Power Electronic project

....-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 = - = - = 05455V, 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.

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 Xn• 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 - a3x3 (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