70451. The Boolean expression for the circuit of the given figure
70452. What will be BCD number when the output is 0.37 V?
70453. The first machine cycle of an instruction is always
70454. A counter has N flip flops. The total number of states are
70455. Out of S, R, J, K, Preset, Clear inputs to flip flops, the synchronous inputs are
70456. The Boolean expression A ⊕ B is equivalent to
70457. Which of these are two state devices?
70458. What will be minimum conversion rate in 6 bit dual slope A/D converter uses a reference of -6v and a 1 MHz clock. It uses a fixed count of 40 (101000).
70459. The minimum number of NAND gates required to implement the Boolean function A +AB + AB C is equal to
70460. For the K map of the given figure, the simplified Boolean expression is
70461. The dual of A + [B + (AC)] + D is
70462. A divide by 78 counter can be obtained by
70463. The initial state of MOD-16 down counter is 0110. What state will it be after 37 clock pulses?
70464. The number of address lines in EPROM 4096 x 8 is
70465. If the inputs to a 3 bit binary adder are 1112 and 1112, the output will be 1102
70466. Which display device resembles vacuum tube?
70467. The number of inputs and outputs of a full adder are
70468. A VF display operates on the principle of a vacuum diode.
70469. In a 3 input NAND gate, the number of states in which output is 1 equals
70470. In a mod-12 counter the input clock frequency is 10 kHz. The output frequency is
70471. The total number of input words for 4 input OR gate is
70472. In digital circuits Schottky transistors are preferred over normal transistors because of their
70473. A flip flop is a
70474. A 4 bit parallel type A/D converter uses a 6 volt reference. How many comparators are required and what is the resolution in volts?
70475. Quantization error occurs in
70476. Which of these are universal gates?
70477. Out of latch and flip flop, which has clock input?
70478. A mod 4 counter will count
70479. In the given figure shows a 4 bit serial in parallel out right shift register. The initial contents as shown are 0110. After 3 clock pulses the contents will be
70480. Which of the following finds application in pocket calculators?
70481. Octal number 12 is equal to decimal number
70482. Hexadecimal number E is equal to binary number
70483. Decimal number 46 in excess 3 code =
70484. Logic hardware is available only in NAND and NOR.
70485. A 4 input AND gate is equivalent to
70486. The circuit of the given figure is
70487. A half adder adds
70488. The no. of comparators required in a 3 bit comparator type ADC is
70489. If the ladder reference voltage is 2 V, then minimum comparator resolution required is
70490. In a 7 segment LED display, the minimum number of segments is activated when the input decimal number is
70491. A 4 bit down counter starts counting from 1111 irrespective of modulus.
70492. The number of counter states which an 8 bit stair step A/D converter has to pass through before conversion takes place is equal to
70493. 74 HC series can sink upto 4 mA. The 74 LS series has IIL(max) - 0.4 mA. How many 74 LS inputs be driven by 74 HC output?
70494. An 8 bit transistor register has output voltage of low-high-low-high-low-high-low-high. The decimal number stored is
70495. The number of cells in a 4 variable K map is
70496. Find the FSV (full scale voltage) in a 6 bit R-2R ladder D/A converter has a reference voltage of 6.5 V.
70497. Parallel adder is
70498. An 8 bit binary number is to be entered into an 8 bit serial shift register. The number of clock pulses required is
70499. In the given figure, A = B = 1 and C = D = 0. Then Y =
70500. Assertion (A): CMOS devices have very high speed.Reason (R): CMOS devices have very small physical size and simple geometry.
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