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201. |
Complement of F’ gives back |
A. | f’ |
B. | f |
C. | ff |
D. | ff’ |
Answer» B. f |
202. |
Decimal digit in BCD can be represented by |
A. | 1 input line |
B. | 2 input lines |
C. | 3 input lines |
D. | 4 input lines |
Answer» D. 4 input lines |
203. |
The number of logic gates and the way of their interconnections can be classified as |
A. | logical network |
B. | system network |
C. | circuit network |
D. | gate network |
Answer» A. logical network |
204. |
EPROM uses an array of |
A. | p-channel enhancement type mosfet |
B. | n-channel enhancement type mosfet |
C. | p-channel depletion type mosfet |
D. | n-channel depletion type mosfet |
Answer» B. n-channel enhancement type mosfet |
205. |
The EPROM was invented by |
A. | wen tsing chow |
B. | dov frohman |
C. | luis o brian |
D. | j p longwell |
Answer» B. dov frohman |
206. |
Address decoding for dynamic memory chip control may also be used for |
A. | chip selection and address location |
B. | read and write control |
C. | controlling refresh circuits |
D. | memory mapping |
Answer» A. chip selection and address location |
207. |
Which of the following describes the action of storing a bit of data in a mask ROM? |
A. | a 0 is stored by connecting the gate of a mos cell to the address line |
B. | a 0 is stored in a bipolar cell by shorting the base connection to the address line |
C. | a 1 is stored by connecting the gate of a mos cell to the address line |
D. | a 1 is stored in a bipolar cell by opening the base connection to the address line |
Answer» C. a 1 is stored by connecting the gate of a mos cell to the address line |
208. |
The check sum method of testing a ROM |
A. | allows data errors to be pinpointed to a specific memory location |
B. | provides a means for locating and correcting data errors in specific memory locations |
C. | indicates if the data in more than one memory location is incorrect |
D. | simply indicates that the contents of the rom are incorrect |
Answer» D. simply indicates that the contents of the rom are incorrect |
209. |
The initial values in all the cells of an EPROM is |
A. | 0 |
B. | 1 |
C. | both 0 and 1 |
D. | alternate 0s and 1s |
Answer» B. 1 |
210. |
To store 0 in such a cell, the floating point must be |
A. | reprogrammed |
B. | restarted |
C. | charged |
D. | power off |
Answer» C. charged |
211. |
The major disadvantage of RAM is? |
A. | its access speed is too slow |
B. | its matrix size is too big |
C. | it is volatile |
D. | high power consumption |
Answer» C. it is volatile |
212. |
Which one of the following is used for the fabrication of MOS EPROM? |
A. | tms 2513 |
B. | tms 2515 |
C. | tms 2516 |
D. | tms 2518 |
Answer» C. tms 2516 |
213. |
How many addresses a MOS EPROM have? |
A. | 1024 |
B. | 512 |
C. | 2516 |
D. | 256 |
Answer» C. 2516 |
214. |
ROMs retain data when |
A. | power is on |
B. | power is off |
C. | system is down |
D. | all of the mentioned |
Answer» D. all of the mentioned |
215. |
When a RAM module passes the checker board test it is |
A. | able to read and write only 0s |
B. | faulty |
C. | probably good |
D. | able to read and write only 1s |
Answer» C. probably good |
216. |
What is the difference between static RAM and dynamic RAM? |
A. | static ram must be refreshed, dynamic ram does not |
B. | there is no difference |
C. | dynamic ram must be refreshed, static ram does not |
D. | sram is slower than dram |
Answer» C. dynamic ram must be refreshed, static ram does not |
217. |
How many natural states will there be in a 4-bit ripple counter? |
A. | 4 |
B. | 8 |
C. | 16 |
D. | 32 |
Answer» C. 16 |
218. |
A ripple counter’s speed is limited by the propagation delay of |
A. | each flip-flop |
B. | all flip-flops and gates |
C. | the flip-flops only with gates |
D. | only circuit gates |
Answer» A. each flip-flop |
219. |
One of the major drawbacks to the use of asynchronous counters is that |
A. | low-frequency applications are limited because of internal propagation delays |
B. | high-frequency applications are limited because of internal propagation delays |
C. | asynchronous counters do not have major drawbacks and are suitable for use in high- and low-frequency counting applications |
D. | asynchronous counters do not have propagation delays, which limits their use in high- frequency applications |
Answer» B. high-frequency applications are limited because of internal propagation delays |
220. |
Internal propagation delay of asynchronous counter is removed by |
A. | ripple counter |
B. | ring counter |
C. | modulus counter |
D. | synchronous counter |
Answer» D. synchronous counter |
221. |
How many flip-flops are required to construct a decade counter? |
A. | 4 |
B. | 8 |
C. | 5 |
D. | 10 |
Answer» A. 4 |
222. |
The terminal count of a typical modulus-10 binary counter is |
A. | 0000 |
B. | 1010 |
C. | 1001 |
D. | 1111 |
Answer» C. 1001 |
223. |
A ripple counter’s speed is limited by the propagation delay of |
A. | each flip-flop |
B. | all flip-flops and gates |
C. | the flip-flops only with gates |
D. | only circuit gates |
Answer» A. each flip-flop |
224. |
A 4-bit counter has a maximum modulus of |
A. | 3 |
B. | 6 |
C. | 8 |
D. | 16 |
Answer» D. 16 |
225. |
UP-DOWN counter is a combination of |
A. | latches |
B. | flip-flops |
C. | up counter |
D. | up counter & down counter |
Answer» D. up counter & down counter |
226. |
UP-DOWN counter is also known as |
A. | dual counter |
B. | multi counter |
C. | multimode counter |
D. | two counter |
Answer» C. multimode counter |
227. |
In an UP-counter, each flip-flop is triggered by |
A. | the output of the next flip-flop |
B. | the normal output of the preceding flip-flop |
C. | the clock pulse of the previous flip-flop |
D. | the inverted output of the preceding flip-flop |
Answer» B. the normal output of the preceding flip-flop |
228. |
In DOWN-counter, each flip-flop is triggered by |
A. | the output of the next flip-flop |
B. | the normal output of the preceding flip-flop |
C. | the clock pulse of the previous flip-flop |
D. | the inverted output of the preceding flip-flop |
Answer» D. the inverted output of the preceding flip-flop |
229. |
Binary counter that count incrementally and decrement is called |
A. | up-down counter |
B. | lsi counters |
C. | down counter |
D. | up counter |
Answer» A. up-down counter |
230. |
Once an up-/down-counter begins its count sequence, it |
A. | starts counting |
B. | can be reversed |
C. | can’t be reversed |
D. | can be altered |
Answer» D. can be altered |
231. |
In 4-bit up-down counter, how many flip-flops are required? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 5 |
Answer» C. 4 |
232. |
A modulus-10 counter must have _ |
A. | 10 flip-flops |
B. | 4 flip-flops |
C. | 2 flip-flops |
D. | synchronous clocking |
Answer» B. 4 flip-flops |
233. |
Which is not an example of a truncated modulus? |
A. | 8 |
B. | 9 |
C. | 11 |
D. | 15 |
Answer» A. 8 |
234. |
The designation means that the |
A. | up count is active-high, the down count is active-low |
B. | up count is active-low, the down count is active-high |
C. | up and down counts are both active-low |
D. | up and down counts are both active-high |
Answer» A. up count is active-high, the down count is active-low |
235. |
The full form of SIPO is |
A. | serial-in parallel-out |
B. | parallel-in serial-out |
C. | serial-in serial-out |
D. | serial-in peripheral-out |
Answer» A. serial-in parallel-out |
236. |
How can parallel data be taken out of a shift register simultaneously? |
A. | use the q output of the first ff |
B. | use the q output of the last ff |
C. | tie all of the q outputs together |
D. | use the q output of each ff |
Answer» D. use the q output of each ff |
237. |
What is meant by parallel load of a shift register? |
A. | all ffs are preset with data |
B. | each ff is loaded with data, one at a time |
C. | parallel shifting of data |
D. | all ffs are set with data |
Answer» A. all ffs are preset with data |
238. |
After three clock pulses, the register contains |
A. | 01110 |
B. | 00001 |
C. | 00101 |
D. | 00110 |
Answer» C. 00101 |
239. |
What will be the 4-bit pattern after the second clock pulse? (Right-most bit first) |
A. | 1100 |
B. | 0011 |
C. | 0000 |
D. | 1111 |
Answer» C. 0000 |
240. |
In digital logic, a counter is a device which |
A. | counts the number of outputs |
B. | stores the number of times a particular event or process has occurred |
C. | stores the number of times a clock pulse rises and falls |
D. | counts the number of inputs |
Answer» B. stores the number of times a particular event or process has occurred |
241. |
A counter circuit is usually constructed of |
A. | a number of latches connected in cascade form |
B. | a number of nand gates connected in cascade form |
C. | a number of flip-flops connected in cascade |
D. | a number of nor gates connected in cascade form |
Answer» C. a number of flip-flops connected in cascade |
242. |
How many types of the counter are there? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 5 |
Answer» B. 3 |
243. |
A decimal counter has states. |
A. | 5 |
B. | 10 |
C. | 15 |
D. | 20 |
Answer» B. 10 |
244. |
Ripple counters are also called |
A. | ssi counters |
B. | asynchronous counters |
C. | synchronous counters |
D. | vlsi counters |
Answer» B. asynchronous counters |
245. |
Synchronous counter is a type of |
A. | ssi counters |
B. | lsi counters |
C. | msi counters |
D. | vlsi counters |
Answer» C. msi counters |
246. |
Three decade counter would have |
A. | 2 bcd counters |
B. | 3 bcd counters |
C. | 4 bcd counters |
D. | 5 bcd counters |
Answer» B. 3 bcd counters |
247. |
BCD counter is also known as |
A. | parallel counter |
B. | decade counter |
C. | synchronous counter |
D. | vlsi counter |
Answer» B. decade counter |
248. |
The parallel outputs of a counter circuit represent the |
A. | parallel data word |
B. | clock frequency |
C. | counter modulus |
D. | clock count |
Answer» D. clock count |
249. |
The time from the beginning of a read cycle to the end of tACS/tAA is called as |
A. | write enable time |
B. | data hold |
C. | read cycle time |
D. | access time |
Answer» D. access time |
250. |
Why did PROM introduced? |
A. | to increase the storage capacity |
B. | to increase the address locations |
C. | to provide flexibility |
D. | to reduce the size |
Answer» C. to provide flexibility |
251. |
Which of the following is programmed electrically by the user? |
A. | rom |
B. | eprom |
C. | prom |
D. | eeprom |
Answer» C. prom |
252. |
PROMs are available in |
A. | bipolar and mosfet technologies |
B. | mosfet and fet technologies |
C. | fet and bipolar technologies |
D. | mos and bipolar technologies |
Answer» D. mos and bipolar technologies |
253. |
Which of the following best describes the fusible-link PROM? |
A. | manufacturer-programmable, reprogrammable |
B. | manufacturer-programmable, one-time programmable |
C. | user-programmable, reprogrammable |
D. | user-programmable, one-time programmable |
Answer» D. user-programmable, one-time programmable |
254. |
How can ultraviolet erasable PROMs be recognized? |
A. | there is a small window on the chip |
B. | they will have a small violet dot next to the #1 pin |
C. | their part number always starts with a “u”, such as in u12 |
D. | they are not readily identifiable, since they must always be kept under a small cover |
Answer» A. there is a small window on the chip |
255. |
Which part of a Flash memory architecture manages all chip functions? |
A. | program verify code |
B. | floating-gate mosfet |
C. | command code |
D. | input/output pins |
Answer» B. floating-gate mosfet |
256. |
How much locations an 8-bit address code can select in memory? |
A. | 8 locations |
B. | 256 locations |
C. | 65,536 locations |
D. | 131,072 locations |
Answer» B. 256 locations |
257. |
What is a fusing process? |
A. | it is a process by which data is passed to the memory |
B. | it is a process by which data is read through the memory |
C. | it is a process by which programs are burnout to the diode/transistors |
D. | it is a process by which data is fetched through the memory |
Answer» C. it is a process by which programs are burnout to the diode/transistors |
258. |
Fusing process is |
A. | reversible |
B. | irreversible |
C. | synchronous |
D. | asynchronous |
Answer» B. irreversible |
259. |
The cell type used inside a PROM is |
A. | link cells |
B. | metal cells |
C. | fuse cells |
D. | electric cells |
Answer» C. fuse cells |
260. |
How many types of fuse technologies are used in PROMs? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 5 |
Answer» B. 3 |
261. |
Metal links are made up of |
A. | polycrystalline |
B. | magnesium sulphide |
C. | nichrome |
D. | silicon dioxide |
Answer» C. nichrome |
262. |
Silicon links are made up of |
A. | polycrystalline silicon |
B. | polycrystalline magnesium |
C. | nichrome |
D. | silicon dioxide |
Answer» A. polycrystalline silicon |
263. |
During programming p-n junction is |
A. | avalanche reverse biased |
B. | avalanche forward biased |
C. | zener reverse biased |
D. | zener reverse biased |
Answer» A. avalanche reverse biased |
264. |
The full form of FAMOS is |
A. | floating gate avalanche injection mos |
B. | float gate avalanche injection mos |
C. | floating gate avalanche induction mos |
D. | float gate avalanche induction mos |
Answer» A. floating gate avalanche injection mos |
265. |
PROM is programmed by |
A. | eprom programmer |
B. | eeprom programmer |
C. | prom programmer |
D. | rom programmer |
Answer» C. prom programmer |
266. |
The PROM starts out with |
A. | 1s |
B. | 0s |
C. | null |
D. | both 1s and 0s |
Answer» B. 0s |
267. |
For implementation of PROM, which IC is used? |
A. | ic 74187 |
B. | ic 74186 |
C. | ic 74185 |
D. | ic 74184 |
Answer» B. ic 74186 |
268. |
IC 74186 is of |
A. | 1024 bits |
B. | 32 bits |
C. | 512 bits |
D. | 64 bits |
Answer» C. 512 bits |
269. |
How many memory locations are addressed using 18 address bits? |
A. | 165,667 |
B. | 245,784 |
C. | 262,144 |
D. | 212,342 |
Answer» C. 262,144 |
270. |
How many address bits are needed to operate a 2K * 8-bit memory? |
A. | 10 |
B. | 11 |
C. | 12 |
D. | 13 |
Answer» B. 11 |
271. |
What is the bit storage capacity of a ROM with a 1024 × 8 organization? |
A. | 1024 |
B. | 4096 |
C. | 2048 |
D. | 8192 |
Answer» D. 8192 |
272. |
The logical sum of two or more logical product terms is called |
A. | sop |
B. | pos |
C. | or operation |
D. | nand operation |
Answer» A. sop |
273. |
The expression Y=AB+BC+AC shows the operation. |
A. | ex-or |
B. | sop |
C. | pos |
D. | nor |
Answer» B. sop |
274. |
The expression Y=(A+B)(B+C)(C+A) shows the operation. |
A. | and |
B. | pos |
C. | sop |
D. | nand |
Answer» B. pos |
275. |
The canonical sum of product form of the function y(A,B) = A + B is |
A. | ab + bb + a’a |
B. | ab + ab’ + a’b |
C. | ba + ba’ + a’b’ |
D. | ab’ + a’b + a’b’ |
Answer» B. ab + ab’ + a’b |
276. |
A variable on its own or in its complemented form is known as a |
A. | product term |
B. | literal |
C. | sum term |
D. | word |
Answer» B. literal |
277. |
Canonical form is a unique way of representing |
A. | sop |
B. | minterm |
C. | boolean expressions |
D. | pos |
Answer» C. boolean expressions |
278. |
There are Minterms for 3 variables (a, b, c). |
A. | 0 |
B. | 2 |
C. | 8 |
D. | 1 |
Answer» C. 8 |
279. |
Why is a demultiplexer called a data distributor? |
A. | the input will be distributed to one of the outputs |
B. | one of the inputs will be selected for the output |
C. | the output will be distributed to one of the inputs |
D. | single input gives single output |
Answer» A. the input will be distributed to one of the outputs |
280. |
Most demultiplexers facilitate which type of conversion? |
A. | decimal-to-hexadecimal |
B. | single input, multiple outputs |
C. | ac to dc |
D. | odd parity to even parity |
Answer» B. single input, multiple outputs |
281. |
In 1-to-4 demultiplexer, how many select lines are required? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 5 |
Answer» A. 2 |
282. |
In a multiplexer the output depends on its |
A. | data inputs |
B. | select inputs |
C. | select outputs |
D. | enable pin |
Answer» B. select inputs |
283. |
In 1-to-4 multiplexer, if C1 = 1 & C2 = 1, then the output will be |
A. | y0 |
B. | y1 |
C. | y2 |
D. | y3 |
Answer» D. y3 |
284. |
How many select lines are required for a 1-to-8 demultiplexer? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 5 |
Answer» B. 3 |
285. |
How many AND gates are required for a 1-to-8 multiplexer? |
A. | 2 |
B. | 6 |
C. | 8 |
D. | 5 |
Answer» C. 8 |
286. |
Which IC is used for the implementation of 1-to-16 DEMUX? |
A. | ic 74154 |
B. | ic 74155 |
C. | ic 74139 |
D. | ic 74138 |
Answer» A. ic 74154 |
287. |
The word demultiplex means |
A. | one into many |
B. | many into one |
C. | distributor |
D. | one into many as well as distributor |
Answer» D. one into many as well as distributor |
288. |
Why is a demultiplexer called a data distributor? |
A. | the input will be distributed to one of the outputs |
B. | one of the inputs will be selected for the output |
C. | the output will be distributed to one of the inputs |
D. | single input to single output |
Answer» A. the input will be distributed to one of the outputs |
289. |
In a multiplexer the output depends on its |
A. | data inputs |
B. | select inputs |
C. | select outputs |
D. | enable pin |
Answer» B. select inputs |
290. |
In 1-to-4 multiplexer, if C1 = 0 & C2 = 1, then the output will be |
A. | y0 |
B. | y1 |
C. | y2 |
D. | y3 |
Answer» B. y1 |
291. |
In 1-to-4 multiplexer, if C1 = 1 & C2 = 1, then the output will be |
A. | y0 |
B. | y1 |
C. | y2 |
D. | y3 |
Answer» D. y3 |
292. |
What is the addition of the binary numbers 11011011010 and 010100101? |
A. | 0111001000 |
B. | 1100110110 |
C. | 11101111111 |
D. | 10011010011 |
Answer» C. 11101111111 |
293. |
Perform binary addition: 101101 + 011011 = ? |
A. | 011010 |
B. | 1010100 |
C. | 101110 |
D. | 1001000 |
Answer» D. 1001000 |
294. |
Binary subtraction of 100101 – 011110 is |
A. | 000111 |
B. | 111000 |
C. | 010101 |
D. | 101010 |
Answer» A. 000111 |
295. |
Perform multiplication of the binary numbers: 01001 × 01011 = ? |
A. | 001100011 |
B. | 110011100 |
C. | 010100110 |
D. | 101010111 |
Answer» A. 001100011 |
296. |
100101 × 0110 = ? |
A. | 1011001111 |
B. | 0100110011 |
C. | 101111110 |
D. | 0110100101 |
Answer» C. 101111110 |
297. |
On multiplication of (10.10) and (01.01), we get |
A. | 101.0010 |
B. | 0010.101 |
C. | 011.0010 |
D. | 110.0011 |
Answer» C. 011.0010 |
298. |
Divide the binary numbers: 111101 ÷ 1001 and find the remainder |
A. | 0010 |
B. | 1010 |
C. | 1100 |
D. | 0011 |
Answer» D. 0011 |
299. |
Divide the binary number (011010000) by (0101) and find the quotient |
A. | 100011 |
B. | 101001 |
C. | 110010 |
D. | 010001 |
Answer» A. 100011 |
300. |
Binary subtraction of 101101 – 001011 = ? |
A. | 100010 |
B. | 010110 |
C. | 110101 |
D. | 101100 |
Answer» A. 100010 |
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