ECE 533 HIF P g l o f 6

ECE 533: Power Electronics

Quiz 2 - Undergraduate StudentsNovember 27 r2006

l. Policy: closed book, calculators allowed. Show all work, partial credit will begiven.

2. Work in the provided space.

3. The quiz has 2 problems and 6 pages, including the cover page.

3 c tN a m e : \ / ) -

/ [ nStudent Number: /u

Problem 1 [50 ptsl

Problem 2 [50 ptsl

Total [100 ptsl:

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Problem 1, [50 points]:

The converter of Fig.l operates such that the current through the primary side

magnetrzing inductance L^ is continuous. During the subinterval 0 < t < DZ" switch I

conducts and switch 2 is open. For D4 < t < 4 the switch I is open and switch 2

conducts.

+

vz(t)

Fig. 1

(a) t6 pointsl Draw equivalent converter circuits for the both portions of a switching

interval 7". Each drawing should show an ideal transformer, a magnetizing inductance, as

well as the voltage across Zr.

Lrn

.t

DIs <t < Is

tur,, LPr1Jeu, r'U"h-1p

L

v w v h I' t . <

.^- , + V\'

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(b) t20 pointsl Determine analytical expression for the conversion rctio M(D) t VlVs ,

also find the dc value of magnetizing current as a function of n, D, Vs, and R.

<Ur-n>.r-- o : b [t5 -v\ + f] t-nVr\ t3 Vg g-n$ \ - lV

Hb\=+=D-v.U\

o <h< bTs

4c k \: \tn - tfrs

( [ . ] r : = o =

\ L $ : [ l -

\{tu\\

I

DTs <.{- < Ts

l,ctt\ = "W

\ - e P etn t -uzf -

v t b -n$ \R.ua

D lc,a -Y"

I

II

D ur_*- lua

(c) [8 points] Sketch M(D) for 0 < D <1, whgn n = 3.

=> ftn

D

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(d) t l6points] Forn =3,Vs:1V, D=0.5,fr=500kH2, and L^: l FrH,R-2 f l , and

large C sketch the waveforms,of the input current and the voltage across switch 2 (see

Fig.l). Clearly label the maximum and minimum values of the waveforms.

4sa

tg* -zir3hrSA ' ,

r n t rYo:r11rp 1..:-.-.-::

jv1 g = l M

1I:-=v*oF !-tv-\\.2V" .r,h

\/"sili6,4 Z

-lhr-, ir {re

'l,X== -3irt't

.^ tlwM(orpt ,/ i,| ,\^1

fov h.5 \o,ri ={ u - u 4u-s Vqlu .

b

= -2-/�

pz- K

Vou\f : -2-V "

.|ol

,r T [Vr^v\rrpl = :_ DTS:ZL

<.

T

.\EA

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uo,f = 0154

Problem 2 [50 pointsl

Fig.2 shows a buck converter operating in CCM. All losses but the diode conduction

losses can be neglected. The diode has forward voltage dtop Vp.

Q , L

Fig.2

(a) [15 points] Determine the nonlinear averaged equations of this converter.

l -c l5 <b <TslL

r)X

rr ' ,Anru L = L : i = . 1 [ t )u r ' '

r ^ - ^ d l "l c - - C - , - = ( [ t _ )

crI

{t" -t. lut

o t . - r d i ruL= t G

=_(T>_Vp

, \ c= 0 i I = < iu ) - { '. d r K1t' c

<1r>

_ <1J>- b

Ir_

-> L d < i u =<1t> -J Vt

C t ,! '

d< u>trl

< iL)

J < ,i.>

o<\< [-[:

l-

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O) t20 pointsl Construct small-signal ac equations

\ , t - t \ '( b " ' \\' l ̂ i 'tu'*J ) Vs

tr- ttlt\

s_t.t $ it3to\1 = (on]\ (vr-&)

r d { t N AL tT= Pub+

Ajv '

nt .

- \- w + v ) -

t jV '

'\r tV

t f

A At , ' l -D{L + dJ .u

. dct K

^ ^- , i r y, u r _ l L

t \

n d'\f

d+

i

K

Attx

l\*

(c) [15 points] Construct an equivalent circuit that corresponds to the small-signal

equations of part (b).A'\t L I{

rlt-

r \ l \

tblJ\ t.t ' i,-\ :

A Ai h -

{ \ = D,tu

nd Wqtvt\

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