Charge, delay, and a switching budget

Prerequisites: Voltage, current, capacitance, exponentials, and scientific notation.

Learning objectives

• Calculate an RC threshold-crossing time.
• Distinguish capacitor energy from supply energy.
• Estimate dynamic power with an explicit activity convention.

A wire does not charge instantly

Model a switching output as a voltage source V driving a capacitor C through resistance R. Starting from zero, the capacitor voltage is v(t)=V(1−exp(−t/RC)). The time constant τ=RC describes the charging timescale. After one time constant, the voltage reaches about 63.2% of its final value. This is an approximation to a transistor and interconnect, not a complete device simulation.

Suppose R=2 kΩ, C=20 fF, and V=1 V. The time constant is 40 ps. A receiver that recognizes a high above 0.7 V needs 0.7=1−exp(−t/40 ps), giving t=−40 ln(0.3)≈48.16 ps. Merely crossing half the supply would take about 27.73 ps. Therefore a delay figure is incomplete unless its measurement threshold and load are stated.

If a second identical load doubles capacitance, the estimated threshold-crossing time doubles. Increasing transistor width may reduce effective resistance but usually adds capacitance elsewhere. Optimizing one isolated gate can move cost into its driver, its wiring, or its neighbors.

Account for where the energy goes

The capacitor’s stored energy at full voltage is CV²/2. Charging it from an ideal constant-voltage supply draws CV², half stored in the capacitor and half dissipated in resistance. Discharging to zero dissipates the stored half. Thus an ideal full charge-discharge cycle dissipates CV². Leakage and short-circuit current add costs omitted by this model.

For C=20 fF and V=1 V, one full cycle costs 20 fJ. If an output experiences 100 million charging events each second, this mechanism contributes 2 μW. A convenient expression is P_dynamic=αCV²f when α is defined as the average number of 0→1 charging events per clock cycle. Some conventions define activity differently, so copy the definition along with the equation.

Voltage reduction changes two budgets

At fixed C, α, and f, reducing supply from 1 V to 0.8 V multiplies dynamic power by 0.8²=0.64. That calculation does not establish safe operation at the original clock. Real transistor drive weakens as operating voltage changes, propagation delay increases, and noise margins may shrink. Validate the timing path and electrical operating region before claiming the energy saving.

The useful engineering question is constrained: minimize energy while meeting a required error rate, throughput, latency, and temperature envelope. Your notebook should separate parameters taken from measurements from parameters invented for a teaching model. A precise result computed from an inaccurate model is still inaccurate.

Primary reading

• Microelectronic Devices and Circuits (MIT OpenCourseWare).

Practice problem

A toy gate drives 30 fF through 3 kΩ at a 1.2 V supply. Find its time constant and time to reach 60% of the final voltage. It charges on 0.2 of cycles at 500 MHz. Estimate dynamic power using the stated charging-event convention. Recalculate power at 0.9 V while keeping the other parameters fixed, then state the missing timing check.

Worked solution

The time constant is 3000×30×10^−15=90 ps. The 60% crossing time is −90 ln(0.4)≈82.47 ps. Dynamic power is 0.2×30×10^−15×1.2²×500×10^6=4.32 μW. At 0.9 V it becomes 2.43 μW, a factor of 0.5625. This comparison assumes unchanged capacitance, activity, and frequency. It does not establish that the effective resistance or transistor switching delay remains unchanged. Recheck the worst timing path, setup margin, and valid logic levels at the lower supply and operating temperature.

Check your understanding

For the same R and threshold fraction, what happens if C doubles?

• The RC crossing time doubles.
• The crossing time is halved.
• Only leakage changes.
• The threshold crossing becomes instantaneous.

Explanation

The time is −RC ln(1−fraction), so it scales linearly with C under this approximation.

Why is supply energy for one charge larger than stored capacitor energy?

• Capacitance increases during every charge.
• The voltage source recovers all charge immediately.
• Energy is also dissipated in the charging resistance.
• The stored-energy formula omits voltage.

Explanation

A constant-voltage supply delivers CV²; half becomes stored energy and half is dissipated during charging.

Primary reading

• Microelectronic Devices and Circuits: MIT OpenCourseWare

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