power electronics typical solved problems

8/10/2019 power electronics typical solved problems

1/10

EENG441 SOLVED PROBLEMS

P1-)The transistor Q in the chopper shown below has a maximum (dv/dt) rating of 50 V/s. It is used

to supply a resistive load of RL = 100 . Find the snubber capacitance Cs required to protect the

transistor for dv/dt, when it is turned off. Assume that the transistor is turned off at t=0 , and vc(0)= 0.

Solution:

The solution for the capacitor voltage after Q is turned off:

/( ) (1 ) ( )tc s s L s

v t V e R R C

/ /( ) 1

t tc s Lc s s L c s

s L s L

dv V Ri C e v t V R i V e

dt R R R R

/ 6

2 2

max

9.876550 V/s 50 10

0.1

V/s( ) ( )

975 F

tL Ls s

s s L s s

s

L s

R Rdv dvV e V

dt C R R dt C R C

C

R

Alternatively, the voltage across Q can be solved directly from the eqn. /( ) ( ) (0) ( ) tx t x x x e

/ /

( ) ( ) (0 ) (0 ) ( )

( ) 1

sc s s c s s L s

s L

t ts Ls s s s

s L s L

Vv v V v R i R R R C

R RR R

v t V V V e V eR R R R

+

_

+ _v

Q

Rs= 5

Vs = 500 V

_

Cs

+

RL

vc


2/10

P2-)In the circuit shown below, transistor Q is turned off at t = 0. The transistor current falls linearly

after it is turned off, as shown below. The fall time is tf= 2 s. Diode Dsis ideal. (Note: thefreewheeling diode DFdoes not turn on until the capacitor voltage reaches Vs).

(a) Find and sketch the voltage vacross the transistor until steady state is reached.(b) Find the maximum dv/dtacross the transistor.

Cs

Ds

R

I0

DFVs

+

_

+ _v

Q

C s= 0.2 F V s= 200 V

I0=10 A

iQ

I0

t

iQ

tf0

Solution:

(a) After Q is turned off, the capacitor current is equal toI0-iQ:

0 1 2 3 4 5 t (s)

tf

50

100

150

200

v (V)

slope =I0 /C

(b) The maximum dv/dt across the transistor is

0

6

max

1050 V/ s

0.2 10

Idv

dt C

0 0 0

0

20 0 0

0

0 0

0

1 ( ) ( ) '

( ) '. ' 0 ( ) 50 V2 2

1 ( ) ( ) + ( ) 50

2f

t

c Q c c

f

t

f f f

f f

t

f f f f

t

t dvi I i I i C v t v t i dt

t dt C

I I Iv t t dt t t t v t t

Ct Ct C

I It t v t v t I dt t t t

C C C

50( 2) V ; in st t


3/10

P3-)In the circuit shown below, transistor Q is turned off at t = 0. The transistor current falls linearly

after it is turned off, as shown below. The fall time tf= 2 s. Diode Dsis ideal.

(a) Find and sketch the voltage across the transistor until steady state is reached.(b) Find the maximum dv/dtacross the transistor.(c) Find R so that the initial current through Q is limited to 40A when it is turned on again.

Cs

Ds

R

I0

DFVs

+

_

+ _v

Q

C s= 0.2 F V s= 200 V

I0=10 A

iQ

I0

tft

iQ

Solution:

(a) Same as in P1.(b)Same as in P1.(c) When Q turns on, its current is the sum of the load current and the snubber capacitor

discharging current iC. Let Q be turned on at t = 0+. Then,

0

(0 ) 200(0 ) (0 ) (0 )

200 200(0 ) 10 40 6.67

30

CQ C C

Q

vi I i i

R R

i RR

Cs

Ds

R

I0

Vs

+

_

+ _v

Q

iQ

iC


4/10

P4-)In the circuit shown below, transistor Q is turned off at t = 0. The transistor turns off with its

current falling to zero instantly. Diode Dsis ideal. (Note: the freewheeling diode DFdoes not turn onuntil the capacitor voltage reaches Vs).

(a) Find the capacitance Csso that the maximum dv/dt across the transistor is 100 V/s.

(b) With Cs= 0.25 F, find and sketch the voltage vacross the transistor for 0 5 st . Indicate all

time and voltage values.

Vs= 200 V

I 0= 20 A

Cs

Ds

R

I0

DFVs

+

_

+ _v

QiQ

Solution:

(a) After Q is turned off, the capacitor current becomes equal toI0:

(b)

0

s 1

0

( ) the capacitor charges linearly.

0.25 200It becomes equal to at 2.5 s

20

s

s s

Iv t tC

C VV t

I

00

100 V/ s 0.2 Fc s s

s

Idv dvi I C C

dt dt C

0 1 2 3 4 5 t (s)

50

100

150

200

v (V)

slope =I0/C


5/10

P5-)The step-down chopper shown below is operated at the switching frequencyfs = 10 kHz.

(a) Find the duty ratio kso that the average load currentIa= 2 A.

(b) Find the range of k in which the load current iois continuous ( i.e. kk1, find k1).

D

+

Vs

+voR

L

+-

io

E

S

V s = 100 V

E = 40 V

L = 5 m H

R = 5

Solution:

(a) 5 2 40 50 V 0.5a aa a

s

V E VI V k

R V

(b) Conduction is continuous if I1> 0

40.1

3

1 10ln 1 ( 1) 0.1 10 ln 1 0.4( 1) 0.412

10

z

s

E Tk e z k e

z V


6/10

P6-) The step-up dc-dc converter shown below is operated at a switching frequency of fs= 20 kHz.

(a) For R = 20 find the duty ratio k so that the average power supplied to the load is

Pav= 500 W.

(b) For k=0.7 find the maximum value of the load resistance R so that the source current is is

continuous.

+

-

C

io

isvo

Vs

L D

Q

Vs = 40 V

L = 500 H

fs= 20 kHzR

Solution:

(a)2 2

2. 500

1 )0.6

(1

dc s sav dc dc dc

V V VP V I V

R k R kk

(b) When the switch Q is on

1 1

2 1 2

0 ( ) where (0)

where ( )

s ss s s s s

s

s s s

di VL V t kT i t I kT I i

dt L

V

I I kT I i kTL

Power balance of the converter:2

,dc

in out s s av

VP P V I

R

, 1 2 1 1

2

1 12 2

1 12

2 2 2

2 (1 ) (1 ) 2

s s

s av s s

s s s s

s s s

V VI I I I kT I kT

L L

V V V V V I kT I kT

L R k R k L

For continuous source current,3 6

1 2 2 22 2 20 10 500 100 317.46(1 ) 2 (1 ) (0.3) 0.7

s s ssV V f LI kT R

R k L k k


7/10

P7-)The step-up converter shown below is operated with a duty cycle k= 0.75. The minimum value

of the source currentis is I1= 10 A . Assume that is decreases linearly when transistor Q is turned off.Find the switching frequency of the converter.

+

-

C

io

isvo

Vs

L D

Q R

Vs = 20 V L = 5 mH

R = 20 C F

Solution:

Power balance:

2

, , 216 A

(1 )

a sin out s s av s av

V VP P V I I

R R k

, 1 2 2

1( ) 32 10 22 A

2s avI I I I

In 0 < t< kT1 1

( ) where (0)s ss s s s

di VL V i t I kT I i

dt L

2 1s

s

VI I kT

L

3

2 1

1 0.75 20250 Hz

( ) 5 10 12

ss

s

kVf

T L I I

I2

I1

t

is

kT T


8/10

P8-) The buck regulator shown below is operated at the switching frequencyfs = 10 kHz.

(a) Find the average transistor current IQ,avas a function of the duty ratio k,assuming that the

inductor current iLis continuous.

(b) Find the maximum duty ratio kmaxif Q has a maximum average current rating of 4 A.

(c) Find the maximum duty ratio kcfor whichiLis continuous.

Solution:

(a)2

2

, ,. o s

in out s Q av o s Q av

V VP P V I V kV I k

R R

(b) 2, max

4 A 4 A 0.6325 0.6325sQ av

VI k k k

R

(c) When the switch Q is on,

1 1

2 1 2

0 ( ) where (0)

(1 ) (1) where ( )

s oLs o s L s L

so s s L s

V VdiL V V t kT i t I kT I idt L

V kV kV I I kT I i kT

L

2

, 1 2 1 2 1

4 6

1

21( ) (1) (1 )

2 2

2 2 10 200 10For continuous conduction 0 1 1 0.2

5

s s s s

Q av

s

s

c

V V V V I I I k k I I k I k k

R R R f L

f LI k k k

R

+

-

io

vo

+

Vs

-

iQ

Q L

RD C

Vs = 50 V

C = 100 F

L = 200 H

R = 5

fs= 10 kHz

iL


9/10

P9-) The step-down chopper shown below is operated at the switching frequency fs = 10 kHz.

(a) Find the minimum value of the duty ratio kso that the load current iois continuous.

(b) Find the duty ratio kso that the average load current is 20 A.

(c)

Find the average value (IQ,av) of the transistor current iQfor k= 0.5.

Solution:

(a) The minimum value of the load current is

- (1- ) -

1 -

1 0.1; 2 ms 0.1 ms 0.05

1- 2

k a a

s

a

s

V e e E T LI a T a

R e R R f

0.05

1

For continuous conduction

1 1 500 ln 1 ( 1) ln 1 ( 1)

0.05 20.

002547a

s

EI k e e

a V

(b) Average load current

20 5 5020

200.7

05a sa

V E kV E I k

R R

(c) iQ = iowhen Q is on.

/ /1In 0 , ( ) ( ) (0) ( ) t ts s

o o o o

V E V E t kT i t i i i e I e

R R

0.5 0.05 0.05/ /

1 0.05

200 509.75 A ( ) 30 (9.75 30). 30 20.25 A

5 1 5

t t

o

e eI i t e e

e

/ 0.5 0.05

,

0

1 ( ). 30 20.25 51 15 20.25 20 1 .0 A

kT

kT

Q av oI i t dt k e e

T T

D

QiQ+

Vs

+voR

L

+-

io

E

V s = 200 V

E = 50 V

L = 10 m H

R = 5

fs= 10 kHz


10/10

P10-) Below is shown a step-down dc-dc converter with an LC filter at its output. Diode D is ideal.

Switch S is operated with duty ratio k= 0.4 at the switching frequency fs= 10 kHz. Assume that theoutput voltage vo= Vais constant and also the output current io= Iais constant. Given that Va= kVs,

Vs= 100 V, L = 250 H and C = 100 F

(a) Find the minimum and maximum values of the inductor current iL for Ia= 10A. Showallyour calculations.

(b) Sketch the source current is, and find its average value.

Solution:

(a)

1 2 10 : ( ) ( )s a s a

L L

V V V V t kT i t I t i kT I I kT

L L

(1)

2 1 20 (1 ) : ( ) (1 ) (1 )a a

L L

V Vt k T i t I t i k T I I k T

L L

(2)

, 1 2 1 2

1( ) 2

2L av a a

I I I I I I I (3)

1 2(1 )s

a s

VV kV I I k k T

L (4)

(3)&(4) 1 2(1 ) (1 )2 2

s sa a

V VI I k k T I I k k T

L L

4

1 24

10010 (0.6)(0.4)10 5.2 A 14.8 A

5 10I I

(b)

, 1 2

1( ) 4 A

2s av a

I k I I kI

+

-

L

C

io

is

fsfsS

D

vo

LoadVs

kT T t

1

0

Switching function of S:iL

kT T t0

Vs

iL

vD

I2

I1

tt'

iL

I2

I1

tkT T

power electronics typical solved problems

Documents