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ASSIGNMENT Answer the following question. a) With reference to a p-n-p transistor, explain briefly what is meant by the term transistor action and why a bipolar junction transistor is so named. b)...

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ASSIGNMENT
  1. Answer the following question.



  1. a) With reference to a p-n-p transistor, explain briefly what is meant by the term transistor action and why a bipolar junction transistor is so named.

b) With the aid of a circuit diagram, explain how the input and output characteristics of an n-p-n transistor having a common-base configuration can be obtained.
c) An n-p-n transistor has the following characteristics, which may be assumed to be linear between the values of collector voltage stated.

  1. a) The output voltage from an amplifier is 4 V. If the voltage gain is 27 dB, calculate the value of the input voltage assuming that the amplifier input resistance and load resistance are equal.

b) Explain how thermal runaway might be prevented in a transistor.
c) A differential amplifier has an open-loop voltage gain of 120. The input signals are 2.45 V and 2.35 V. Calculate the output voltage of the amplifier.
d) A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and the CMRR.
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Submission deadline: 05 February, 2013 ASSIGNMENT Answer the following question. a) With reference to a p-n-p transistor, explain briefly what is meant by the term transistor action and why a bipolar junction transistor is so named. b) With the aid of a circuit diagram, explain how the input and output characteristics of an n-p-n transistor having a common-base configuration can be obtained. c) An n-p-n transistor has the following characteristics, which may be assumed to be linear between the values of collector voltage stated. a) The output voltage from an amplifier is 4 V. If the voltage gain is 27 dB, calculate the value of the input voltage assuming that the amplifier input resistance and load resistance are equal. b) Explain how thermal runaway might be prevented in a transistor. c) A differential amplifier has an open-loop voltage gain of 120. The input signals are 2.45 V and 2.35 V. Calculate the output voltage of the amplifier. d) A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and the CMRR. What to Submit: Your answers should be in a report format. Your report should not exceed 30 pages (A4). You may use an Appendix for any additional but not directly related information. You may visit the following IET website for a help in report writing format:  HYPERLINK "http://www.theiet.org/students/resources/final-year-project/writing-reports.cfm" http://www.theiet.org/students/resources/final-year-project/writing-reports.cfm List of references (preferably in IEEE numeric style format). Wikipedia must not be used as a reference. See the link below for details:  HYPERLINK "http://http-server.carleton.ca/~nartemev/IEEE_style.html" http://http-server.carleton.ca/~nartemev/IEEE_style.html 1 page (A4) self-evaluation your work identifying strengths, weaknesses and areas for...

Answered Same Day Dec 22, 2021

Solution

David answered on Dec 22 2021
124 Votes
ASSIGNMENT

Solution: 1(a) The various rules are:
Rule1: p.q Rule2: r.s
Rule3: (p ⊕ q).(r + s) Rule4: IF s.q = 1 THEN f = 0.
The truth table for the given set of rules is as tabulated below:
Temp.
Sensor 1
Temp.
Sensor 2
Motion
Detector
Sound
Detector

Output

Note
p q r s f
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1 Motion and sound
0 1 0 0 0
0 1 0 1 0 Temp and sound BUT excluded
0 1 1 0 1 Temp and motion
0 1 1 1 0 Motion and sound BUT excluded
1 0 0 0 0
1 0 0 1 1 Temp and sound
1 0 1 0 1 Temp and motion
1 0 1 1 1 Motion and Sound
1 1 0 0 1 Both temp
1 1 0 1 0 Both temp BUT excluded
1 1 1 0 1 Both temp
1 1 1 1 0 Both temp BUT excluded
Solution: 1(b) The Boolean expression for the above truth table is as expressed below:
f = p’q’rs + p’qr’s + p’qrs’ + pq’r’s + pq’rs’ + pq’rs + pqr’s’ + pqrs’
Solution:1(c) Let us draw the K-MAP for the above function:
f= qrs’ + pqs’ + q’rs + pq’s + prs’
Solution: 1(d) Either one of the following diagrams is fine.


Solution: 2(a) For p-n-p transistor as shown in fig. 2a(i), the emitter is relatively heavily doped with
holes. When we make the emitter terminal sufficiently positive with respect to the base, the
majority ca
iers in the emitter i.e. holes start drifting from the emitter to the base due to forward
iasing of base-emitter junction.

Fig. 2a(i)
Base of the transistor is relatively lightly doped with electrons and also the base width is smaller
than the diffusion length of the electrons. Hence only a few electron-hole recombinations take
place, around 0.5%, and hence most of the holes entering the base, do not combine with electrons.

The base-collector junction is RB to e- in the base, but FB to holes in the base region. As the base
width is very small and is packed with holes presently, these holes pass through the base-emitter
junction and then move towards the negative potential of the collector terminal. Hence, the control
of cu
ent is independent of the collector-base voltage and almost governed by the emitter-base
voltage. The essence of transistor action is this cu
ent control by means of the base-emitter voltage.

Fig. 2(ii)
The reason for its name as Bipolar Junction Transistor is because herein a transistor both types of
charge ca
iers (holes and electrons) take part in the cu
ent flow and hence the name has the term
ipolar. The transistor also comprises two p-n junctions and hence it is a junction transistor; hence
the name – bipolar junction transistor.

Solution: 2(b) The circuit diagram for an n-p-n transistor connected in common-base configuration
to obtain the input and output characteristics is as...
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