Current in a Single-phase Full-wave Rectifier
Consider the following single-phase full-wave rectifier circuit shown in Figure 1. Figure
2 shows an equivalent model of the rectifier shown in Figure 1. The equivalent model
can be utilized to analyze output voltage and current.
Figure 1: A single-phase full-wave rectifier
vAN
N
vBN
-
+
vo
io
L
R
VC
Figure 2: An equivalent model of a single-phase full-wave rectifier
v i = 2V1 sin(ωt ) , v AN =
2V1
2V1
sin(ωt ) and v BN = −
sin(ωt )
n
n
First thing that we have to determine is: whether the output current is continuous or
discontinuous. To do this we have to analyze the output current during the first halfcycle (for a half-wave rectifier we analyzed the output current during the first cycle) i.e.,
when the green thyristor in the equivalent model is conducting.
i=
2V
 R  V
sin(ωt − φ ) + A exp − t  − C
z
 L  R
where z =
 ωL 
2
R 2 + (ωL) and φ = tan −1 

 R 
At ωt = α , i = 0
0=
2V
 R α V
sin(α − φ ) + A exp − ⋅  − C
z
 L ω R
2V
 R α V
A exp − ⋅  = C −
sin(α − φ )
z
 L ω R
V

2V
R α
A= C −
sin(α − φ ) exp ⋅ 
z
 R

 L ω
To determine whether the output current is continuous or discontinuous, check the value
of the first half-cycle current at ωt = π + α i.e., when the thyristor (red in the equivalent
model) in the next branch is fired.
Find: i =
2V
 R π + α  VC
sin(π + α − φ ) + A exp − ⋅
−
z
ω  R
 L
If i ≤ 0, then the current is discontinuous.
If i > 0, then the current is continuous.
If the current becomes zero before the conduction of the thyristor (red) in the next halfcycle, we can conclude that the wave shape of all subsequent halves will be identical to
the first half. However, if the current is greater than zero when the thyristor (red) is fired
at ωt = π + α , it will take several half-cycles for the current to settle to a steady-state
wave shape.
If the current is continuous then each half-cycle current at steady-state will begin with the
same initial positive value (say, k).
i=
2V
 R  V
sin(ωt − φ ) + A exp − t  − C
z
 L  R
At ωt = α , i = k
........ (1)
k=
2V
 R α V
sin(α − φ ) + A exp − ⋅  − C
z
 L ω R


2V
V
R α
A = k + C −
sin(α − φ ) exp ⋅  ......... (2)
R
z
 L ω


For a continuous current, it again becomes equal to k at ωt = π + α i.e.,
k=
2V
 R π + α  VC
sin(π + α − φ ) + A exp − ⋅
........ (3)
−
z
ω  R
 L
Solve Equations (2) and (3) for A and k.
k=
2V
2V
 R π V
sin(π + α − φ ) −
sin(α − φ ) exp − ⋅  + C
z
z
 L ω R
 R π
1 − exp − ⋅ 
 L ω
  R π 
exp − L ⋅ ω  − 1
 
 
v
2V
vAN
vBN
t
0
iG1
t
iG2
t
iO
t
vO
VC
t
0
Figure 3: Discontinuous Current
v
2V
vAN
vBN
t
0
iG1
t
iG2
t
iO
t
vO
VC
t
0
Figure 4: Continuous Current
Case 1
. For the single-phase controlled rectifier shown in Figure 5 , L = 10 mH, R = 3 Ω,
Vc = 48 V and α = 400. (a) Determine whether the load current is continuous or
discontinuous.
v i = 240 2 sin 120π t V
+
1.4 : 1
R
vi
io
vo
L
+
Vc
-
-
Figure 5: A single-phase controlled rectifier.
ω := 377
V :=
X := ω ⋅L
240
1.4
z :=
Vc := 48
2
R + ( X)
−3
R := 3
L := 10 ⋅10
 X

R
2
φ := atan
φ ⋅57.3 = 51.493
First half-cycle calculations :
α :=
40
 Vc

2 ⋅V
R α
−
⋅sin( α − φ )  ⋅exp ⋅ 
z
R

L ω
A := 
φ = 0.899
57.3
R
= 300
L
A = 45.357
Vc
R
= 16
2 ⋅V
= 50.319
z
Vc 
 2 ⋅V
 R θ
⋅sin( θ − φ ) −
 + A⋅exp − ⋅ 
R
 z
 L ω
i( θ ) := 
i( π + α ) = −3.839
Therefore the current is discontinuous. Now you have to find out when the
current drops to zero.
θ := 3.14
β := root ( i( θ ) , θ )
β = 3.764
β ⋅57.3 = 215.669
Case 2
. For the single-phase controlled rectifier shown in Figure 6 , L = 10 mH, R = 3 Ω,
Vc = 24 V and α = 400. (a) Determine whether the load current is continuous or
discontinuous.
v i = 240 2 sin 120π t V
+
1.4 : 1
R
vi
io
vo
L
+
Vc
-
-
Figure 6: A single-phase controlled rectifier.
ω := 377
X := ω ⋅L
V :=
240
1.4
z :=
Vc := 24
2
R + ( X)
2
R := 3
 X

R
φ := atan
−3
L := 10 ⋅10
φ ⋅57.3 = 51.493
First half-cycle calculations :
α :=
40
57.3
A = 31.414
 Vc

2 ⋅V
R α
−
⋅sin( α − φ )  ⋅exp ⋅ 
R
z


L ω
φ = 0.899
A := 
Vc
R
= 300
L
R
=8
Vc 
 2 ⋅V
 R θ
⋅sin( θ − φ ) −
 + A⋅exp − ⋅ 
z
R


 L ω
i( θ ) := 
2 ⋅V
= 50.319
z
i( π + α ) = 3.505
Therefore the current is continuous.
After several half-cycles at steady-state, each half-cylce current starts with
an initial value of k
2 ⋅V
⋅sin( π + α − φ ) −
z
k :=
2 ⋅V
 R π  Vc   R π  
⋅sin( α − φ ) ⋅exp − ⋅  +
⋅ exp − ⋅  − 1 
z
 L ω R   L ω 
 R π
1 − exp − ⋅ 
 L ω
k = 3.818
At steady-state, the constant A has a different value:


A :=  k +
Vc
R
A = 38.068
−

2V
R α
⋅sin( α − φ )  ⋅exp ⋅ 
z

L ω