Determine the Boolean expressions, in terms of Q3, Q2, Q1, Q0 that will cause LE

Question

Determine the Boolean expressions, in terms of Q3, Q2, Q1, Q0 that will cause LED 0 and LED 1 turn on. 6. +5 1/2 HC139 1/2 HC139 01 Q3 EN 2 LED 1X 2 LED 0 LED 1 A 4-bit binary signed (2's complement) ripple adder (with a carry-out output and an overflow output) is used to add -4 +-6, answer the following questions: 7. What will the binary outputs , , 0 reveal as a result of the signed addition? Will the carry-out output be active as a result of the signed addition? Will the overflow output be active as a result of the signed addition? 8. Calculate the minimum value of R for the following interface circuit. (Assume . that the output logic level voltage (Vout) must remain valid and VLED-3.0V.) +5V +5 V Vout 74HCO4 specs: VoL(/max)-0.33V lo(max)-4.0mA VOtH(min) -3.84V IoR(max)-4.0mA

Explanation / Answer

Answer:- For an LED to glow, LED must be connected in forward bias. LED zero will glow when pin 2 of first decoder will be at logic zero.

Since decoder here has active low output pins hence when input of decoder U1a will be 10 i.e Q0 = 0 and Q1 = 1, output pin2 will be zero and LED zero will glow. Also chip enable pin i.e Q2 must be logic zero since the enable pin is active low. So for LED zero we can write-

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LED0 = Q2.Q1.Q0 Here Q3 has no effect for LED0.

For LED1 to glow, output from OR gate must be a logic high since other side of LED is connceted to ground.
OR gate output will be high when any one input to OR gate is high.

Note here for chip U1b to enable logic low is needed and same zero is given to pin1 i.e input pin of chip U1b. It means on output side either pin zero or pin2 will active, and pin1 and pin3 will never be selected as output. So in this case LED1 will never glow.

LED1 = It will never glow with this circuitry.

NOTE:- In above question, I have assumed that input pin1 of decoder IC is the LSB and input pin2 is the MSB. If reverse is chosen then answer will change and accordingly.

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