8/10/2019 Final Exam 2005 Solutions

1/18

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Name:

\O\rsdtbn

student#:

Section:

CARLETON

UNIVERSITY

FINAL

EXAMINATION

i|29,2005

Duration:

3

hours

Department name and course number:

Course

Instructor(s):

R. Mason.

A.

Steele

AUlHORIZED MLMORANDA

NON-PROGRAMMAB LE

CALC ULATOR

Students MUST

count the

number

of

pages

in this

examination paper

before

beginning

to

write,

and

report

any discrepancies

immediately

to a

proctor.

This

question

paper

has 17

pages.

This

examination

question paper

MAY

NOT be taken

from

the examination room.

This

exam consists

of

5

questions,

which should be answered on this

exam

paper

in

the space

provided.

Attempt all

questions.

Marks allocated

to each

question

are indicated

(totalmarks:

100).

Note: The

solution must

be

clearly indicated. Multiple

solutions

or

solutions that are not

clearly

identified, will

be

marked incorrect. Using approximate relations

(unless

they

are

given

below or

specified in

a

question)

is not accepted.

Clearly

state

all

assumptions made.

Clearly

mark the

units

for

all final answers. Clearly indicate axis/units for any

graphs.

SHOW

YOUR

WORK

Ftno.\

\i

i"ter

Z

00?,

Electron

ics

ELEC-2507

(A,B,C)

Number

of

students: 220

Diode:

Forward

current:

I

o

x

I

r(ert,t,rr

)

Small signalresistance:

,u

=*

Ir,

KT

V.t

=

-

=,/,)l7ll

at

rOOm

temperatgre

q

Bipolar

Transistor:

Active

mode

operation:

VeE:0.7V

Saturation mode operation: V6pru1: 0.2V

i,.

=

Fto

i,

=d,,

i,

=iu+i,

s

= _

,=F ,=Vo

m

v,

'1r

q I.

ornl

o=

F

F+l

Operational Amplifier:

V. =

A(Vu

-V_);

Ri =

co. Ro

-

Q

MOSFET:

I

,,

=

k'Yl

(r,,,

-

v,)v,,

-+1,

\

rL',^ _

J

I

t''W

(Vcs

r'

\2

I

DS,ro,

=

o'

iY(l

+

il'p5):

'

k,w

k'=VC^;

K=

L

Vrr.,o,

=V,t,

-V,

g,t,

=

xg.

r.

=

k,Yr(v,,,*v,)

=

r

'e -

o

Dnt

I/

/t

T

Question

NIax

Marks

Score

I l0

2

20

3

tT-..i"t^

vo

(r^r. Loaa)

=.-urg.

ilyi

leL,.;

)

='_K3.96

rztzo)is:rr;;i

"::

,:;

ue

i.;

-6.Oqq.V

r*;

arcwod.sott$tov1

r,r,"tnl]

ncrh

ana 5s,\

/cswttr

in,

Ir-=0.34{g

-.

\o'8oqoo)^

'o

-8loq4V

iv)

For rhe nominat t

2v

suppty, compute

the

minimum

value

of

R1 so

*l?t.

;ilt}

fr

nrh;f*.m

T^

relatively

constant output.

For

-rL

n,r,.t.v.,,-

,^1,t,-

"l

R.u

[wor"ta"-)

+L

a-1/v-

o4n,r.-gJ

hw

?,

be

e

4u

e4-

"1.

+\r.

brdcdr.n

ejbn

,ILt

V.;

lz

-

TzK:

o,Svnfl

t/or*=

LJzo +L(z

Jzr"=

Z

KJ.9L

+40X

o'5tA

=7'9LLV

Lr11=

\t-

?.

t

L?

=-W

4,o+8

tn&

=

Tr*

-\tr,.

=

Z.,C+i-n

-o

'SY^

(3

Marks)

T^

-

R

Lr^i",'

tK-

Voo.in

f

er

Lv.,in

RvL:

-

12

+

lx

"Lz

+L(7

fa

*2

t/26=

6

fr=

I

l-1u1.

-

iz-z(s.9e).

=

3.92.

n)It,r.

tr v/

t

l

T.x

fu

anlrvnp*r.'-

"

*[

lUlu

ar.

ON

;s

Va,{f.A.\

ii)

Find

the change

in V6 if

the

l2V

power

supply

were

to vary

by

+lV.

, |

(2

Marks)

a,trow-

Ocrc,

j.*

lh-*

Linc--

R4.gwta*'r'"'

-1"

&.

be

a/a4teA

=

:v)*_

aVs

R+2G

4o

+

bVo

\,K+ZLz-o)

AV.

:

0.0385

bVs

A.gtr

Tenuf

s

,^rr;ch

a

Rr

3.

5?8

v"A

RL*in:1

&/

'-#"

.L

R

l.

\-Yrs6

-

3't7gr"

=2.tt4\

8/10/2019 Final Exam 2005 Solutions

7/18

r"*.,

So\,^*rbn

Student

#:

Section:

ELEC-2507 Final Examination, April 29, 2005 Page 7

of

17

c)

A

square

wave input

as shown

below

is applied

to Vi of

the ideal diode

circuit in

Fig.

2.3.

Fig.2.3

Sketch the resulting waveform

at

Vo

assuming the

initial

value

of Vo is

0V.

Make

sure to include

scales on the

voltage

and time axis

(4

Marks)

-l5rn---

1i

I

a?'

i....i....

.i----j..--

.i

;

8/10/2019 Final Exam 2005 Solutions

8/18

(Page 7 of 17) Q2_part C answer:

The time constant for this RC circuit is 1 1 1msecRC k

The pulse width of the input signal is 10msecW

T

1) for 15msect : since the input source is zero 0iV ; the voltage at output is zero 0oV ;

2) At 15msect : the capacitor will see an instant change in input voltage (from 0-to-5V) and

(instantly) acts as a short circuit. This means that,o

V will follow the input 5Vo V

3) 15msec 25msect : at 15msec the capacitor starts charging, and continues charging as time

passes. The equivalent circuit is shown to the right.

15 25m t m : 1 sec5a at t

m

O mV V e e

Say:

at 16msect ,capacitor charged for 1msec

1 sec ( 16 sec 15 sec)a

t m m m , then the output voltage is1 sec

1 sec5 5 0.37 1.84

m

m

oV e V

Similarly, say:

at

2 sec

1 sec17 sec 17 sec 15 sec 2 sec 5 1m

m

a ot m t m m m V e V

at

3 sec

1 sec18 sec 18 sec 15 sec 3 sec 5 0.25m

m

a ot m t m m m V e V

at4 sec

1 sec19 sec 19 sec 15 sec 4 sec 5 0.09m

m

a ot m t m m m V e V

at5 sec

1 sec20 sec 20 sec 15 sec 5 sec 5 0.034 0m

m

a ot m t m m m V e V

This means, the capacitor has been fully charged to 5V and will stay charged until25msec.

Since in this circuitW

T ; it does not behave as a regular diode clamper circuit.

4) at 25msect : The input instantly decreases by 5V and falls to 0V, as seen below the charge

across the capacitor appears right across the diode and this causes the diode to conduct heavily

and to quickly discharge the capacitor.

1k0i

V

5c

V V

1

25t msec

25t msec

0o

V

Equivalent t o

5) at 25msect : The input source 0

iV and capacitor has been already discharged and 0

cV ,

therefore the voltage at the output will be (and stay at) zero, 0o

V .

1

R

k

?o

V

5i

V V

1C

8/10/2019 Final Exam 2005 Solutions

9/18

Student #:

ELEC.25O7

Q3:

MOSFET

Final Examination, .lpril 29, 2005

The

MOSFET circuit shown

in

Fig,

3.1 is biased by

a constant

current source.

*".",

So

lrr*ion

+1

0v

2Mf)

-10v

Section:

Page

8

of

l7

VD

0.6mA

whiL

abcrY

8/10/2019 Final Exam 2005 Solutions

10/18

Name: S",tr,

^*fb

n

Student #:

Final

Examination,

April

29, 2005

Section:

ELEC:.2507

ii. Calculate

g,

&

ro

given

V1:

80V.

X

\Zo7n

[

\o

X

o'6n

) '/'

tlu

:

133.33

l(

8/10/2019 Final Exam 2005 Solutions

11/18

Name:

So\r,*tnn

Student

#: Section:

Final

Examination,

April 29, 2005

+1 0V

1

oko

-10v

Fig. 3.2

(Assume

all

capacitors

have

an

infinite

value).

t

*:

ELEC-2507

Page

l0

of

17

\OK

(4

Marks)

Vo

I

lu'

i.

Sketch the

equivalent

hybrid-n

small

signal circuit model for this circuit.

2I(?.RG\

?-

Fr=

7ft6.+n*"

5onrtr

-

lJzREac\-"

-r'

iii. Sketch

the

response

of

the maglrltude of

the

gain

(in

dB)

with

respect

to

angular

frequency

(radians

Persecond)

\-\ \ \=zo\oa\lltql\

zoro3\h\*zo\"j\*t*-o.\ - l\iffi\

''Marks)

t'^^U\r^3\ AAb

J

t=

uo

K

+-

zol'J

\

#\=2o/5

\

=P

trB

io

?=oK

+

?e,E\#\=

zo'u

4"

f"o.1L

=.l'l...15

\ffiFl

=dodr}

:

"

T(It

\o

5Lt,

SoLf

?r(.w

END

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zR

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zR

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