NEET Semiconductor Physics MCQs & PYQs

Question 1

A p–n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :

Answer

Energy of EM wave E = hυ

⇒ 2 eV = 6.626 × 10^-34×υ

⇒ \[v= 5\times 10_{14} Hz\]

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Question 2

The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform.

Answer

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Question 3

To get an output 1 from the circuit shown in the figure, the input may be :

Answer

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Question 4

The combination of ‘NAND’ gates shown here under (figure) are equivalent to

Answer

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Question 5

In the following circuit the equivalent resistance between A and B is

Answer

Both the diodes are reversed biased so can be replaced by open circuit.

Req= 8+2+6= 16 ohm

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Question 6

GaAs (with a band gap = 1.5 eV) as an LED can emit :

Answer

GaAs (Gallium Arsenide) with a band gap of 1.5 eV can emit light in the infrared region of the electromagnetic spectrum.

The energy of a photon emitted by an LED is related to its wavelength by the equation:

$E = \frac{hc}{\lambda}$

Solving we get

$λ= 826 nm which lies in infrared region.💬 Discuss this QuestionQuestion 27: moderate If break down voltage of zener diode is 6 V then value of I2 will be : 1. 18 mA 2. 24 mA 3. 6 mA 4. Zero View Answer Voltage across 500 ohm resistor = 12 Volt I = 12/500 = 24 mA I 1 = 6/1000= 6mA I2= I - I1= (24-6 )mA= 18 mA 💬 Discuss this QuestionQuestion 28: moderate The following configuration of gate is equivalent to 1. NAND 2. XOR 3. OR 4. None of these View Answer 💬 Discuss this QuestionQuestion 29: moderate Determine V0 for the network 1. 6.6 V 2. 9 V 3. -9 V 4. -6.6 V View Answer Both the diodes are forward biased so current will flow through the circuit i= 10/4.5= 20/9 mA V0= i R= (20/9)×3.3 Volt= 6.6 volt 💬 Discuss this QuestionQuestion 30: moderate The following figure shows a logic gate circuit with two inputs A and B and the output Y. The voltage waveforms of A, B, and Y are as given: The logic gate is : 1. OR 2. AND 3. NAND 4. NOR View Answer 💬 Discuss this Question« Previous 12345 … 13 Next »{"@context":"https://schema.org","@type":"FAQPage","mainEntity":[{"@type":"Question","name":"A p–n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :","acceptedAnswer":{"@type":"Answer","text":"Energy of EM wave E = hυ\r\n\r\n⇒ 2 eV = 6.626 × 10^-34×υ\r\n\r\n⇒ \\[v= 5\\times 10_{14} Hz\\]"}},{"@type":"Question","name":"In the following circuit the equivalent resistance between A and B is","acceptedAnswer":{"@type":"Answer","text":"Both the diodes are reversed biased so can be replaced by open circuit.\r\n\r\nReq= 8+2+6= 16 ohm"}},{"@type":"Question","name":"GaAs (with a band gap = 1.5 eV) as an LED can emit :","acceptedAnswer":{"@type":"Answer","text":"GaAs (Gallium Arsenide) with a band gap of 1.5 eV can emit light in the infrared region of the electromagnetic spectrum.\r\n\r\nThe energy of a photon emitted by an LED is related to its wavelength by the equation:\r\n\r\n \r\n\r\nE=hcλE = \\frac{hc}{\\lambda}\r\n\r\nSolving we get\r\n\r\n λ= 826 nm which lies in infrared region. \r\n\r\n\r\n\r\n "}},{"@type":"Question","name":"If break down voltage of zener diode is 6 V then value of I2 will be :\r\n\r\n ","acceptedAnswer":{"@type":"Answer","text":"Voltage across 500 ohm resistor = 12 Volt\r\n\r\nI = 12/500 = 24 mA\r\n\r\nI 1 = 6/1000= 6mA\r\n\r\nI2= I - I1= (24-6 )mA= 18 mA"}},{"@type":"Question","name":"Determine V0 for the network","acceptedAnswer":{"@type":"Answer","text":"Both the diodes are forward biased so current will flow through the circuit\r\n\r\ni= 10/4.5= 20/9 mA\r\n\r\nV0= i R= (20/9)×3.3 Volt= 6.6 volt"}}]}document.addEventListener('DOMContentLoaded', function () {
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