
Fig. uses
• expensive transformer
¾¾ v
2 is 180 out of phase o with v
1
• two diodes only
• higher operating efficiency
• When v
1
+ ve, i
1
¾¾ D
1
, no current flows through D
2
• When v
2
+ ve, i
2
¾¾ D
2
, no current flows through D
1
• The current in R ( I = i
1
+ i
2
)
∴ I dc
=2I m
/ π

Chapter (1)
Signal Processing Circuits
Filters
•
Rectification means direct current (dc)
• The output current of the half-wave and full-wave rectifier contain large ac components
• There is ripple voltage across R
L i t half-wave rectifier full-wave rectifier

Chapter (1)
Signal Processing Circuits
Capacitor Filter
How?
• The capacitor is shunted ¾¾ between R
L and diode
• The charge store in the capacitor ¾¾ when D conducting
¾¾ during the positive period
• The charge release from the capacitor ¾¾ when D nonconducting
¾¾ during the negative period

Chapter (1)
Signal Processing Circuits
At the steady state
.
•
• At t = 0 ¾¾ v i c
=
At t = t
• At t = t
2
1
Cdv/dt
¾¾ s
¾¾ v
= 0  D ¾¾ off  i
V s
L
> V
= V
C m
= V
L
 D ¾¾ off  i
D
 D
D
¾¾
= 0 Â v s
= 0 Â v s on sin ω t
¾¾ during ¾¾ t
1
< V
C
V
Â
= i
D
> t > t
2
= V
• During the discharging period,
¾¾ t
2
> t > t
3
¾¾ exponentially
• At t = t
3
¾¾ vs < VC = V L (C-dis)
¾¾ the cycle repeats
L
V
< V
C
= V
L
(C-dis) rises to satisfy
L
V
2 e
(C-charges)
(C-dis)
C
= V
− ( t − t
L
2
) decays
/ R
L
C

Chapter (1)
Signal Processing Circuits
V
L
From this equation :
= V
2 e
( t
t
2
) / R
L
C
• T = RC is the periodic time ¾¾ deepens on the values of R and C
• i
L never goes to zero it α v
L
• The dc components ¾¾ I dc and V dc are larger as compared to the half-wave rectifier alone
• The ripple voltage ¾¾ V r use of the capacitor
¾¾ voltage conducting diode ¾¾ reduced by
• V r
¾¾ very small when RC > T (the period of the supply voltage ¾¾ the decay in
V
C
= V
L small ¾¾ approximately straight line

Chapter (1)
Signal Processing Circuits
Capacitor Filter – Approximate Analysis
ASSUMPTION
• RC large ¾¾ v r small ¾¾ t
2
– t
1 small ¾¾ v c
≅ constant
All the current ¾¾ supplied by the capacitor
• The charge transferred to R
L
∆ q v r
=
= I dc
T
I dc
T
C
=
= C ∆ v c
I dc fC
=
= Cv r
V dc fR
L
C

Chapter (1)
Signal Processing Circuits

Chapter (1)
Signal Processing Circuits
HOMEWORK
1.
Redraw the circuit in Ex. 8 p. 82 when a capacitor of 1F is shunted between the diode and R
L
2. Calculate the ripple voltage of this circuit
3.Derive an equation to solve the ripple voltage for the full-wave rectifier circuit
4. Practice Problem 3-8 . Drawing and calculations