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Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Electronic: Material, Devices And Simple Circuits4 Marks
Question

When the diode is forward biased, it is found that beyond forward voltage V = Vk, called knee voltage, the conductivity is very high. At this value of battery biasing for p-n junction, the potential barrier is overcome and the current increases rapidly with increase in forward voltage.

When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.

  1. In which of the following figures, the p-n diode is forward biased.
  1.  

  1.  

  1.  

  1.  

  1. Based on the V-I characteristics of the diode, we can classify diode as:
  1. Bi-directional device.
  2. Ohmic device.
  3. Non-ohmic device.
  4. Passive element.
  1. The V-Icharacteristic of a diode is shown in the figure. The ratio of forward to reverse bi as resistance is:

  1. 100
  2. 106
  3. 10
  4. 10-6
  1. In the case of forward biasing of a p-n junction diode, which one of the following figures correctly depicts the direction of conventional current (indicated by an arrow mark)?
  1.  

  1.  

  1.  

  1.  

  1. If an ideal junction diode is connected as shown, then the value of the current I is:

  1. 0.013A
  2. 0.02A
  3. 0.01A
  4. 0.1A
✓

Answer

  1. (c)

Explanation:

The p-n diode is forward biased when p-side is at a higher potential than n-side.

  1. (c) Non-ohmic device.
  2. (d) 10-6

Explanation:

Forward bias resistance,

$\text{R}_1=\frac{\triangle\text{V}}{\triangle\text{I}}=\frac{0.8-0.7}{(20-10)\times10^{-6}}$

$=\frac{0.1}{10\times10^{-3}}=10$

Reverse bias resistance, $\text{R}_2=\frac{10}{1\times10^{-6}}=10^7$

Then, the ratio of forward to reverse bias resistance,

$\frac{\text{R}_1}{\text{R}_2}=\frac{10}{10^7}=10^{-6}$

  1. (d) 

Explanation:

In p-region the direction of conventional current is same as flow of holes. In n-region, the direction of conventional current is opposite to the flow of electrons.

  1. (c) 0.01A

​​​​​​​​​​​​​​Explanation:

In the given circuit the junction diode is forward biased and offers zero resistance.

$\therefore$ The current, $\text{I}=\frac{\text{3V-1V}}{200\Omega}=\frac{\text{2V}}{200\Omega}=0.01\text{A}.$​​​​​​​

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