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MCQOPTIONS
This section includes 521 Mcqs, each offering curated multiple-choice questions to sharpen your Physical Electronics Devices and ICs knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
In an unbiased p-n junction, the junction current at equilibrium is |
| A. | Due to diffusion of majority carriers |
| B. | Due to diffusion of minority carriers |
| C. | Holes are urged to move by the carrier potential |
| D. | The free-electron in the n-region attract the holes |
| Answer» D. The free-electron in the n-region attract the holes | |
| 152. |
Avalanche photodiodes are preferred over PIN diodes in optical communication systems because of |
| A. | Speed of operation |
| B. | Higher sensitivity |
| C. | Larger bandwidth |
| D. | Large power handling capacity |
| Answer» E. | |
| 153. |
For the given transistors |
| A. | 1, 2, 4, 3 |
| B. | 1, 2, 3, 4 |
| C. | 2, 1, 3, 4 |
| D. | 2, 1, 4, 3 |
| Answer» E. | |
| 154. |
The diffusion potential across a p-n junction |
| A. | Decreases with increasing doping concentration |
| B. | Increases with decreasing band gap |
| C. | Does not depend on doping concentration |
| D. | Increases with increasing doping concentration |
| Answer» E. | |
| 155. |
The drift velocity of electrons, in silicon |
| A. | Is proportional to the electric field for all values of electric field |
| B. | Is independent of the electric field |
| C. | Increases at low values of the electric field and decreases at high values of electric field |
| D. | Increases linearly with electric field at low values of electric field and gradually saturates at higher values of electric field |
| Answer» E. | |
| 156. |
The cross sectional area of silicon bar is 100 m |
| A. | 2.58 m |
| B. | 11.36 k |
| C. | 1.36 m |
| D. | 24.8 k |
| Answer» C. 1.36 m | |
| 157. |
A silicon semiconductor sample at T = 300 K is doped with phosphorus atoms at a concentrations of 10 |
| A. | 0.3 eV |
| B. | 0.2 eV |
| C. | 0.1 eV |
| D. | 0.4 eV |
| Answer» B. 0.2 eV | |
| 158. |
In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base-emitter junction |
| A. | Doubles |
| B. | Halves |
| C. | Increases by about 20 mV |
| D. | Decreases by about 20 mV |
| Answer» D. Decreases by about 20 mV | |
| 159. |
A diffused resistor in an IC |
| A. | Is fabricated before transistor diffusion |
| B. | Is fabricated after transistor diffusion |
| C. | Can be fabricated with precision for any resistance value |
| D. | Is formed along with fabrication of transistors |
| Answer» E. | |
| 160. |
Epitaxial growth is a |
| A. | Layer formation of silicon or germanium atoms upon semiconductor wafer when it is heated in an atomsphere containing semiconductor atoms |
| B. | A process in which small pellets of semiconductor are melted into a semiconductor wafer |
| C. | A technique in which holes are etched in a wafer before the allying process |
| D. | An extension of microalloy technique with impurities diffused into base region |
| Answer» B. A process in which small pellets of semiconductor are melted into a semiconductor wafer | |
| 161. |
Epitaxial growth is used in integrated circuit |
| A. | Because it produces low parasitic capacitance |
| B. | Because it yields back-to-back isolating junctions |
| C. | To grow single crystal n-doped silicon on a single crystal p-type substrate |
| D. | To grow selectively single-crystal p-doped silicon of one resistivity on p-type substrate of a different resistivity |
| Answer» D. To grow selectively single-crystal p-doped silicon of one resistivity on p-type substrate of a different resistivity | |
| 162. |
Substrate in a monolithic IC chip has thickness of about |
| A. | 1 mm |
| B. | 5 mm |
| C. | 50 mm |
| D. | 200 mm |
| Answer» C. 50 mm | |
| 163. |
The photoetching process consists in |
| A. | Remove of photoresist |
| B. | Curbing lines on the waver before dicing |
| C. | Diffusing impurities |
| D. | Removed of layer from selected portion |
| Answer» E. | |
| 164. |
A silicon sample doped n-type at 10 |
| A. | 43.2% |
| B. | 78.1% |
| C. | 96.3% |
| D. | 54.3% |
| Answer» C. 96.3% | |
| 165. |
Monolithic integrated circuit system offer greater reliability than discrete-component systems because |
| A. | There are fewer interconnections |
| B. | High-temperature metalizing is used |
| C. | Electric voltage are low |
| D. | Electric elements are closely matched |
| Answer» B. High-temperature metalizing is used | |
| 166. |
The sheet resistance of a semiconductor is |
| A. | An important characteristic of a diffused region especially when used to form diffused resistors |
| B. | An undesirable parasitic element |
| C. | A characteristic whose value determines the required area for a given value of integrated capacitance |
| D. | A parameter whose value is important in a thin-film resistance |
| Answer» B. An undesirable parasitic element | |
| 167. |
The p-type substrate in a monolithic circuit should be connected to |
| A. | Any dc ground point |
| B. | The most negative voltage available in the circuit |
| C. | The most positive voltage available in the circuit |
| D. | No where, i.e. be floating |
| Answer» C. The most positive voltage available in the circuit | |
| 168. |
Silicon dioxide layer is used in IC chips for |
| A. | Providing mechanical strength to the chip |
| B. | Diffusing elements |
| C. | Providing contacts |
| D. | Providing mask against diffusion |
| Answer» E. | |
| 169. |
In the voltage regular circuit shown below the maximum load current i |
| A. | 1.4 mA |
| B. | 2.3 mA |
| C. | 1.8 mA |
| D. | 2.5 mA |
| Answer» B. 2.3 mA | |
| 170. |
If P is passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process is |
| A. | S - R - Q - P |
| B. | R - P - S - Q |
| C. | Q - S - R - P |
| D. | P - Q - R - S |
| Answer» C. Q - S - R - P | |
| 171. |
The equation governing the diffusion of neutral atom is |
| A. | <table><tr><td rowspan="2"></td><td rowspan="2"></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> N</center></td><td rowspan="2">= D<br></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> <sup>2</sup>N</center></td><td rowspan="2"><br></td></tr><td align="center"> t</td><td align="center"> x<sup>2</sup></td></table> |
| B. | <table><tr><td rowspan="2"></td><td rowspan="2"></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> N</center></td><td rowspan="2">= D<br></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> <sup>2</sup>N</center></td><td rowspan="2"><br></td></tr><td align="center"> x</td><td align="center"> t</td></table> |
| C. | <table><tr><td rowspan="2"></td><td rowspan="2"></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> <sup>2</sup>N</center></td><td rowspan="2">= D<br></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> N</center></td><td rowspan="2"><br></td></tr><td align="center"> t<sup>2</sup></td><td align="center"> x</td></table> |
| D. | <table><tr><td rowspan="2"></td><td rowspan="2"></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> <sup>2</sup>N</center></td><td rowspan="2">= D<br></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> N</center></td><td rowspan="2"><br></td></tr><td align="center"> x<sup>2</sup></td><td align="center"> t</td></table> |
| Answer» D. <table><tr><td rowspan="2"></td><td rowspan="2"></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> <sup>2</sup>N</center></td><td rowspan="2">= D<br></td><td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> N</center></td><td rowspan="2"><br></td></tr><td align="center"> x<sup>2</sup></td><td align="center"> t</td></table> | |
| 172. |
Almost all resistor are made in a monolithic IC |
| A. | During the base diffusion |
| B. | During the collector diffusion |
| C. | During the emitter diffusion |
| D. | While growing the epitaxial layer |
| Answer» B. During the collector diffusion | |
| 173. |
The false statement is |
| A. | Capacitor of thin film capacitor made with proper dielectric is not voltage dependent |
| B. | Thin film resistors and capacitor need to be biased for isolation purpose |
| C. | Thin film resistors and capacitor have smaller stray capacitances and leakage currents |
| D. | None of the above |
| Answer» C. Thin film resistors and capacitor have smaller stray capacitances and leakage currents | |
| 174. |
Isolation in ICs is required |
| A. | To make it simpler to test circuits |
| B. | To protect the transistor from possible thermal run away |
| C. | To protect the components mechanical damage |
| D. | To minimize electrical interaction between circuit components |
| Answer» E. | |
| 175. |
In a monolithic-type IC |
| A. | Each transistor is diffused into a separate isolation region |
| B. | All components are fabricated into a single crystal of silicon |
| C. | Resistors and capacitors of any value may be made |
| D. | All isolation problems are eliminated |
| Answer» C. Resistors and capacitors of any value may be made | |
| 176. |
The main purpose of the metalization process is |
| A. | To act as a heat sink |
| B. | To interconnect the various circuit elements |
| C. | To protect the chip from oxidation |
| D. | To supply a bonding surface for mounting the chip |
| Answer» C. To protect the chip from oxidation | |
| 177. |
Increasing the yield of an IC |
| A. | Reduces individual circuit cost |
| B. | Increases the cost of each good circuit |
| C. | Results in a lower number of good chips per wafer |
| D. | Means that more transistor can be fabricated on the same size wafer |
| Answer» B. Increases the cost of each good circuit | |
| 178. |
Silicon dioxide is used in integrated circuits |
| A. | Because of its high heat conduction |
| B. | Because it facilitates the penetration of diffusants |
| C. | To control the location of diffusion and to protect and insulate the silicon surface |
| D. | To control the concentration of diffusants |
| Answer» D. To control the concentration of diffusants | |
| 179. |
In n-well CMOS fabrication substrate is |
| A. | Lightly doped n-type |
| B. | Lightly doped p-type |
| C. | Heavily doped n-type |
| D. | Heavily doped p-type |
| Answer» C. Heavily doped n-type | |
| 180. |
The material popularly used for contacts and interconnections in IC's is |
| A. | Copper |
| B. | Aluminium |
| C. | Silver |
| D. | Zinc |
| Answer» C. Silver | |
| 181. |
The typical number of diffusions used in making epitaxial-diffused silicon integrated circuit is |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| 182. |
The voltages at V |
| A. | 6 V and 5.4 V |
| B. | 5.4 V and 6 V |
| C. | 3 V and 5.4 V |
| D. | 6 V and 3 V |
| Answer» B. 5.4 V and 6 V | |
| 183. |
Silicon diode is less suited for low voltage rectifier operation, because |
| A. | It can withstand high temperature |
| B. | Its reverse saturation current is low |
| C. | Its cut-in voltage is high |
| D. | Its breakdown voltage is high |
| Answer» E. | |
| 184. |
In the voltage regulator circuit shown below. The power rating of zener diode is 400 mW. The value of RL that will establish maximum power in Zener diode is |
| A. | 5 k |
| B. | 2 k |
| C. | 10 k |
| D. | 8 k |
| Answer» C. 10 k | |
| 185. |
In the voltage regulator shown below the power dissipation in the zener diode is |
| A. | 1 mV |
| B. | 1.5 mV |
| C. | 2 mV |
| D. | 0.5 mV |
| Answer» E. | |
| 186. |
The emitter current in a junction transistor with normal bias |
| A. | May be greatly increased by a small change in collector bias |
| B. | Is equal to the sum of the base current and collector current |
| C. | Is approximately equal to the base current |
| D. | Is designated as I |
| E. | <sub>CO</sub> |
| Answer» C. Is approximately equal to the base current | |
| 187. |
One way in which the operation of an NPN transistor differes from that of a PNP transistor is that |
| A. | The emitter junction is reverse-biased in the NPN |
| B. | The emitter injects minority carriers into the base region of the PNP and majority carriers in the base region of the NPN |
| C. | The emitter injects holes into the base region of the PNP and electrons into the base region of the NPN |
| D. | The emitter injects electrons into the base region of the PNP and holes into the base region of the NPN |
| Answer» D. The emitter injects electrons into the base region of the PNP and holes into the base region of the NPN | |
| 188. |
A small increase in the collector reverse-bias will cause |
| A. | A large increase in emitter current |
| B. | A large increase in collector current |
| C. | A large decrease in collector current |
| D. | Very small change in collector reverse saturation current |
| Answer» E. | |
| 189. |
The p-n-p transistor circuit as shown in the above figure is in saturation. Given that V |
| A. | 4.7 mA |
| B. | 5.3 mA |
| C. | 8.6 mA |
| D. | 10 mA |
| Answer» B. 5.3 mA | |
| 190. |
Consider the following semiconductor diodes |
| A. | 1, 2, 3, 4 |
| B. | 3, 4, 1, 2 |
| C. | 3, 1, 4, 2 |
| D. | 1, 3, 2, 4 |
| Answer» D. 1, 3, 2, 4 | |
| 191. |
The g |
| A. | 39.6 mA/V |
| B. | 3.96 mA/V |
| C. | 1.98 mA/V |
| D. | 0.99 mA/V |
| Answer» C. 1.98 mA/V | |
| 192. |
In figure both transistors are identical and have a high value of beta. Take the do base emitter voltage drop as 0.7 volt and kT/q = 25 mV. The small signal low frequency voltage gain (V |
| A. | 0 |
| B. | |
| C. | 1.2 |
| D. | 1.2 |
| Answer» E. | |
| 193. |
The equivalent circuit of a CE transistor is represented by |
| A. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/143oa.png%20"> |
| B. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/143ob.png%20"> |
| C. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/143oc.png%20"> |
| D. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/143od.png%20"> |
| Answer» C. <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/143oc.png%20"> | |
| 194. |
The h parameter equivalent circuit of a junction transistor is valid for |
| A. | High frequency, large signal operation |
| B. | High frequency, small signal operation |
| C. | Low frequency, small signal operation |
| D. | Low frequency, large signal operation |
| Answer» D. Low frequency, large signal operation | |
| 195. |
Consider the following statements associated with bipolar junction transistor and junction gate FET |
| A. | 1 and 2 are correct |
| B. | 2 and 3 are correct |
| C. | 3 and 4 are correct |
| D. | 1 and 4 are correct |
| Answer» C. 3 and 4 are correct | |
| 196. |
In figure, all transistors are identical and have a high value of beta ( ). The voltage V |
| A. | 10 V |
| B. | 0 V |
| C. | 15 V |
| D. | 5 V |
| Answer» E. | |
| 197. |
Consider the circuit shown in Figure (i). If the diode used here has the V-I characteristic as in Figure (ii), then the output waveform V |
| A. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/oa.png"> |
| B. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/ob.png"> |
| C. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/oc.png%20"> |
| D. | <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/od.png"> |
| Answer» D. <img src="http://images.interviewmania.com/wp-content/uploads/2019/05/od.png"> | |
| 198. |
The built-in potential (Diffusion Potential) in a p-n junction |
| A. | Is equal to the difference in the Fermi level of the two sides, expressed in volts |
| B. | Increases with the increase in the doping levels of the two sides |
| C. | Increases with the increase in temperature |
| D. | Is equal to the average of the Fermi levels of the two sides |
| E. | A, B and C |
| Answer» F. | |
| 199. |
For the circuit shown in the figure the voltage gain is | VO/ VS | = AV Given gm = 1 = 100, AV is given by 25 |
| A. | 100 |
| B. | 800 |
| C. | <table><tr><td rowspan="2">80 x</td> <td style="border-bottom:1px solid #000000;vertical-align:bottom;padding-bottom:2px;"><center> 10</center></td><td rowspan="2"><br></td></tr><td align="center">11</td></table> |
| D. | 40 |
| Answer» E. | |
| 200. |
In the common emitter amplifier with RL = 4000 , given hfe = 100, hoe =1, hie = 1000, the current gain | AI | is given by 36 103 |
| A. | 400 |
| B. | 25 |
| C. | 90 |
| D. | 100 |
| Answer» D. 100 | |