Semiconductor Electronics Material Devices and Simple Circuits
150707
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of these
Explanation:
B - In reverse biasing of \(\mathrm{P}-\mathrm{N}\) junction width of the depletion layer increases hence potential barrier increases.
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150724
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of the above
Explanation:
B The p-type material connected to negative terminal pulls the holes away from the junction. The n-type material connected to the positive terminal pulls the electron away from the junction due to this depletion region widens. Hence, reverse bias applied to p-n junction raises potential barrier.
CG PET- 2014
Semiconductor Electronics Material Devices and Simple Circuits
150730
If the forward voltage in a semiconductor diode is changed from \(0.5 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), then the forward current changes by \(1.0 \mathrm{~mA}\). The forward resistance of diode junction will be
1 \(100 \Omega\)
2 \(120 \Omega\)
3 \(200 \Omega\)
4 \(240 \Omega\)
Explanation:
C Given that, Change in forward voltage in a semiconductor diode is \((\Delta \mathrm{V})=(0.7-0.5) \mathrm{V}\) And, change in forward current \((\Delta \mathrm{I})=1.0 \mathrm{~mA}=1 \times 10^{-3}\) A Hence, forward resistance are - \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{0.7-0.5}{1 \times 10^{-3}}=\frac{0.2}{1 \times 10^{-3}}=200 \Omega\)
BITSAT-2020
Semiconductor Electronics Material Devices and Simple Circuits
150738
With in depletion region of \(p-n\) junction diode :
1 p-side is positive and n-side is negative
2 p-side is negative and n-side is positive
3 both sides are positive or both negative
4 both sides are neutral
Explanation:
B With in depletion region of \(p-n\) junction diode \(\mathrm{p}\)-side is negative and \(\mathrm{n}\)-side is positive. The depletion layer of \(p-n\) junction diode has immobile holes and immobile electrons. The holes are positively charged are called positive ions. The free electrons negatively charged and are called the negative ions due to polarization of charge carrier.
Semiconductor Electronics Material Devices and Simple Circuits
150707
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of these
Explanation:
B - In reverse biasing of \(\mathrm{P}-\mathrm{N}\) junction width of the depletion layer increases hence potential barrier increases.
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150724
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of the above
Explanation:
B The p-type material connected to negative terminal pulls the holes away from the junction. The n-type material connected to the positive terminal pulls the electron away from the junction due to this depletion region widens. Hence, reverse bias applied to p-n junction raises potential barrier.
CG PET- 2014
Semiconductor Electronics Material Devices and Simple Circuits
150730
If the forward voltage in a semiconductor diode is changed from \(0.5 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), then the forward current changes by \(1.0 \mathrm{~mA}\). The forward resistance of diode junction will be
1 \(100 \Omega\)
2 \(120 \Omega\)
3 \(200 \Omega\)
4 \(240 \Omega\)
Explanation:
C Given that, Change in forward voltage in a semiconductor diode is \((\Delta \mathrm{V})=(0.7-0.5) \mathrm{V}\) And, change in forward current \((\Delta \mathrm{I})=1.0 \mathrm{~mA}=1 \times 10^{-3}\) A Hence, forward resistance are - \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{0.7-0.5}{1 \times 10^{-3}}=\frac{0.2}{1 \times 10^{-3}}=200 \Omega\)
BITSAT-2020
Semiconductor Electronics Material Devices and Simple Circuits
150738
With in depletion region of \(p-n\) junction diode :
1 p-side is positive and n-side is negative
2 p-side is negative and n-side is positive
3 both sides are positive or both negative
4 both sides are neutral
Explanation:
B With in depletion region of \(p-n\) junction diode \(\mathrm{p}\)-side is negative and \(\mathrm{n}\)-side is positive. The depletion layer of \(p-n\) junction diode has immobile holes and immobile electrons. The holes are positively charged are called positive ions. The free electrons negatively charged and are called the negative ions due to polarization of charge carrier.
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Semiconductor Electronics Material Devices and Simple Circuits
150707
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of these
Explanation:
B - In reverse biasing of \(\mathrm{P}-\mathrm{N}\) junction width of the depletion layer increases hence potential barrier increases.
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150724
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of the above
Explanation:
B The p-type material connected to negative terminal pulls the holes away from the junction. The n-type material connected to the positive terminal pulls the electron away from the junction due to this depletion region widens. Hence, reverse bias applied to p-n junction raises potential barrier.
CG PET- 2014
Semiconductor Electronics Material Devices and Simple Circuits
150730
If the forward voltage in a semiconductor diode is changed from \(0.5 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), then the forward current changes by \(1.0 \mathrm{~mA}\). The forward resistance of diode junction will be
1 \(100 \Omega\)
2 \(120 \Omega\)
3 \(200 \Omega\)
4 \(240 \Omega\)
Explanation:
C Given that, Change in forward voltage in a semiconductor diode is \((\Delta \mathrm{V})=(0.7-0.5) \mathrm{V}\) And, change in forward current \((\Delta \mathrm{I})=1.0 \mathrm{~mA}=1 \times 10^{-3}\) A Hence, forward resistance are - \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{0.7-0.5}{1 \times 10^{-3}}=\frac{0.2}{1 \times 10^{-3}}=200 \Omega\)
BITSAT-2020
Semiconductor Electronics Material Devices and Simple Circuits
150738
With in depletion region of \(p-n\) junction diode :
1 p-side is positive and n-side is negative
2 p-side is negative and n-side is positive
3 both sides are positive or both negative
4 both sides are neutral
Explanation:
B With in depletion region of \(p-n\) junction diode \(\mathrm{p}\)-side is negative and \(\mathrm{n}\)-side is positive. The depletion layer of \(p-n\) junction diode has immobile holes and immobile electrons. The holes are positively charged are called positive ions. The free electrons negatively charged and are called the negative ions due to polarization of charge carrier.
Semiconductor Electronics Material Devices and Simple Circuits
150707
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of these
Explanation:
B - In reverse biasing of \(\mathrm{P}-\mathrm{N}\) junction width of the depletion layer increases hence potential barrier increases.
Manipal UGET -2020
Semiconductor Electronics Material Devices and Simple Circuits
150724
The effect of reverse bias in a junction diode on its potential barrier is
1 increases
2 decreases
3 remains same
4 None of the above
Explanation:
B The p-type material connected to negative terminal pulls the holes away from the junction. The n-type material connected to the positive terminal pulls the electron away from the junction due to this depletion region widens. Hence, reverse bias applied to p-n junction raises potential barrier.
CG PET- 2014
Semiconductor Electronics Material Devices and Simple Circuits
150730
If the forward voltage in a semiconductor diode is changed from \(0.5 \mathrm{~V}\) to \(0.7 \mathrm{~V}\), then the forward current changes by \(1.0 \mathrm{~mA}\). The forward resistance of diode junction will be
1 \(100 \Omega\)
2 \(120 \Omega\)
3 \(200 \Omega\)
4 \(240 \Omega\)
Explanation:
C Given that, Change in forward voltage in a semiconductor diode is \((\Delta \mathrm{V})=(0.7-0.5) \mathrm{V}\) And, change in forward current \((\Delta \mathrm{I})=1.0 \mathrm{~mA}=1 \times 10^{-3}\) A Hence, forward resistance are - \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{0.7-0.5}{1 \times 10^{-3}}=\frac{0.2}{1 \times 10^{-3}}=200 \Omega\)
BITSAT-2020
Semiconductor Electronics Material Devices and Simple Circuits
150738
With in depletion region of \(p-n\) junction diode :
1 p-side is positive and n-side is negative
2 p-side is negative and n-side is positive
3 both sides are positive or both negative
4 both sides are neutral
Explanation:
B With in depletion region of \(p-n\) junction diode \(\mathrm{p}\)-side is negative and \(\mathrm{n}\)-side is positive. The depletion layer of \(p-n\) junction diode has immobile holes and immobile electrons. The holes are positively charged are called positive ions. The free electrons negatively charged and are called the negative ions due to polarization of charge carrier.
