

MCQOPTIONS
Saved Bookmarks
This section includes 6 Mcqs, each offering curated multiple-choice questions to sharpen your Linear Integrated Circuit knowledge and support exam preparation. Choose a topic below to get started.
1. |
For a full wave rectification, in a low voltage ac voltmeter, the meter current can be expressed a? |
A. | I<sub>o</sub> = (1.9√óV<sub>in</sub>)/R<sub>1</sub> |
B. | I<sub>o</sub> = (3.9√óV<sub>in</sub>)/R<sub>1</sub> |
C. | I<sub>o</sub> = (0.9√óV<sub>in</sub>)/R<sub>1</sub> |
D. | I<sub>o</sub> = (2.9√óV<sub>in</sub>)/R<sub>1</sub> |
Answer» D. I<sub>o</sub> = (2.9‚Äö√†√∂‚àö‚â•V<sub>in</sub>)/R<sub>1</sub> | |
2. |
The current to voltage converter photosensitive device can be used as |
A. | Light intensity meter |
B. | Light radiating meter |
C. | Light deposition meter |
D. | None of the mentioned |
Answer» B. Light radiating meter | |
3. |
If the input applied to DAC using current to voltage converter is 10110100, determine the reference voltage (Assume Io= 2mA and R1=1.2kΩ)$ |
A. | 53.1v |
B. | 3.41v |
C. | 9.21v |
D. | 67.34v |
Answer» C. 9.21v | |
4. |
Which cell can be used instead of a photocell to obtain active transducer in photosensitive devices? |
A. | Photovoltaic cell |
B. | Photo diode |
C. | Photo sensor |
D. | All of the mentioned |
Answer» B. Photo diode | |
5. |
Determine the maximum value of output current of the DAC in MC1408? |
A. | 0.773√ó(V<sub>ref</sub>/R<sub>1</sub>) |
B. | 0.448√ó(V<sub>ref</sub>/R<sub>1</sub>) |
C. | 0.996√ó(V<sub>ref</sub>/R<sub>1</sub>) |
D. | 0.224√ó(V<sub>ref</sub>/R<sub>1</sub>) |
Answer» D. 0.224‚Äö√†√∂‚àö‚â•(V<sub>ref</sub>/R<sub>1</sub>) | |
6. |
The output current equation for MC1408 digital to analog converter would be |
A. | I<sub>o</sub>= -(V<sub>ref</sub>/R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)]. |
B. | I<sub>o</sub>= (V<sub>ref</sub>/2R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)]. |
C. | I<sub>o</sub>= (V<sub>ref</sub>/R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)]. |
D. | I<sub>o</sub>= -(V<sub>ref</sub>/2R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)]. |
Answer» D. I<sub>o</sub>= -(V<sub>ref</sub>/2R<sub>1</sub>)‚Äö√†√∂‚àö‚â•[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)]. | |