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प्रश्न
The width of depletion region of p-n junction diode is _______.
(A) 0.5 nm to 1 nm
(B) 5 nm to 10 nm
(C) 50 nm to 500 nm
(D) 500 nm to 1000 nm
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उत्तर
500 nm to 1000 nm
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संबंधित प्रश्न
With the help of a neat circuit diagram, explain the working of a photodiode.
Colour of light emitted by LED depends upon__________________ .
- its forward bias
- its reverse bias
- the band gap of the material of semiconductor
- its size
Meeta's father was driving her to school. At the traffic signal, she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.
Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father?
(ii) What answer did Meeta's father give?
(iii) What are the tiny lights in traffic signals called and how do these operate?
If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased
If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is(ii) reveres biased?
Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in
Forward biasing
How are these characteristics made use of in rectification?
Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in
Reverse biasing
How are these characteristics made use of in rectification?
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true?
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
State its any ‘two’ uses of photodiode.
Describe, with the help of a circuit diagram, the working of a photodiode.
How is a photodiode fabricated?
Draw V − I characteristics of a p-n junction diode. Answer the following questions, giving reasons:
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(ii) Why does the reverse current show a sudden increase at the critical voltage?
Name any semiconductor device which operates under the reverse bias in the breakdown region.
Draw the circuit diagram of a full wave rectifier using p-n junction diode.
Explain its working and show the output, input waveforms.
Show the output waveforms (Y) for the following inputs A and B of (i) OR gate (ii) NAND gate ?
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What happens to the width of depletion player of a p-n junction when it is (i) forward biased, (ii) reverse biased?
Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?
Explain the formation of depletion layer and potential barrier in a p−n junction.
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Write the important considerations which are to be taken into account while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED, if it is required to emit light in the visible range? Draw a circuit diagram and explain its action.
Draw the V-I characteristics of an LED. State two advantages of LED lamps over convertional incandescent lamps.
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Draw solar cell V-I characteristics.
Explain photodiode.
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Name the device which converts the change in intensity of illumination to change in electric current flowing through it. Plot I-V characteristics of this device for different intensities. State any two applications of this device.
Pressure P varies as P = `alpha/beta "exp" (- (alpha x)/"k"_"BT")`, where x denotes the distance, kB is the Boltzmann's constant, T is the absolute temperature and α and β are constant. The dimension of β is ______.
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. lt can detect a signal of wavelength ______.
Which one of the following is not the advantage of LED?
In Figure, assuming the diodes to be ideal ______.
If the resistance R1 is increased (Figure), how will the readings of the ammeter and voltmeter change?
Read the following paragraph and answer the questions.
|
LED is a heavily doped P-N junction which under forward bias emits spontaneous radiation. When it is forward-biased, due to recombination of holes and electrons at the junction, energy is released in the form of photons. In the case of Si and Ge diode, the energy released in recombination lies in the infrared region. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV. The compound semiconductor Gallium Arsenide – Phosphide is used for making LEDs of different colours.
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- Why are LEDs made of compound semiconductor and not of elemental semiconductors?
- What should be the order of bandgap of an LED, if it is required to emit light in the visible range?
- A student connects the blue coloured LED as shown in the figure. The LED did not glow when switch S is closed. Explain why?
OR
iii. Draw V-I characteristic of a p-n junction diode in
(i) forward bias and (ii) reverse bias
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Briefly explain how emf is generated in a solar cell.
Draw solar cells of I-V characteristics.
Draw the circuit diagram of an illuminated photodiode and its I-V characteristics.
The value of power dissipated across the Zener diode (Vz = 15 V) connected in the circuit as shown in the figure is x × 10–1 watt. The value of x, to the nearest integer, is ______.
What energy conversion takes place in a solar cell?
Identify the special purpose diodes 'A', 'B' and 'C' from the given symbols.
Consider the following statements A and B and identify the correct answer:
- For a solar-cell the I-V characteristics lies in the (IV) quadrant of the given graph.
- In a reverse biased p-n junction diode, the current measured in (μA), is due to majority charge carriers.
Consider the following statements A and B and identify the correct answer:
- For a solar-cell, the I-V characteristics lies in the IV quadrant of the given graph.
- In a reverse biased pn junction diode, the current measured in (A), is due to majority charge carriers.
