Transcript الاسبوع الثاني

Single phase full wave rectifier .
Ripple factor calculation .
Power conversion efficiency .
Filter type .
Examples .
A. One , four
B. Two times
C. 48%
Vin
Vin
Vm
Vs
t
VL
π
t
₂π
T=π
Where Vs: the voltage on the secondary
Vin : the voltage on the primary
Vo or VL : the voltage on the load
𝐼𝑟𝑚𝑠 =
1 𝜋
𝜋 0
=
=
=
=
=
𝐼𝑚 sin 𝜃
𝐼𝑚
𝜋
2
𝐼𝑚
𝜋
2
2
𝐼𝑚
2𝜋
2
𝐼𝑚
2𝜋
2
𝐼𝑚
2𝜋
2 𝑑𝜃
𝜋
2 𝑑𝜃
sin
𝜃
0
1
2
1
2
𝜋 1−cos 2𝜃
𝑑𝜃
0
2
𝜋
(1 −
0
𝜃−
1
2
cos 2𝜃)𝑑𝜃
sin 2𝜃
2
(𝜋 − 0) −
1
2
1
2
sin 2𝜋−sin 0
2
𝑰𝒓𝒎𝒔 =
1
2
=
𝑰𝒎
𝟐
1
2
𝐼𝑚 2
2
∴
𝑰𝒅𝒄 =
𝑰𝒎
𝝅
Ex: A bridge rectifier is used to supply a 500 ohm
load resistor , transformer
with turn ratio N2/N1 = 2/5 is used to couple the bridge to the main supply
find the D.C current in the load if the maximum voltage across the
transformer secondary is 68v , also find the RMS voltage of the main supply .
what is the PIV on the diode .
Solution :Idc=
𝑽𝒅𝒄
𝑹𝑳
∴ 𝑰𝒅𝒄 =
𝒏𝟏
𝒏𝟐
𝒗𝟏
𝑽𝒅𝒄 = 𝟐
𝟒𝟑.𝟑
𝟓𝟎𝟎
𝑽𝒎
𝝅
=𝟐
𝟔𝟖
𝝅
= 𝟒𝟑. 𝟑𝑽
= 𝟖𝟔. 𝟔𝟔𝒎𝑨
𝟓
𝒗𝟏
𝟓∗𝟔𝟖
=
=
∴ 𝒗𝟏 =
= 𝟏𝟕𝟎
𝒗𝟐
𝟐
𝟔𝟖
𝟐
𝑽𝒓𝒎𝒔 = 𝟐 ∗ 𝑽𝒎 = 𝟐 ∗ 𝟏𝟕𝟎 = 𝟐𝟒𝟎 𝑽
PIV = 68 V
Fo=2*Fin = 100HZ
Filters type :
1- capacitive filter :
R
~
FW
C
Rect.
Filter
RL
XL
~
C
RL
Filter
∵ the output of rectifier contain d.c component (wanted) and a.c
component (unwanted )
* The coil will pass d.c component easily to load
XL Hi for a.c XL=2*𝛑 ∗ 𝐟 ∗ 𝐋
XL Lo for d.c
𝟏
Xc Lo for a.c Xc=
𝟐∗𝛑∗𝐟∗𝐜
Xc Hi for d.c
∴ coil pass d.c and imped a.c . Capacitor imped d.c and pass a.c .
D.c component on load may be calculated from
𝑽𝑫𝑪 = 𝑽𝒅𝒄
𝑹𝑳
𝑹+𝑹𝑳
(VDC)- D.C voltage across RL
(Vdc) - d.c voltage out put from rectifier
( R ) - coil resistance .
( RL ) – load resistance .
* When frequency is zero then R and RL become in series forming potential divider.
* R usually much less than RL , therefor most the d.c component appears on load .
 Examples :-
Ex1: H.W rectifier is fed from 24 V/50 Hz . A load of 100Ω resistance is connected on
the rectifier output after using a filtering capacitor of (100µF ) . Calculate :1- time constant during charging assuming that forward resistance of diode is
(30Ω).
2- time constant during discharge .
Solution :
𝝉 = 𝑪. 𝑹
𝝉𝒄𝒉 = 𝑹𝑫 . 𝑪 = 𝟑𝟎 ∗ 𝟏𝟎𝟎 ∗ 𝟏𝟎−𝟔 = 𝟑 𝒎𝒔
𝝉𝒅𝒊𝒔𝒄𝒉 = 𝑹𝑫 . 𝑪 = 𝟏𝟎𝟎 ∗ 𝟏𝟎𝟎 ∗ 𝟏𝟎−𝟔 = 𝟏𝟎 𝒎𝒔
 Ex2:- For the circuit shown :
Calculate the Dc load voltage if 𝑽𝒅𝒄 output from rectifier is 16.4 V a RL=25 Ω
10H
Vin
Solution :
𝑽𝑫𝑪 = 𝑽𝒅𝒄 𝑹
𝑹𝑳
𝑳 +𝑹
𝟕𝟓𝟎
= 𝟏𝟔. 𝟒 𝟕𝟓𝟎+𝟐𝟓 = 𝟏𝟓. 𝟗 𝑽
500µF
16.4V
750Ω
V
‫االختبار ألبعدي‬
𝑽𝟐
𝑽𝟏
=
PIV = 55 V
Fo=2*Fin = 100HZ
𝑵𝟐
𝑵𝟏
𝑽𝟐
𝟐𝟐𝟎
𝟏
=
V2=55 V
𝟒
𝑽𝒎
𝟓𝟓
𝑽𝒅𝒄 = 𝟐
=𝟐
= 𝟑𝟓. 𝟎𝟑 𝑽
𝝅
𝝅
𝑽𝒎
𝟐𝟖.𝟑
𝑽𝒅𝒄 = 𝟐
=𝟐
= 𝟏𝟖. 𝟎𝟐 𝑽
𝝅
𝝅
𝑽𝒓𝒎𝒔 =Vm/1. 𝟒 =28.3/1.4=20.2 V
∴ 𝑰𝒅𝒄 =
𝟐𝟎.𝟐
𝟏𝟎𝟎
= 𝟎. 𝟐𝒎𝑨
PIV = 28.3 V
Fo=2*Fin = 100HZ