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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:
I of6
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\'
2 o f 6
(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
3 of6
(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
4 o f 6
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-
5 of6
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\
6 of6
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