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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.