Dr.

/ Abdelhady Ghanem

Power Electronics (1) – ELE221

Dr./ Abdelhady Ghanem

Lec. №: (2)
 Dr./ Abdelhady Ghanem

Chapter (3)

Single Phase Uncontrolled Half Wave Rectifiers

 Dr./ Abdelhady Ghanem

Introduction
A rectifier is an electrical device that converts alternating current (AC) to direct
current (DC), which flows in only one direction. The process is known as rectification.

There are many applications for rectifiers. Some of them are: variable speed dc drives, battery
chargers, DC power supplies and Power supply for a specific application like electroplating.
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(1) Resistive Load +
Vd
-

➢ A basic half-wave rectifier with a resistive load is shown in Fig. a. i +

The source is AC, and the objective is to create a load voltage that vs = V m sin (t ) R Vo
has a nonzero dc component. The diode is a basic electronic
switch that allows current in one direction only. -

Vm
(a)
➢ For the positive half-cycle of the source in this circuit, the diode is
on (forward-biased). Considering the diode to be ideal, the Vs
ωt
 2
voltage across a forward-biased diode is zero and the current is
positive. -Vm

Vm

Vo
➢ For the negative half-cycle of the source, the diode is reverse-  2 ωt
biased, making the current zero. The voltage across the reverse-
ωt
biased diode is the source voltage, which has a negative value. Vd  2
-Vm
(b)
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

Vd
(1) Resistive Load 𝑣𝑠 = 𝑉𝑚 sin(𝜔𝑡) + -

The dc component Vo of the output voltage is the i +

average value of a half-wave rectified sinusoid vs = V m sin (t ) R Vo
1 𝜋 𝑉𝑚 -
𝑉𝑑𝑐 = 𝑉𝑜 = 𝑉𝑎𝑣𝑔 = න 𝑉𝑚 sin(𝜔𝑡) 𝑑(𝜔𝑡) =
2𝜋 0 𝜋 (a)

Vm
The dc component of the current for the purely resistive load is
𝑉𝑜 𝑉𝑚 Vs
 2
ωt
𝐼𝑜 = =
𝑅 𝜋𝑅 -Vm

Vm

The rms values of Vo and Io can be written as Vo

 2 ωt
𝜋
1 2
𝑉𝑚 𝑉𝑚 ωt
𝑉𝑟𝑚𝑠 = න 𝑉𝑚 sin(𝜔𝑡) 𝑑(𝜔𝑡) = 𝐼𝑟𝑚𝑠 = Vd  2
2𝜋 0 2 2𝑅 -Vm
(b)
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(1) Resistive Load +
Vd
-
The Average output dc power is +
i
2 𝑉 2
𝑑𝑐 𝑉𝑚
2
vs = V m sin (t ) R Vo
𝑃𝑑𝑐 = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 𝐼𝑑𝑐 𝑅= = 2
𝑅 𝜋 𝑅 -
The rms output dc power is
2 (a)
2 𝑅 =
𝑉𝑟𝑚𝑠 𝑉𝑚2 Vm
𝑃𝑎𝑐 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 =
𝑅 4𝑅
Vs
ωt
Example: For the shown half-wave rectifier, the source is a sinusoid  2
of 120 Vrms at a frequency of 50 Hz. The load resistor is 5 Ω. -V m

Determine (a) the average load current, (b) the dc and ac power V m

absorbed by the load and (c) the power factor of the circuit. V o

 2 ωt
(a) the average load current 𝑉𝑚 = 120 2 = 169.7 𝑉
ωt
Vd  2
𝑉𝑜 𝑉𝑚 120 2
𝐼𝑜 = = = = 10.8 𝐴 -Vm
𝑅 𝜋𝑅 5𝜋 (b)
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(b) the dc and ac power absorbed by the load
2 2 2
2 𝑉𝑑𝑐 𝑉𝑚 120 2
𝑃𝑑𝑐 = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 𝐼𝑑𝑐 𝑅= = 2 = = 583.61 𝑊𝑎𝑡𝑡𝑠
𝑅 𝜋 𝑅 5𝜋 2

1 𝜋 2
𝑉𝑚 120 2
𝑉𝑟𝑚𝑠 = න 𝑉 sin(𝜔𝑡) 𝑑(𝜔𝑡) = = = 84.85 𝑉
2𝜋 0 𝑚 2 2

2 2
𝑉𝑟𝑚𝑠 84.85
𝑃𝑎𝑐 = = = 1440 𝑊𝑎𝑡𝑡𝑠
𝑅 5

(c) the power factor of the circuit

𝑉𝑚 120 2 𝑃𝑎𝑐 𝑃𝑎𝑐 1440

𝐼𝑟𝑚𝑠 = = = 16.97 𝐴 𝑃𝐹 = = = = 0.707
2𝑅 10 𝑆 𝑉𝑟𝑚𝑠−𝑠𝑢𝑝 𝐼𝑟𝑚𝑠 120 × 16.97
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

Vs
(2) R-L Load io

Vd
Industrial loads typically contain + - ωt
+  2
inductance as well as resistance. As the i +
source voltage goes through zero, R VR
vs = V m sin (t ) -

Vo
becoming positive in the circuit of Fig. a, +
L VL Vo
the diode becomes forward-biased. The - ωt
Kirchhoff voltage law equation that
-  2
(a)
describes the current in the circuit for
the forward-biased ideal diode is: 
VR
ωt
 2
𝑑𝑖(𝑡)
𝑉𝑚 sin(𝜔𝑡) = 𝑅𝑖 𝑡 + 𝐿 → (1)
𝑑𝑡  
The dc component of the output voltage is VL
ωt
2
𝑉𝑚 𝛽 𝑉𝑚
𝑉𝑑𝑐 = න sin(𝜔𝑡) 𝑑(𝜔𝑡) = (1 − cos 𝛽)
2𝜋 0 2𝜋 
Vd
ωt
 2
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(2) R-L Load The dc component of the output current is
𝑉𝑚
𝐼𝑑𝑐 = (1 − cos 𝛽)
2𝜋𝑅

The solution of equation (1) can be obtained by expressing the current as the sum of the forced
response and the natural response:
𝑖 𝑡 = 𝑖𝑓 𝑡 + 𝑖𝑛 𝑡
Vd
+ -
The forced response for this circuit is the current that exists after the +
i
natural response has decayed to zero. In this case, the forced R
+
VR
response is the steady-state sinusoidal current that would exist in the vs = V m sin (t ) Vo -
+
circuit if the diode were not present. This steady-state current can be L VL
-
found from phasor analysis, resulting in: -
(a)
𝑉𝑚 sin(𝜔𝑡) 𝑉𝑚
𝑖𝑓 𝑡 = = sin(𝜔𝑡 − 𝜃)
𝑍 𝑍
Where:
𝜔𝐿
𝑍= 𝑅2 + (𝜔𝐿)2 𝜃 = tan−1
𝑅
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(2) R-L Load
The natural response is the transient that occurs when the load is energized. It is the solution
to the homogeneous differential equation for the circuit without the source or diode.
𝑑𝑖(𝑡)
𝑅𝑖 𝑡 + 𝐿 =0
𝑑𝑡
For this first-order circuit, the natural response has the form: 𝑖𝑛 𝑡 = 𝐴𝑒 −𝑡Τ𝜏
Where: A = Constant 𝐿
𝜏=
𝑅
Adding the forced and natural responses gets the complete solution:

𝑉𝑚
𝑖 𝑡 = 𝑖𝑓 𝑡 + 𝑖𝑛 𝑡 = sin 𝜔𝑡 − 𝜃 + 𝐴𝑒 −𝑡Τ𝜏 → (2)
𝑍
The constant A is evaluated by using the initial condition for current: 𝑡 = 0 → 𝑖 𝜔𝑡 = 0
Using the initial condition and equation (2) to evaluate A yields:
𝑉𝑚 𝑉𝑚 𝑉𝑚
𝑖 0 =0= sin 0 − 𝜃 + 𝐴𝑒 0 𝐴 = − sin 0 − 𝜃 = sin 𝜃
𝑍 𝑍 𝑍
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(2) R-L Load Substituting for A in equation (2) gives:

𝑉𝑚 Τ𝜏 𝑉𝑚 𝑉𝑚
𝑖 𝑡 = sin 𝜔𝑡 − 𝜃 + 𝐴𝑒 −𝑡 = sin 𝜔𝑡 − 𝜃 + sin 𝜃 𝑒 −𝑡Τ𝜏
𝑍 𝑍 𝑍
𝑉𝑚 𝑉𝑚
∴ 𝑖 𝜔𝑡 = sin 𝜔𝑡 − 𝜃 + sin 𝜃 𝑒 −𝑡 Τ𝜏
= sin 𝜔𝑡 − 𝜃 + sin 𝜃 𝑒 −𝜔𝑡Τ𝜔𝜏 → (3)
𝑍 𝑍
The point when the current reaches zero in Eq. (3) occurs when the diode turns off. The first
positive value of 𝜔t in Eq. (3) that results in zero current is called the extinction angle 𝛽.

To find 𝛽, substitute 𝜔t= 𝛽 in Eq. (3) 𝑉𝑚

𝑖 𝛽 =0= sin 𝛽 − 𝜃 + sin 𝜃 𝑒 −𝛽Τ𝜔𝜏
𝑍
Which reduces to sin 𝛽 − 𝜃 + sin 𝜃 𝑒 −𝛽Τ𝜔𝜏 = 0

There is no closed-form solution for 𝛽, and some numerical method is required.

 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(2) R-L Load
To summarize, the current in the half-wave rectifier circuit with R-L load is expressed as:

𝑉𝑚
sin 𝜔𝑡 − 𝜃 + sin 𝜃 𝑒 −𝜔𝑡Τ𝜔𝜏 𝑓𝑜𝑟 0 ≤ 𝜔𝑡 ≤ 𝛽
𝑍
𝑖 𝜔𝑡 =

0 𝑓𝑜𝑟 𝛽 ≤ 𝜔𝑡 ≤ 2𝜋
𝜔𝐿 𝐿
𝑍= 𝑅2 + (𝜔𝐿)2 −1
𝜃 = tan 𝜏=
𝑅 𝑅
1 𝛽
The dc component of the output current is 𝐼𝑜 = න 𝑖(𝜔𝑡) 𝑑(𝜔𝑡)
2𝜋 0
Or, It can be found as:
𝑉𝑚 𝛽 𝑉𝑚 𝑉𝑚
𝑉𝑑𝑐 = න sin(𝜔𝑡) 𝑑(𝜔𝑡) = (1 − cos 𝛽) 𝐼𝑑𝑐 = 𝐼𝑜 = (1 − cos 𝛽)
2𝜋 0 2𝜋 2𝜋𝑅
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(2) R-L Load The rms value of Io can be written as:

1 2𝜋 2 1 𝛽2
𝐼𝑟𝑚𝑠 = න 𝑖 𝜔𝑡 𝑑(𝜔𝑡) = න 𝑖 𝜔𝑡 𝑑(𝜔𝑡)
2𝜋 0 2𝜋 0

Or it can be written as:

1 𝛽 2
𝑉𝑚2 1
𝑉𝑟𝑚𝑠 = න 𝑉𝑚 sin(𝜔𝑡) 𝑑(𝜔𝑡) = 𝛽 − sin 2𝛽
2𝜋 0 4𝜋 2

𝑉𝑟𝑚𝑠 𝑉𝑟𝑚𝑠 1 𝑉𝑚2 1

𝐼𝑟𝑚𝑠 = = = 𝛽 − sin 2𝛽
𝑍 𝑅2 + (𝜔𝐿)2 𝑅 2 + (𝜔𝐿)2 4𝜋 2
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

Example: For the R-L half-wave rectifier, R=100Ω, L=0.1 H, 𝜔=377 rad/s, and Vm=100 V.
Determine (a) an expression for the current in this circuit, (b) the average current, (c) the rms
current, (d) the power absorbed by the R-L load, and (e) the power factor.

𝑍= 𝑅 2 + (𝜔𝐿)2 = 1002 +(377 × 0.1)2 = 106.87Ω

𝜔𝐿 377 × 0.1
𝜔𝐿 377 × 0.1 𝜔𝜏 = = = 0.377 𝑟𝑎𝑑
𝜃 = tan −1
= tan−1
= 20.7° = 0.36 𝑟𝑎𝑑 𝑅 100
𝑅 100

(a) an expression for the current in this circuit

𝑓𝑜𝑟 0 ≤ 𝜔𝑡 ≤ 𝛽
𝑉𝑚 Τ𝜔𝜏 100
𝑖 𝜔𝑡 = sin 𝜔𝑡 − 𝜃 + sin 𝜃 𝑒 −𝜔𝑡
= sin 𝜔𝑡 − 0.36 + sin 0.36 𝑒 −𝜔𝑡Τ0.377
𝑍 106.87
𝑎𝑡 𝜔𝑡 = 𝛽 sin 𝛽 − 𝜃 + sin 𝜃 𝑒 −𝛽Τ𝜔𝜏 = 0 → sin 𝛽 − 0.36 + sin 0.36 𝑒 −𝛽Τ0.377 = 0

∴ 𝛽 = 3.501 𝑟𝑎𝑑 = 201°

 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(b) the average current
𝑉𝑚 100 𝑉𝑑𝑐 30.82
𝑉𝑑𝑐 = 1 − cos 𝛽 = 1 − cos 3.5 = 30.82 𝑉 𝐼𝑑𝑐 = 𝐼𝑜 = = = 0.308 𝐴
2𝜋 2𝜋 𝑅 100
(c) the rms current

1 𝑉𝑚2 1 1 1002 1
𝐼𝑟𝑚𝑠 = 𝛽 − sin 2𝛽 = 3.5 − sin 7 = 0.457 𝐴
𝑅2 + (𝜔𝐿)2 4𝜋 2 106.87 4𝜋 2

2
(d) the power absorbed by the R-L load 𝑃 = 𝐼𝑟𝑚𝑠 𝑅 = 0.4572 × 100 = 21 𝑊

(e) the power factor

𝑃 𝑃 21
𝑃𝐹 = = = = 0.65
𝑆 𝑉𝑟𝑚𝑠−𝑠𝑢𝑝 𝐼𝑟𝑚𝑠 100ൗ × 0.457 Note that the power factor is not cosθ.
2
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(3) R-L Load with Freewheeling Diode
D1
+
iD1 io
iD2 R
➢ A freewheeling diode D2, can be connected across an R-L load vs = V m sin (t ) D2 Vo
as shown in Fig. a.
L
➢ Both diodes cannot be forward-biased at the same time. Diode
-
D1 will be ON when the source is positive, and diode D2 will (a)
be ON when the source is negative. + +
io io
For a positive source voltage, R R
➢ D1 is on. Vo=Vs Vo=0

➢ D2 is off. L L
➢ The equivalent circuit is the same as that of Fig. b. - -
➢ The voltage across the R-L load is the same as the source. (b) (c)

For a negative source voltage,

➢ D1 is off.
➢ D2 is on.
➢ The equivalent circuit is the same at that of Fig. c.
➢ The voltage across the R-L load is zero
 Dr./ Abdelhady Ghanem

Single Phase Uncontrolled Half Wave Rectifiers

(3) R-L Load with Freewheeling Diode
D1
+
iD1 io
iD2 R
➢ Since the voltage across the R-L load is the same as the vs = V m sin (t ) D2 Vo
source voltage when the source is positive and is zero L
when the source is negative, the load voltage is a half- -
(a)
wave rectified sine wave. Steady-state load, source, and Vo + +
diode currents are shown in the figure. io
io io
R R
Vo=Vs Vo=0
L L
Example: Determine the average load voltage and ωt
current for the circuit, where R=2 Ω and L=25mH, Vm is  -
2
(b)
-
(c)
100 V, and the frequency is 50 Hz.

𝑉𝑚 100 iD1
𝑉𝑜 = = = 31.8 𝑉 ωt
𝜋 𝜋  2
𝑉𝑜 31.8
𝐼𝑜 = = = 15.9 𝐴
𝑅 2 iD2
ωt
 2
Dr./ Abdelhady Ghanem