- 1(current)
Q. No.Zener diodes with breakdown voltage ranging over \((2~\text{V}\text-200~\text{V})\) are commercially available. The breakdown voltage of a Zener diode:
| 1. | increases with increasing doping concentration. |
| 2. | decreases with increasing doping concentration. |
| 3. | does not depend on doping concentration. |
| 4. | may increase or decrease with increasing doping concentration. |
Subtopic: Â Applications of PN junction |
From NCERT
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The input to a Zener diode circuit is an AC-voltage along with a DC: \(V_i=(20~\text V)\sin\omega t+(20~\text V)~\)
Current flows through the diode:
Current flows through the diode:
| 1. | continuously, in both directions |
| 2. | continuously, only in one direction |
| 3. | intermittently, in both directions |
| 4. | intermittently, only in one direction |
Subtopic: Â Applications of PN junction |
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The voltage across \(AB\) i.e., \(V_{AB}\) is varied from \(0\) V to \(12\) V. The current through branch \(XY~(i)\) is measured. Then:
| 1. | \(i=0.05\) A if \(V_{AB}\geq10\) V and \(i=\dfrac{V_{AB}}{200~\Omega}\) otherwise |
| 2. | \(i=0.05\) A if \(V_{AB}\leq10\) V and \(i=\dfrac{V_{AB}}{200~\Omega}\) otherwise |
| 3. | \(i=0.05\) A |
| 4. | \(i=\dfrac{V_{AB}}{200~\Omega}\) |
Subtopic: Â Applications of PN junction |
From NCERT
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\(D_1,D_2\) are ideal diodes and the output \((V_{AB})\) of the transformer is \(10\) V (RMS). The current through the resistor is:
1. zero
2. \(20\) mA
3. \(40\) mA
4. \(20\sqrt2\) mA
1. zero2. \(20\) mA
3. \(40\) mA
4. \(20\sqrt2\) mA
Subtopic: Â Applications of PN junction |
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Choose the incorrect statement about light-emitting diodes (LED):
| 1. | This is a \(\mathrm{p\text-n}\) junction in forward biasing. |
| 2. | Usually, it is prepared from Gallium-arsenide-phosphide. |
| 3. | It can be prepared from \(\mathrm{Si}\) or \(\mathrm{Ge}\). |
| 4. | \(\mathrm{p}\)-layer is kept in the uppermost portion. |
Subtopic: Â Applications of PN junction |
From NCERT
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Consider the circuit shown in the figure given below. The input voltage is \(V_i\), the current through the diode is \(I_z\), and the current through the load is \(I_L\).
| 1. | \(\text{If} ~{V_i}= 15~\text{V}~\text{and}~{R_L}= 40~\Omega, ~\text{and}~{I_z}= 0.375~\text{A}\) |
| 2. | \(\text{If} ~{V_i}= 10~\text{V}~\text{and}~{R_L}= 5~\Omega, ~\text{and}~{I_L}=1~\text{A}\) |
| 3. | \(\text{If} ~{V_i}= 20~\text{V}~\text{and}~{R_L}= 20~\Omega, ~\text{and}~{I_z}= 0.25~\text{A}\) |
| 4. | \(\text{If} ~{V_i}= 20~\text{V}~\text{and}~{R_L}= 10~\Omega, ~\text{and}~{I_z}= 0.5~\text{A}\) |
Subtopic: Â Applications of PN junction |
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From NCERT
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- 1(current)
Q. No.
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