=========================preview======================
(ELEC102)final_pastd.pdf
Back to ELEC102 Login to download
======================================================
(300) (17)

Find Vs(t) in terms of C, R, and Vo(t). 1 If C = 1mF and R = 2kW, find and plot Vs(t) . Assume the op amp is
VS dVO
ideal. apply KCL . @-C (6)
(17) R dt CR = 0.5mF *2kW= 1ms
RC

dVO 8V --8V
\V @-CR =-2ms * =-8V

S dt 4ms
VS(t) VO(t)
(7)
VO(t)


VS(t)
8V
6ms t

(4)
0 Vs(t)2ms 8V -8V

t 0 -8V


(17)


a. In the circuit, replace the 741 op amp by the op amp circuit model and redraw the circuit.

2

b. Draw the circuit of an op amp voltage follower and


name two advantages of the circuit. (9)

Given that Ri = 2MW, Ro = 75W, A = 200k.
(8)
R1
R2

VS

(4)
voltage follower

1. Very high input resistance
(5)

2. Very low output resistance
R2

VO
VS
(8)



(13)

Given that VS (t) = Vm coswtV and VO/ V1 = 21 , find V1/ VS in terms of R, C and jw. If C = 1mF and R = 1kW, find the complex transfer function G (= VO / VS ). Is the op amp circuit a low pass filter? Assume the op amp is ideal.
(13)
C
SV
R
VO

V1

1kW 20kW


R 1
V = V * = V *
1 SS
11

R + 1+
jwCjwCR 1 21V

V = 21V = 21*V = S

O 1 S
11
1+ 1+
jwCR jw(1mF *1kW)

VO 21
G ==
VS 1
1+
jw *1ms
(10)
(3)

Circuit is a high pass filter
(21)

In an ideal op amp filter circuit, the complex transfer function G (= VO / VS ) is given as. -11 (b) 1+ jwCR
C = 1x10-8 F and R = 1kW.

11
(a) Show that the cut off frequency wO is 100k rad/s.

11
G
(b) Plot the magnitude of G ( | G | ) versus w . Show clearly the



(8)
value of | G| when w = 0 and w = wO in your plot.

2
(c)
If VS (t) = 2cos100kt ,V find VO(t) .

(21)



w in rad/s



wO =100k
(a)
(c)
11
wO == = 100krad /s
-11 -11
CR 10nF *1kW
VO = 20* =
o
1 + j
245 (8)
(5)
22 o
V (t) =-
cos( 100 kt -45 )V
O 2
(22)



In the ideal op amp circuit, the diode has the forward, reverse and breakdown characteristics as shown.
(a)
Draw the circuit model of the diode at breakdown.

(b)
If I = 20mA, show that Vo = -16.5V. Hence find ID.


(22)
I

V1
325W

VO
ID
ID

VD


-0.05A
-0.1A
(a)

model


VZK = 6V
(7)
rZ = 10W
(b) When I = 10mA VO =-I * (500W+ 325W) =-16.5V (3)
\Vin = V1 =-10V
(4)
Dis breakdown (2) -10V --16.5V -6V
ID =-=-0.05A
10O (6)
(13)

(a) In the diode circuit, if V1 = C7V, R1 = 18kW, R2 = 2kW,
find IO . Given that the diode equation is
I = (1pA) exp [(VD/25mV) C 1]. (13)
R1
VO
V1
D


R2
R2 2kW
V = V * = -7V * =-0.7V
O1 R1 + R2 18kW+ 2kW
(5)
diode is ON
0.7V
\IO = IS (e25mV -1)
700m
25m (8)
= 1x10-12A (e -1) @ 1.45A
(15)

(b) In the diode circuit, find Vo when Vi = 18V. Assume the diodes are ideal. (15)
D1 D2

Vi = 18V, D1 and D2 ON,
(4)
Vi -Vo 20 -Vo Vo
\+ =
1kW 1kW 2kW \Vi -Vo + 20 -Vo = 0.5Vo Vi + 20 18V + 20 (11)
\Vo == = 15.2V
2.5 2.5
(10)

(a) Given that Vi (t) = 6 cos 2p1kt V, plot and label c