2. (25 pts) You have just finished building a register using 74LS74 D\' flip-flo

Question

2. (25 pts) You have just finished building a register using 74LS74 D' flip-flops. To debug your circuit, you manually load a number into this register and measure the following voltages on the Q outputs of the flip-flops: QA-3.3V, QB-3.3V, C-0V, QD 0.2V, QE-3.3V, QF-0.2V, QG-0.1V, QH-3.6V A) (10 pts) Assuming QA holds the Least Significant Bit (LSB) and QH holds the Most Significant Bit (MSB), what is the decimal equivalent of the number you measured? B) (15 pts) Based on the measured inputs to the flip-flops, you're absolutely sure the bits you loaded into this register represent the hexadecimal number $97, Additionally, your lab partner notices that one of the chips is getting warm. What is the most likely problem if any? Explain your answer in detail for full credit! Use the back of this page if you need more space.

Explanation / Answer

1)let's consider the inputs

QH-MSB most significant bit

QA-LSB most significant bit

QH QG QF QE QD QC QB QA

3.6 0.1 0.2 3.3 0.2 0 3.3 3.3

1 0 0 1 0 0 1 1

for binary to decimal conversion

(1*2^7)+(0*2^6)+(0*2^5)+(1*2^4)+(0*2^3)+(0*2^2)+(1*2^1)+(1*2^0)= 147d

the decimal equivalent for the input given is 147D.

b)

The HEXADECIMAL equivalent for the specified input is 93H

The most possib;e reasons fr the chip getting warme were as follows

1) Ageing

2) High leakage currents.

3) External temperature effect.

4) Wrong connection

Mostly the reason might be due to the high input voltage applied to one of the chips which may result in increase in the current in transistor eventually the tempearature increases which makes the chip getting warm.

Here in this case the voltage to the QH is 3.6v which is the higher voltage for the state 1 tied to a higher voltage of logic level1 this might cause the chip to get warm .

Transistors have gate capacitance. Every flow of electrical energy necessarily has some inefficiency. Energy can't be destroyed, so where does this lost electrical energy go in the form of HEAT.

The other possible factor might be the chip which is getting warm didnt have a properly connected ground.

The main problem with IC operation at high temperatures is the greatly increased leakage current of individual transistors. The leakage current can increase to such an extent that the switching voltage levels of the devices is affected, so that signals can't propagate properly within the chip, and it stops functioning.

They usually recover when allowed to cool down, but that is not always the case.

  

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.