Digital Electronics - Integrated Circuit Technologies - Discussion
Discussion Forum : Integrated Circuit Technologies - General Questions (Q.No. 20)
20.
A certain gate draws 1.8 
A when its output is HIGH and 3.3 µA when its output is LOW. VCC is 5 V and the gate is operated on a 50% duty cycle. What is the average power dissipation (PD)?
A when its output is HIGH and 3.3 µA when its output is LOW. VCC is 5 V and the gate is operated on a 50% duty cycle. What is the average power dissipation (PD)?2.55 W
W1.27 W
W12.75 W
W5 W
WDiscussion:
6 comments Page 1 of 1.
Pavani said:
1 decade ago
Pd=Vcc*Icc(avg)
Icc(avg)=ICCH+ICCL/2
Pd=5*(1.8+3.3)/2=12.75
Icc(avg)=ICCH+ICCL/2
Pd=5*(1.8+3.3)/2=12.75
Praveen said:
8 years ago
Yes, your Explanation is correct. Thanks @Pavani.
(1)
Ahmed said:
7 years ago
What is the answer, If duty cycle 35%?
Amir said:
5 years ago
I don't think there's a gate that 35% efficient.
Ashish said:
5 years ago
Duty cycle is 35% then Vcc is 3.5.
Arijit Chakraboty said:
4 years ago
Average power dissipation = (time of high * voltage * high current) + ( time of low + voltage + low current ).
= (0.5 * 5 * 1.8) + ( 0.5 * 5 * 3.3) [ as 50% duty cycle, so, 50% time to high and rest to low, here 50%].
The average power dissipation = 12.75 μW.
= (0.5 * 5 * 1.8) + ( 0.5 * 5 * 3.3) [ as 50% duty cycle, so, 50% time to high and rest to low, here 50%].
The average power dissipation = 12.75 μW.
(1)
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