IC Power Dissipation from Supply Current — Average ICC method An IC operates from VCC = +5 V with ICC in HIGH state (ICCH) = 10 mA and ICC in LOW state (ICCL) = 23 mA. Assuming a 50% duty cycle, what is the chip’s average power dissipation?

Introduction / Context:Digital ICs often specify supply current in two states: when outputs are HIGH (ICCH) and when outputs are LOW (ICCL). To estimate average power, designers compute the duty-cycle-weighted average current and multiply by the supply voltage.

Given Data / Assumptions:

• VCC = 5 V (constant supply).
• ICCH = 10 mA (average current when outputs are HIGH).
• ICCL = 23 mA (average current when outputs are LOW).
• Outputs spend equal time HIGH and LOW (50% duty).

Concept / Approach:Average current for a two-state system with equal time in each state is: I_avg = (ICCH + ICCL) / 2. The average power is then P_avg = VCC * I_avg.

Step-by-Step Solution:Compute I_avg = (10 mA + 23 mA) / 2 = 33 mA / 2 = 16.5 mA.Compute P_avg = 5 V * 16.5 mA = 82.5 mW.Thus, the average power dissipation is 82.5 mW.

Verification / Alternative check:If duty cycle were not 50%, use I_avg = D * ICCH + (1 − D) * ICCL. For D = 0.5, the calculation above is confirmed.

Why Other Options Are Wrong:

• 50 mW: Corresponds to I_avg = 10 mA, ignoring the higher ICCL contribution.
• 115 mW: Would imply I_avg = 23 mA; that assumes 100% LOW time.
• 165 mW: Would require 33 mA average at 5 V, which is the sum, not the average.

Common Pitfalls:Confusing “sum” with “average,” or forgetting to multiply by VCC to convert current to power. Ensure units are consistent: volts * amperes = watts.

Final Answer:82.5 mW