KULLIYYAH OF ENGINEERING

END-OF-SEMESTER EXAMINATION

SEMESTER 2, 2014/2015 SESSION

Programme : Engineering Level of Study : UG 1

Time : 9:00 am- 12:00 pm Date : 24/05/2015

Duration : 3 Hrs

Course Code : ECE 1312 Section(s) : 1-9

Course Title : Electronics

This Question Paper Consists of 9 (Nine) Printed Pages (Including Cover Page) with 5 (Five) Questions.

INSTRUCTION(S) TO CANDIDATES

DO NOT OPEN UNTIL YOU ARE ASKED TO DO SO

Total mark of this examination is 100. This examination is worth 50 % of the total course assessment. Answer ALL QUESTIONS Only approved calculator with ‘KoE approved’ sticker is allowed (non-programmable and non-

graphical). Marks assigned to each problem are listed in the margins.

Any form of cheating or attempt to cheat is a serious offence which may lead to dismissal.

All electronics gadgets are prohibited in the exam hall / venue.(e.g. mobile / smart phones, smart watches, and smart glasses)

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QUESTION 1 (20 marks)

a) Design a Lithium battery charger using a half-wave rectifier circuit as shown in Fig. 1(a), so

that the peak diode current is 250 mA. Consider the diode piecewise linear model with diode cut-

in voltage V γ=0.65 V and diode forward resistance r f=15 Ω.

(Hint: determine the transformer turns ratio). (6 marks)

b) Calculate the diode reverse biased voltage. (2 marks)

Fig. 1(a)

c) A simple diode circuit is shown in Fig.1(c). If the currents through the diode and resistor R2 are

equal, that is ID = IR2. Assume that the diode cut-in voltage, V γ=0.7 V . Determine the values of

the followings:

i. Resistor R2 (3 marks)

ii. Output voltage VO (3 marks)

Fig. 1(c)

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d) A solar cell is a special type of pn-junction diode. The current equation of a particular solar cell is

represented as follows:

I D=5.0 ×10−13[e(V D

V T)−1]−7.7 ×10−2 (A)

When the cell is irradiated in the sunlight and the cell current I D=0, the voltage V Dat that time of

the solar cell is called the open circuit voltage V OC. Similarly, when the cell is irradiated in the

sunlight and the cell voltage V D=0, the current I Dat that time of the solar cell is called the short

circuit current I SC. If the cell is operating in the room temperature determine the parameters of the

solar cell as follows:

i. Open circuit voltage V OC (4 marks)

ii. Short circuit current I SC (2 marks)

QUESTION 2 (20 marks)

a) The V C−I C characteristics curves and load line of a transistor amplifier are shown in

Fig. 2(a). Assume that the transistor is operating at the room temperature and the collector reverse

saturation current of the transistor I S=7.2 ×10−9 mA.

i. Find the quiescent current I CQ (5 marks)

ii. Determine the Early voltage V A (5 marks)

[ Hint: I C=I S(e

V BE

V T −1)(1+V CE

V A) ]

Fig. 2(a)3

b) A common-emitter amplifier circuit and its voltage transfer characteristic curve are shown in

Fig. 2(b). Determine the minimum input voltage V I at point “X ” as shown in the figure, so that

the transistor goes to V CE ( sat ). Assume that the transistor common emitter current gain β = 150

and then V BE (on )=0.7 V . (5 marks)

Fig. 2(b)

c) A common-emitter amplifier with base to the collector biasing circuit is shown in Fig. 2(c).

Assume that the transistor is working in the room temperature and its common emitter current

gain β = 35 and the base to emitter voltage V BE (on) = 0.7 V respectively. Determine the

operating point (quiescent point) current I BQ , I CQ and V CEQ. (5 marks)

Fig. 2(c)

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QUESTION 3 (20 marks)

a) Draw a circuit diagram for common-collector BJT amplifier circuit. The circuit uses an npn

transistor with a single DC source and voltage divider DC biasing configuration. (2 marks)

b) A transistor amplifier is designed to have gm=95 m Ω−1. If transistor parameters are β=125and

V A=200 V.

i. Find the required collector current (1 Marks)

ii. Determine r π andrO (2 Marks)

c) The transistor parameters for the circuit shown in Fig. 3(c) are

β=120 ,V BE (on )=0.7 V and V A=80 V.

i. Find the quiescent values I BQ , I CQ , IEQ and V CEQ (4 marks)

ii. Find the small-signal parametersgm, r π andrO (3 marks)

iii. Draw the small-signal equivalent circuit. (2 marks)

iv. Determine the amplifier small-signal voltage gain Av (2 marks)

v. Determine the amplifier input resistorRi (2 marks)

vi. Determine the amplifier output resistorRo (2 marks)

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Fig. 3(c)

QUESTION 4 (20 marks)

a) Write four differences between BJT and MOSFET. Sketch a cross section of an n-channel

MOSFET and label the terminals. (4 marks)

b) The transistor in the circuit shown in Fig. 4(b) has parameters VTN = 0.25 V and Kn = 0.03 mA/V2.

The circuit is biased at VDD = 10 V. Assume that R1 + R2 = 500 k. Design the circuit such that ID

= 0.4 mA and VDS = 6 V. Confirm that the transistor is in saturation mode.

(9 marks)

Fig. 4(b)

c) The parameters of an n-channel MOSFET are: VTN = 0.6 V, k’n = 0.1 mA/V2 and  = 0.015 V-1.

The transistor is biased in the saturation region with ID = 1.2 mA.

i. Find the width-to-length (W/L) ratio of the transistor such that the

transconductance, gm = 2.5 mA/V. ( 3 marks)

ii. Calculate the small signal output resistance, ro ( 2 marks)

iii. Calculate the value of VGS ( 2 marks)

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QUESTION 5 (20 marks)

a) Fig. 5(a) is showing a common-source amplifier circuit and its DC output load line. The transistor

parameters are VTN = 0.8 V, Kn = 0.20 mA/V2 and = 0. Assume that VDD = 5 V and

RTH = 200 k. Answer the following questions.

Fig 5(a)

i. Calculate the value of ID (assume the transistor is in saturation) and the values of

RD and gm (4 marks)

ii. Draw the small signal equivalent circuit (3 marks)

iii. Determine the value of the voltage gain AV (3 marks)

b) An ideal Op-amp is shown in Fig. 5(b). Determine the values of the followings :

i. vO, i1, i2, iL and iO (5 marks)

ii. voltage gain Av (3 marks)

Fig. 5(b)7

c) Evaluate the output voltage vo of the circuit as shown in Fig. 5(c). (2 marks)

Fig. 5(c)

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Useful equations for MOSFET:

Kn=Kn

'

2∙ W

L

Useful equations for BJT:

Useful equation for pn- junction:

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I D=Kn [2 (V GS−V TN ) V DS−VDS2 ]

I D=Kn (V GS−V TN )2

ro=V A

I CQ

V A=1λ

gm=2√ Kn I DQ

ro=V A

I CQ

gm=I CQ

V T

r π=βV T

I CQ

gm rπ=β

I D=I s(e

v D

V T −1)

END OF PAPER

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