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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
What will be FSV in 2 bit BCD D/A converter is a weighted resistor type with E = 1V, R = 1 MΩ and R = 10KΩ |
| Answer» FSV for BCD = (10d - 1)VR. = (102 -1) x . | |
| 102. |
With four Boolean variables, how many Boolean expressions can be performed? |
| Answer» 224 = 216. | |
| 103. |
A + (B . C) = |
| Answer» This is distributive law. | |
| 104. |
The number of logic devices required in a Mod-16 synchronous counter are |
| Answer» 4 flip-flops and 2 AND gates. | |
| 105. |
A 2 bit BCD D/A converter is a weighted resistor type with E = 1V, R = 1 MΩ and R = 10KΩ then Resolution in percent and volt is __________ . |
| Answer» Resolution % for BCD = where d is Resolution in volts. Resolution in volts = volts per step = 0.01V 0r 10mV. | |
| 106. |
The gates G and G in the figure have propagation delays of 10 n sec. and 20 n sec. respectively. If the input V makes an output change from logic 0 to 1 at time = then the output wavefrom V is |
| Answer» Truth table are as follows. | |
| 107. |
The maximum counting range of a four stage counter using IC 74193 is |
| Answer» IC 74193 is a divide by 16 counter. Since 4 stages are used, counting range = 164 = 65536. | |
| 108. |
The circuit of the given figure is |
| Answer» | |
| 109. |
A half adder adds |
| Answer» | |
| 110. |
The no. of comparators required in a 3 bit comparator type ADC is |
| Answer» Number of comparators = 2n - 1 = 23 - 1 = 7. | |
| 111. |
If the ladder reference voltage is 2 V, then minimum comparator resolution required is |
| Answer» . | |
| 112. |
In a 7 segment LED display, the minimum number of segments is activated when the input decimal number is |
| Answer» Only 2 segments are lit as seen in | |
| 113. |
The number of bits in EBCDIC is |
| Answer» EBCDIC is an 8 bit code. | |
| 114. |
A 3 bit up-down counter can count from |
| Answer» Since it is an up-down counter it can count upwards or downwards. | |
| 115. |
The density of dynamic RAM is |
| Answer» Dynamic RAM has higher packing density and requires lesser space. | |
| 116. |
To count 1000 bottles in a milk plant, the minimum number of flip flops requires is |
| Answer» 20 = 512 and 210 = 1024 Hence at least 10 flip-flops are required. | |
| 117. |
Decimal 8 in excess-3 code = |
| Answer» Decimal 8 in excess 3 code = decimal 11 = 1011 in binary. | |
| 118. |
The inputs to a 3 bit binary adder are 111 and 110. The output will be |
| Answer» Rules for binary addition. 0 + 0 = 0 0 + 1 = 1 + 0 = 1 1 + 1 = 10 i.e., 1 + 1 equals 0 with a carry of 1 to next higher column. 1 + 1 + 1 = 11 i.e., 1 + 1 + 1 equals 1 with a carry of 1 to next higher column. | |
| 119. |
BCD equivalent of - 8 is |
| Answer» Decimal 8 = 1000 and - 8 = 11000. | |
| 120. |
The number of inputs and outputs in a full adder are |
| Answer» | |
| 121. |
Logic hardware is available only in NAND and NOR. |
| Answer» Other gates are also available. | |
| 122. |
A 4 input AND gate is equivalent to |
| Answer» All the switches have to be closed so that the circuit can be made. In AND gate all the inputs have to be high for output to be high. | |
| 123. |
The function of interface circuit is |
| Answer» Output of driver stage has to be matched with input of load. | |
| 124. |
1274 - 3A7 = __________ . |
| Answer» Convert to decimal, subtract and change the result to hexadecimal. | |
| 125. |
The number of memory locations in which 14 address bits can access is |
| Answer» 214 = 16384. | |
| 126. |
A 6 bit R-2 ladder D/A converter has a reference voltage of 6.5 V. It meets standard linearity then resolution in percent and volts. |
| Answer» R-2R ladder, resolution = Resolution in volts = . | |
| 127. |
What will be output for decimal input 82? |
| Answer» Output for decimal input 82/(10000010) = 0.01 x 82 = 0.82 V. | |
| 128. |
The circuit of the given figure gives the output = |
| Answer» When any input is High, the corresponding transistor conducts and output is Low. Hence it is a NOR gate. | |
| 129. |
Two 2's complement number having sign bits X and Y are added and the sign bit of the result is Z. then, the occurrence of overflow is indicated by the Boolean function. |
| Answer» carry of a one bit full order is given by expression XY + YZ + ZX. | |
| 130. |
If A = 0101, then A' is |
| Answer» A' = 1010 + 1 = 1011. | |
| 131. |
A + A . B = |
| Answer» A.B= 1 or 0, A + A.B = 0 if A is zero and A + A.B = 1 if A = 1. Hence A + A.B = A. | |
| 132. |
The device 'one shot' has |
| Answer» One shot means one stable state. | |
| 133. |
For the logic circuit of the given figure the simplified Boolean expression is |
| Answer» #NAME? | |
| 134. |
If number of information bits is 4, the parity bits in Hamming code are located at bit positions __________ from the LSB. |
| Answer» The number of parity bits is 3. These bits are located at 20, 21, 22, i.e., 1, 2, 4th bits starting from LSB. | |
| 135. |
A certain JK FF has = 12 n sec the largest MOD counter that can be constructed from these FF and still operate up to 10 MHz is |
| Answer» Clock period(t) = = 10-7 sec Number of FF = mod counter 28 = 256. | |
| 136. |
11011 in gray code = __________ . |
| Answer» Change to decimal and then to binary. | |
| 137. |
In the circuit of the given figure, Y = |
| Answer» XX = 0. | |
| 138. |
A . 0 = |
| Answer» 1.0 = 0 and 0.0 = 0. | |
| 139. |
Decimal -90 equals __________ in 8 bit 2s complement |
| Answer» +90 in binary is 01011010. Its 2's complement is 10100110. | |
| 140. |
23.6 = __________ |
| Answer» 23= 10111 and 0.6 =10011 Hence 10111 . 10011 . | |
| 141. |
In a shift left register, shifting a bit by one bit means |
| Answer» In binary shift one bit left means multiplication by 2. | |
| 142. |
If A = B = 1, the outputs P and Q in the given figure are |
| Answer» It is a half adder 1 + 1 = 10. Therefore SUM = P = 0 and Carry = Q = 1. | |
| 143. |
The complement of exclusive OR function is |
| Answer» Complement of XOR is X NOR. | |
| 144. |
In the given figure, Y = |
| Answer» . | |
| 145. |
In a 3 input NOR gate, the number of states in which output is 1 equals |
| Answer» Only one input, i.e., A = 0, B = 0 and C = 0 gives 1 as output. | |
| 146. |
The series 54 H/74 H denotes |
| Answer» It denotes high speed TTL. | |
| 147. |
For a 4096 x 8 EPROM, the number of address lines is |
| Answer» 4096 = 212 . | |
| 148. |
+ C = |
| Answer» A B C D + A B C D = A B D(C + C) = A B D . | |
| 149. |
Find the output voltage for 011100 in a 6 bit R-2 ladder D/A converter has a reference voltage of 6.5 V. |
| Answer» (an-1 2n-1 + an-2 2n-2 + .....a121 + a020) (24 + 23 + 22) = 6.5 x = 2.84 . | |
| 150. |
Binary number 1101 is equal to octal number |
| Answer» 1101 = 13 in decimal = 15 (i.e., 8 + 5) in octal. | |