reference notes455647_1_EE460-Project-131.pdfKing Fahd Un.docx

reference notes/455647_1_EE460-Project-131.pdf
King Fahd University of Petroleum and Minerals
Department of Electrical Engineering
EE Power Electronics Project
Design of a DC Chopper
I. Design of an AC/DC converter with the following the
specifications:
II. Design of step-down DC chopper with the following
specifications:
available is an ac with 230 V

(rms).
V.
-to-
-to-
III. Calculation for both circuits:
(a) Determine the values of Le and Ce for the output LC-filter.
(b) Determine the (peak and rms) voltage ratings and the
(average, rms, and the peak) current for
all components and devices.
(c) Verify your design calculation by using Pspice simulation.
Design AC/DC
Circuit
Design DC-DC
Chopper Circuit
Output Load
The project will be due on Sunday December 22, 2013.

reference notes/455647_2_DC-20Converters-Design (1).pdf
....-ju"ncv
O.
214 Chapter 5 Dc-Dc Converters
Example 5.10
A buck converter is shown in Figure 5.29. The input voltage is
V, == 110 V, the average load
age is Va == 60 V, and the average load current is la == 20 A.
The chopping u
1 == 20 kHz. The peak-to-peak ripples are 2.5% for load
voltage, 5% for load current, and
for filter Le current. (a) Determine the values of L" L, and Ceo
Use PSpice (b) to verify the
suits by plotting the instantaneous capacitor voltage vc, and
instantaneous load current iL ;
(c) to calculate the Fourier coefficients and the input current is.
The SPICE model pax'ameters
the transistor are IS == 6.734f, BF = 416.4, BR == 0.7371, CJC
== 3.638P, CJE::
TR == 239.5N, TF = 30L2P, and that ofthe diode are IS :: 2.2E-
15, BV = 1800V, IT ==

Solution
V, = 110 V, va = 60 V, I. == 20 A.
ay: == 0.025 x Va = 0.025 x 60 = 1.5 V
Va 60
R==-=-=311
10 20
From Eq. (5.48),
Va 60
k = - = - = 05455
V, 110 .
From Eq. (5.49),
Is = kla = 0.5455 x 20 == 10.91 A
alL = 0.05 x I. :: 0.05 x 20 == 1 A

M = 0.1 x 10 == 0.1 x 20 == 2 A
8. From Eq. (5.51), we get the value of L.:
VaWs - Va) 60 X (110 - 60)
Le = MIV, = 2 x 20 kHz x 110 = 681.82 ~H
From Eq. (5.53) we get the value of Ce:
2c == ,11
e ,lV, X 81 1.5 x 8 X 20 kHz == 8.33 ~F
L4
+
+
Vs 110 V
FIGURE 5.29
o~-----------+----------~--------~Buck converter.
5.12 Chopper Circuit Design 215

Vs
L
8
v, OV
O~----------------------------*-------~~------~
(a) Circuit
Vgj
2ov~______________1~________-L____--'
o 27.28 IlS SOIlS
(b) Control voltage
FIGURE 5.30
Buck chopper for PSpice simulation.
Assuming a linear rise of load current iL during the time from t
= 0 to tl = kT, we can
write approximately

Ah Ah
L- = L-- = AVe
t} kT
which gives the approximate value of L:
kTAVc kAVe
L=~= AIJ
(5.128)
0.5454 X 1.5 = 40.91 H
1 X 20kHz f.L
b. k = 0.5455, / = 20 kHz, T = 11/ = 50 f.LS, and ton = k X T =
27.28 f.LS. The buck
chopper for PSpice simulation is shown in Figure 5.30a. The
control voltage Vg is
shown in Figure 5.30b. The list of the circuit file is as follows:
Example 5.10 Buck Converter
vs 1 0 IX: 110V

VY 1 2 IX: OV ; Voltage source to measure input current
Vg 7 3 PULSE (OV 20V 0 O.1NS 0.1NS 27.28US 50US)
RB 7 6 250 ; Transistor base resistance
LE 3 4 681. 8200
216 Chapter 5 Dc-Dc Converters
CE
L
R
VX
J:M
4
4

8
5
o
o
8
5
o
3
8.33UF
40.91UH
3
rx::
DMOD
IC=60V initial voltage

; Voltage source to measure load Cll:rr••" ..
Freewheeling diode
. MODEL DMOD D(IS=2.2E-15 BV=1800V TT=O) ; Diode
model parameters
Q1 2 6 3 QMOD ; BJ'1' switch
.MODEL QMOD NPN (IS=6.734F BF=416.4 BR=.7371
CJC=3.638P
+ CJE=4.493P TR=239.5N TF=301.2P) BJ'1' model parameters
•'!'RAN 1US 1. 6M3 1. 5M3 1US UIC ; Transient analysis
. PROBE ; Graphics postprocessor
.options abstol = 1.00n reltol = 0.01 vntol 0.1 ITL5=50000 ;
convergence
.FOUR 20KHZ I(VY) ; Fourier analysis
• END

The PSpice plots are shown in Figure 5.31, where I(VX) = load
I (Le) = inductor L, current, and V (4) = capacitor voltage.
Using the PSpice
in Figure 5.31 gives v" = Vc = 59.462 V, <1v,. = 1.782 V, M =
2.029 A, I(av)
19.813 A, <1h = 0.3278 A, and fa = 19.8249 A. This verifies the
design; however,
gives a better result than expected.
Example 5.10 A Buck Converter
Temperature: 27.0
80.0V +_------_+------+_------_+_------+_----_+
6O.0V i-------------------"'-----------+
40.0 V +_----_+-----+_-------_+_--------+_------_+
o V (4)
20.0 A +_----_+------+_----_+_------+_----_+
19.6 A +_----_+-----+_----_+_-----+_----_+
01 (VX)

~.OA+_----_+-----+_----_+_-----+_----_+
20.0 A 1------------------------------1-
O.OA +_----_+-------+_--------+---------+_----_+
1.50 ms 1.52 ms 1.54 ms 1.56 ms 1.58 ms 1.60 ms
o I (te)
TIme
FIGURE 5.31
PSpice plots for Example 5.10.
http:TF=301.2P
http:TR=239.5N
5.13 State-Space Analysis of Regulators 217
c. The Fourier coefficients of the input current are
FOURIER COMPONENTS OF TRANSIENT RESPONSE I (VY)
DC COMPONENT = 1.079535E+Ol

HARMONIC FREQUENC'f FOURIER NORMALIZED PHASE
NORMALIZED
m (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 2.000E+04 1.251E+Ol 1.OOOE+OO -1.195E+Ol
O.OOOE+OO
2 4.000E+04 1.769E+OO 1. 415E-Ol 7.969E+01 9.163E+Ol
3 6.000E+04 3.848E+OO 3.076E-Ol -3.131E+01 -1. 937E+Ol
4 8.000E+04 1. 686E+OO 1.348E-Ol 5.500E+01 6.695E+Ol
5 1. OOOE+05 1.939E+OO 1. 551E-01 -S.187E+Ol -3.992E+01
6 1.200E+OS 1.S77E+OO 1.261E-01 3.347E+01 4.S42E+01
7 1.400E+OS 1.014E+OO 8.107E-02 -7.328E+Ol -6.133E+Ol
8 1.600E+OS 1. 43SE+OO 1.147E-Ol 1.271E+Ol 2.466E+01
9 1.800E+OS 4.38SE-Ol 3.506E-02 -9.751E+Ol -8.S56E+01
TOTAL HARMONIC DISTORTION = 4.401661E+Ol
PERCENT

Key Points of Section 5.12
• The design of a dc-dc converter circuit requires (1)
determining the converter
topology, (2) finding the voltage and currents of the switching
devices, (3) finding
the values and ratings of passive elements such as capacitors
and inductors, and
(4) choosing the control strategy and the gating algorithm to
obtain the desired
output.
5.13 STATE-SPACE ANALYSIS OF REGULATORS
Any nth order linear or nonlinear differential equation in one
time-dependent variable
can be written [26] as n first-order differential equation in n
time-dependent variables
Xl through X n• Let us consider, for example, the following
third-order equation:
(5.129)
where y' is the first derivative of y, y' = (dldt) y. Let y be Xl'
Then Eq. (5.129) can be

represented by the three equations
Xl' = X2 (5.130)
X2" = X3 (5.131)
X3" = -aOxl - alxZ - a3x 3 (5.132)
In each case, n initial conditions must be known before an exact
solution can be found.
For any nth-order system, a set of n-independent variables is
necessary and sufficient
to describe that system completely. These variables Xl> X2, ... ,
Xn• are called the state
variables for the system. If the initial conditions of a linear
system are known at time to,
then we can find the states of the systems for all time t > to and
for a given set of input
sources.
reference notes/455647_3_dc-dc-20step-20down-
20converter.pdf

DC-DC Step-Down Converter
Dr. Mahmoud Kassas
Electrical Engineering Dept.
King Fahd University of Petroleum & Minerals
Dhahran, Saudi Arabia

reference notes/455647_4_Filter-20Design-20Diode.pdf
1
3.10 Rectifier Circuit Design:
Objectives:
• Learn about Design Specifications
• Become familiar with C-Filter Design
• Learn about LC-Filter Design
2
The design of a rectifier involves determining
the rating of semiconductor diodes. Rating
specification in terms of:

• Average current
• RMS current
• Peak current
• PIV "peak inverse voltage"
Design Specifications
3
• The output of the rectifiers contains
harmonics.
• Filters can be used to smooth out the dc
output voltage.
R
Le
+
-

V0VL
+
-
L-Filter
R
+
-
V0VL
+
-
C-Filter
Ce R
Le

+
-
V0VL
+
-
LC-Filter
Ce
filters:-DC
filters:-AC
Li
+
-
VLVs

+
-
LC-Filter
Ci Rectifier
4
Full-bridge Rectifier 120 V (rms), 60 Hz R = 500 Ω
• Design C-filter so that RF = 5% (Ripple-
factor).
• With the value of C calculate Vdc.
5
Charging of capacitor When the instantaneous voltage vs is
higher than the capacitor voltage vc, diodes D1 and D2 or D3

and D4 conduct supply voltage thereby charging the
capacitor.
Discharging of capacitor When the instantaneous supply
voltage vs falls below the instantaneous capacitor voltage vc
diodes D1 and D2 or D3 and D4 are reverse biased and the
capacitor discharges through the load resistor.
The capacitor voltage varies between Vcmin and Vcmax and
this is known as the output ripple voltage.
The output voltage waveforms and equivalent circuit on
charging and discharging are shown below.
6
Equivalent Circuit:
0)0(
1
=+=′+∫ LcL
e

RItVdtI
C
eRC
t
m
LmC eR
V
IVtV
−
=⇒==′ )0( eRC
t
mLL eVIRV
−
=⋅ =
The capacitor discharges exponentially through R.

7
eRC
t
mmppr
LLppr
eVVV
ttVttVV
2
)(
21)( )()(
−
−
−

−=
=−==
⇒−≅− ,1 xe x
)1( 2)(
e
mmppr RC
t
VVV −−≅−
2)( tRC
V
V
e
m
ppr ≅−
f
T

2
1
2
=≅
e
m
ppr fRC
V
V
2)(
≅∴ −

=−=−= −
e
e
m
e
m
mpprmdc fRC
fRC
V
fRC
V
VVVV
4
14
42

1
)(
The peak-to-peak ripple voltage can be found from
Since x << 1
If t2
8
22
)( ppr
ac
V
V −≅
( )142
1
4

m
dc
ac
fRC
fRC
fRC
V
fRC
V
V
V
RF
F
RFRf
Ce µ2.1262

1
1
4
1
+
⋅
=
49.158=dcV
Thus the rms output ripple voltage
V

9
( )
∫∫
∑
===
++=
∞
=
ππ
π
ωω
π
ω
π

ωω
00
1
2
.sin
2
2
.)(
2
1
sincos
V
V
tdtVtdtVV
tnbtnaVV

m
mLdc
n nndcL
[ ]
+
+−
+
−
+−−
=

+
+−
+
−
+−−
=
++−=
=
==
∫
∫
∫∫

n
n
n
nV
a
n
tn
n
nV
a
tdtntn
V
a
tdtnt
V
a

tdtntVtdtnVa
m
n
m
n
m
n
m
n
mLn
1
)1cos(
1
1)1cos(
1
)1cos(

1
1)1cos(
)1sin()1sin(
.cos.sin
2
.cos.sin
2
.cos
1
0
0
0
0
2

0
ππ
π
ωπ
π
ωωω
π
ωωω
π
ωωω
π
ωω
π
π
π
π

ππ
Obtain Fourier series for RL-Load:
10
( )( )11
14
+−
−+
=
nn
V
a mn π
( )( ) [ ]
( )( )
( ) ( )

( )( )
( )( ) ( )( )11
4
11
4
11
)1cos()1cos(2)1cos()1cos(
11
)1(1)1cos(11)1cos(
+−
−
=
−+
=
+−
++−−+++−−−

=
+−
−++−+++−−
=
nnnn
nn
nnnnn
nn
nnnn
ππππ
ππ
For n = 1, 3, 5,…………… an = 0
For n = 0, 2, 4,……………………..
Details:

11
)223(............6cos
35
4
4cos
15
4
2cos
3
42
cos.
)1)(1(
142
0.sin.sin
2
.sin
1

4,2
2
0 0
−−−−=
+−
−
+=∴
===
∑
∫ ∫
∞
=
Eqnt

V
t
V
t
VV
V
tn
nn
VV
V
tdtntVtdtnVb
mmmm
L
n
mm
L

mLn
ω
π
ω
π
ω
ππ
ω
ππ
ωωω
π
ωω
π
π π

12
LC-Filter (output)
• Load resistance R = 40 Ω & Load inductance
L = 10 mH
f = 60 Hz RF = 10%
Equivalent circuit for harmonic is:
13
To make it easier for the nth harmonic
ripple current to pass through the filter
capacitor, the load impedance must be
much greater than that of the capacitor
impedance.
This condition is generally satisfied by:
10 times

eCn
LnR
ω
ω
1
)( 22 >>+
eCn
LnR
ω
ω
10
)( 22 =+
14
n
ee

n
e
e
e
on VCLn
V
Cn
Ln
Cn
V
1)(
1
1
1
2 −

−
=
−
−
=
ω
ω
ω
ω
2
1
2

∞
= onnac
VV
Under this condition the effect of the load is negligible
(Voltage divider)
The total amount of ripple voltage due to all harmonics is:
15
π23
4
2
mVV =

π
m
dc
V
V
2
=
2202 1)2(
1
V
CL
VV
ee
ac −
−
==
ω

( )
mHL
CLfV
V
V
V
V
V
RF
FC
C
LnR
e
eedcdc
o

dc
ac
e
e
83.30
1.0
14
1
326
2
10
)(
2
22

22
=∴
=
−
===
=
=+
π
µ
ω
ω
From Eqn (3-22) we find that the second harmonic is the
dominant
one & its rms
&

For n = 2,
3.10 Rectifier Circuit Design: Slide Number 2Slide Number
3Full-bridge Rectifier 120 V (rms), 60 Hz R = 500 Ω Slide
Number 5Equivalent Circuit: Slide Number 7Slide Number
8Slide Number 9Slide Number 10Slide Number 11LC-Filter
(output)To make it easier for the nth harmonic ripple current to
pass through the filter capacitor, the load impedance must be
much greater than that of the capacitor impedance.��This
condition is generally satisfied by:�
10 times Slide Number 14Slide Number 15

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