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Main points of this past exam are: Zener Breakdown, Current Ratios, Bipolar Junction Transistor, Avalanche Breakdown, Operating Point, Transistor Amplifiers, Common Emitter Transistor
Typology: Exams
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Semester 2 Examinations 2010/
Module Code: ELTR
School: ELECTRICAL & ELECTRONIC ENGINEERING
Programme Titles: Bachelor of Engineering in Electronic Engineering Bachelor of Engineering (Honours) in Electronic Systems Engineering
Programme Codes: EELXE_7_Y EELES_8_Y
External Examiner(s): Dr A. Donnellan Mr I. Kennedy Internal Examiner: Mr M. O’Gorman
Instructions: Answer Question 1 and TWO other questions. Question 1 (worth 40 marks) is compulsory. Questions 2, 3, and 4 are each worth 30 marks.
Duration: 2 Hours
Sitting: Summer 2011
Requirements for this examination: N/A
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you are attempting the correct examination paper. If in doubt please contact an Invigilator.
Q. 1 (a) (i) Define the current ratios and as applied to a bipolar junction transistor. [4 marks]
(ii) Determine , , and the base current, IB for a transistor where IC = 3.65mA and IE = 3.7mA. [6 marks] (b) Briefly distinguish between avalanche breakdown and zener breakdown in a reverse-biased pn-junction. [10 marks] (c) With reference to transistor amplifiers, explain what is meant by the term stabilisation of the operating point. [10 marks] (d) Explain why the output voltage of a common emitter transistor amplifier is
Q. 2 (a) Draw the circuit diagram of a simple zener diode voltage regulator and briefly explain how the circuit minimises the effects of supply voltage variations on the load voltage. [8 Marks]
(b) A zener diode has a reverse breakdown voltage of 5.6 V and after this may be
nominal 15 V supply. Draw the circuit diagram and calculate the value of series resistor required. [12 Marks]
(c) If the supply voltage were to increase by 20%, calculate the change in the load voltage. [10 Marks]
Figure 1 (a) Explain the function of each of the five labelled components in the common emitter amplifier of Figure 1. [10 Marks]
(b) The amplifier of Figure 1 operates with quiescent conditions VCE = 5.22 V , and IC = 1.96 mA. The base and emitter voltages measured under these conditions
calculate suitable values for the four labelled resistors. [12 Marks]
(c) Calculate the minimum value of CE required to prevent ac feedback if the lowest input signal frequency is 20 Hz. [8 Marks]
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