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(ELEC101)exam_pastb.pdf
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(16)

Find VO/ VS in terms of R, C and jw. If C = 1F, R1= 1k., R2 = 10k., find the complex transfer function G (= VO / VS ). Is the op amp circuit a low pass filter? Assume the op amp is ideal.
(16)
R1C R2

VO
VO R2
G == 1+
V 1
SR1 +
jC
R 1 R 1

= 1+ 2 = 1+ 2
R1 .1 R11
1+ j 1. j
CR1 *1 *1k. 10

= 1+
1k1. j

(12)
(4)
Circuit is a high pass filter

(21)

In an op amp filter circuit, the complex transfer function G (= VO / VS ) is given as. 20
1+ jCR

(a) Find O and plot the magnitude of G ( | G | ) versus the angular frequency . Show clearly the value of | G | when = and = O in your plot.
G
V (t) = 1cos2 tV

(b)
If SO , find VO(t) .


Given that the op amp is ideal, C = 1x10-4F , R = 1k. , O = 1/CR . (21)

(21)



(a)

(b)
11

O == = 10rad / s
20 20


CR 0.1mF *1k.

V = 10* =
o

O
1 + 2j 563.4 (9)
(3)

20 oVO (t) = cos(10t . 63.4 )V
5


(12)


I D 15V VO 100.
2V -15V


(a) If V2 = -0.62V D is a ON diode V2
I2 . IO exp( )
25mV 620mV
= (1pA)exp( ) . 59mA
25mV VO ..I*100.. 0.62V ..6.52V
(12)
(12)



-VC +

13V

Vi


VO
3V
-10V

(b)
13V 13 3 VO


3V

Vc = -10V

(8)

--10V +

-10V

VO 3V
VO = Vi + Vc = -10 C 10 = -20 V
(4)


3V
-9V
-20V

(13)


(c)

(c) In the diode circuit, sketch and label clearly VO(t) .
D2 is an ideal diode and D1 is an offset diode with VF = 0.7V. (13)

Vi 10k. VO10V
Vi

D1
10k.
0V D2
-6V

5V VO 2.3V 0V

(13)
(22)



(16)


Draw the cross sectional structure of a NPN BJT transistor operated in the amplifier mode, describe the movement of electrons, and explain briefly why IC / .IE .
If IC .IB, find in terms of . (16)
IE IC
EBC

e -

e

IB
VBE

VCB





1.
EB Junction is a forward bias (on) diode and BC is reverse bias (off) diode

2.
E is very heavily doped (N + for NPN). E has many electrons,

3.
B is very thin. So most electrons injected from E (to B) are attracted to C and


IC .IE (10)
IC IC
IE .=IB +IC =+IC

11

hence =+1

=
+1


=
(6)
1 .
7
Given the BJT circuit below and the IC -VCE curve of the BJT. The Q point is chosen as IC = 18mA and VCE = 5V. (a) Draw the load line VCE = VCC -ICRC. (b) Find RB RC. . (b) If vi (t) = 0.01cost V , estimate the voltage gain from the I-V curve and sketch VO(t) . (28) For the BJT, given VBE(ON) = 0.7V , VCESAT = 0.2V , r = 0.


VCC = 9V

IC



RC
40

RB
IB = 0.16mA



VO(t) +

IC (mA)
VCE
vi(t)
-
20
IB = 0.08mA

IB
Vi = 0.74V
VCE (V)
0
5 10
(28)


(a) Draw load line, .=== 25036 9 mA V I VR C CC C .= . = 5000.08 0.70.74 mA VVRB (b) Vin1 = 0.75V IB1= 0.10mA (5) (5)
(8) (c) 1000.750.73 46 = .. .. == VV VV dV dVvoltage gain A in CE v 12 12 . . VV VV inin CECE