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Question
Answer the following giving reasons:
A p-n junction diode is damaged by a strong current.
Solution
A strong current damages a p-n junction diode because it increases the semiconductor's temperature as more power passes through it and some of that energy is converted to heat. The semiconductor then conducts more current as the temperature rises a little further, which raises the temperature even further. Higher temperature results in more current, which increases the temperature. In what is referred to as a thermal runaway, this cycle keeps repeating. The semiconductor (p-n junction) is ultimately destroyed.
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Consider an npn transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
- Electrons crossover from emitter to collector.
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The breakdown in a reverse biased p–n junction diode is more likely to occur due to ______.
- large velocity of the minority charge carriers if the doping concentration is small.
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- strong electric field in the depletion region if the doping concentration is large.
![]() (a) |
![]() (b) |
- Name the type of a diode whose characteristics are shown in figure (A) and figure (B).
- What does the point P in figure (A) represent?
- What does the points P and Q in figure (B) represent?
A Zener of power rating 1 W is to be used as a voltage regulator. If zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of Rs for safe operation (Figure)?
Read the following paragraph and answer the questions that follow.
A semiconductor diode is basically a pn junction with metallic contacts provided at the ends for the application of an external voltage. It is a two-terminal device. When an external voltage is applied across a semiconductor diode such that the p-side is connected to the positive terminal of the battery and the n-side to the negative terminal, it is said to be forward-biased. When an external voltage is applied across the diode such that the n-side is positive and the p-side is negative, it is said to be reverse-biased. An ideal diode is one whose resistance in forward biasing is zero and the resistance is infinite in reverse biasing. When the diode is forward biased, it is found that beyond forward voltage called knee voltage, the conductivity is very high. When the biasing voltage is more than the knee voltage the potential barrier is overcome and the current increases rapidly with an increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current of about a few microamperes which almost remains constant with bias. This small current is a reverse saturation current. |
- In the given figure, a diode D is connected to an external resistance R = 100 Ω and an emf of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:
(a) 40 mA
(b) 20 mA
(c) 35 mA
(d) 30 mA - In which of the following figures, the pn diode is reverse biased?
(a)
(b)
(c)
(d) - Based on the V-I characteristics of the diode, we can classify the diode as:
(a) bilateral device
(b) ohmic device
(c) non-ohmic device
(d) passive element
OR
Two identical PN junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be:
(b) in the circuits (2) and (3)
(c) in the circuits (1) and (3)
(d) only in the circuit (1)
The V-I characteristic of a diode is shown in the figure. The ratio of the resistance of the diode at I = 15 mA to the resistance at V = -10 V is
(a) 100
(b) 106
(c) 10
(d) 10-6