University of Malaya Continuous Assessment I –Marking Scheme HRK/2010

1

Electronic Circuits I

KEEE 1125

Name : _________________________ Peer Name : _____________________

Student ID : _________________________ Peer ID : _____________________

Marks : _________________________

Learning Outcomes Assessed:

Analyze diode circuits, half wave and full wave rectifier and their applications as rectifiers, clippers and clampers.

1,2,3

Describe the characteristic of Bipolar Junction Transistor (BJT). NiL

Perform analysis of BJT circuits, its load lines and bias configurations NiL

Analyze Field Effect Transistor (FET) NiL

1. Given that VT = 25 mV. A particular diode conducts 1 A at a junction voltage of 0.65 V and 2

A at a junction voltage of 0.67 V. What are its values of n and IS?.

[1 Marks] VDnVT

D SFrom I I e= :

( )VD1nVT

D1 SI I e 1= …

( )VD2nVT

D2 SI I e 2= …

Dividing (2) by (1) :

( )1V -VD1 D2D2 nVT

D1

I= e

I

Substituting ID1 = 1 A, ID2 = 2 A, VD1 = 0.65 V and VD2 = 0.67 V

( ) ( )2 1ln 0.67 0.65

1 n 25mV

n 0.866

= −

=

From (1)

S 0.65

0.866 25mV

14

1I

e9.144 10 A

×

−

=

= ×

University of Malaya Continuous Assessment I –Marking Scheme HRK/2010

2

2. For the circuits below, assuming ideal diodes, find the voltages and currents indicated.

a.

( )5 5

I 1mA10

V 5V

− −= =

= −

[0.5 Marks]

b.

I 0 A

V 5V

== +

[0.5 Marks]

c.

( )5 5

I 1mA10

V 5V

− −= =

= +

[0.5 Marks]

d.

I 0 A

V 5V

== −

[0.5 Marks]

[2 Marks]

University of Malaya Continuous Assessment I –Marking Scheme HRK/2010

3

3. For the circuits below, assuming ideal diodes, utilize Thevenin’s theorem to simplify the

circuits and thus find the values of the labeled currents and voltages.

a.

Rth

thR 10k ||10k 20k

80k

3

= Ω Ω + Ω

= Ω

Vth

Voltage Divider rule

th

20V 15V

3010V

= ×

= +

th

80R k

3= Ω

10 3

I mA80 8k3

= =Ω

3

V mA 20k87.5V

= × Ω

=

[1 Marks]

University of Malaya Continuous Assessment I –Marking Scheme HRK/2010

4

b.

Rth

thR 10k ||10k 10k ||10k

10k

= Ω Ω + Ω Ω= Ω

Applying Superposition Theorem

a. Eliminating 10 V Supply

th1

10V 15

2015

V2

= − ×

= −

b. Eliminating 15 V Supply

th 2

10V 10

205V

= + ×

=

University of Malaya Continuous Assessment I –Marking Scheme HRK/2010

5

Vth

th th1 th 2V V V

155

25

V2

= +

= − +

= −

thR 10k= Ω

th

5V V

2= −

I 0A=

Applying KVL around Loop 1

V 5 7.5 0

V 2.5V

+ − + == −

[1 marks]