Capacitor Under Load Formula: Sizing for Voltage Droop and Hold-Up

When a power source drops out or a heavy transient load kicks in, a capacitor acts as a temporary battery. The capacitor under load formula calculates exactly how much the voltage will sag (droop) when a specific current is drawn over a specific time, or conversely, how much capacitance you need to survive a brownout. For a constant current draw, the core relationship is C = (I × Δt) / ΔV. If you are sizing bulk storage for a motor startup, a camera flash, or a microcontroller hold-up circuit, this linear approximation is your first-line design tool.

The Core Formula & Symbol Matrix

The fundamental equation governing a capacitor discharging under a constant current load is derived from the definition of capacitance (I = C × dV/dt). When we move from calculus to algebra for discrete time steps, we get the linear droop formula.

Rearranged Forms

Depending on what you are trying to find on the bench, you will need to isolate different variables. Here is the formula solved for each parameter:

• Solve for Capacitance (Sizing): C = (I × Δt) / ΔV
• Solve for Current (Max Load): I = (C × ΔV) / Δt
• Solve for Time (Hold-up Duration): Δt = (C × ΔV) / I
• Solve for Voltage Droop (Sag): ΔV = (I × Δt) / C

Assumptions, Limits, and Unit Traps

Before you plug numbers into a calculator, you must understand the boundaries of this formula. It is not a universal law for all circuits; it is a specific model for constant current discharge.

When the Formula Applies (and When It Doesn't)

This linear formula assumes the load draws a constant current regardless of the dropping voltage. This is highly accurate for constant-current LED drivers, linear regulators feeding a steady digital load, and active electronic loads.

However, if your load is a pure resistor (like a heating element or a simple incandescent bulb), the current will drop as the capacitor voltage drops. In that case, the linear formula will overestimate the voltage droop. For purely resistive loads, you must use the exponential RC decay formula: V(t) = V₀ × e^(-t/RC).

The Unit Mistakes That Break the Math

The most common reason a hold-up capacitor design fails on the first prototype is a unit conversion error. The formula demands base SI units (Farads, Amperes, Seconds). If you feed it microfarads and milliseconds without converting, your answer will be off by orders of magnitude.

What a Realistic Answer Magnitude Looks Like

A 1 Farad capacitor is physically massive (a supercapacitor). Standard electrolytic capacitors range from 10 μF to 10,000 μF. If your calculation spits out 'C = 45 F' for a standard PCB decoupling task, you have likely forgotten to convert microfarads to Farads, or you are asking a tiny capacitor to do the job of a battery bank.

Solved Problems with Unit Tracking

Let's walk through two distinct design scenarios, tracking every unit conversion to ensure the math holds up to physical reality.

Problem 1: Microcontroller Brownout Ride-Through

Scenario: An ESP32-S3 development board draws a steady 80 mA during heavy WiFi transmission. The 5V USB supply is known to experience 15 ms brownouts. The onboard LDO requires a minimum of 4.2V to maintain a stable 3.3V rail. We need to size a bulk hold-up capacitor on the 5V rail to ride through the dropout.

1. Identify Knowns:
   I = 80 mA = 0.08 A
   Δt = 15 ms = 0.015 s
   ΔV = 5.0V (nominal) - 4.2V (minimum) = 0.8 V
2. Select Formula: C = (I × Δt) / ΔV
3. Substitute Base Units: C = (0.08 A × 0.015 s) / 0.8 V
4. Calculate: C = 0.0012 / 0.8 = 0.0015 Farads
5. Convert to Practical Units: 0.0015 F × 1,000,000 = 1,500 μF

Result: A standard 1,500 μF or 2,200 μF low-ESR electrolytic capacitor placed across the 5V rail will prevent the ESP32 from resetting during a 15 ms USB brownout.

Problem 2: Solenoid Valve Firing on a Shared 24V Rail

Scenario: A pneumatic solenoid valve draws 2.5 A when energized. It takes 50 ms for the valve to fully actuate and the current to drop to its holding value. The 24V DC power supply cannot handle the 2.5 A transient without dipping. We want to limit the voltage droop on the 24V rail to a maximum of 3V.

1. Identify Knowns:
   I = 2.5 A
   Δt = 50 ms = 0.05 s
   ΔV = 3 V
2. Select Formula: C = (I × Δt) / ΔV
3. Substitute Base Units: C = (2.5 A × 0.05 s) / 3 V
4. Calculate: C = 0.125 / 3 = 0.04167 Farads
5. Convert to Practical Units: 0.04167 F × 1,000,000 = 41,670 μF

Result: You will need a capacitor bank of at least 47,000 μF (a common standard value) rated for at least 35V (applying the 20% voltage derating rule for electrolytics) to handle this solenoid kick.

Real-World Bench Scenario: The Motor Startup Droop

Formulas assume ideal components. Real components have parasitic resistance. Here is a scenario where the math was right, but the physical implementation failed.

The Setup

A robotics hobbyist is building a 12V rover. A 12V DC gearmotor and a logic-level microcontroller share the same 12V battery feed. The logic board uses a buck converter that drops out if the input falls below 8V. To prevent motor noise from resetting the logic, the builder places a 2,200 μF electrolytic capacitor across the 12V rails near the motor.

The Numbers

• Motor Stall Current (I): 6 A
• Motor Startup Time (Δt): 100 ms (0.1 s)
• Capacitance (C): 2,200 μF (0.0022 F)
• Capacitor ESR (Equivalent Series Resistance): 60 mΩ (0.06 Ω)

The Outcome

When the motor starts, the microcontroller instantly resets. The builder measures the 12V rail with an oscilloscope and sees the voltage plummet from 12V down to 4V in the first millisecond, far below the 8V threshold.

What Went Wrong

The builder used the capacitor under load formula to check the capacitive droop but ignored the resistive droop caused by ESR, and miscalculated the total allowable ΔV.

1. The ESR Instant Drop: The moment the motor stalls and pulls 6 A, the capacitor's internal resistance causes an instantaneous voltage drop before any actual discharge happens. V_ESR = I × R = 6 A × 0.06 Ω = 0.36V instant drop.
2. The Capacitive Droop: Using our formula: ΔV = (I × Δt) / C = (6 A × 0.1 s) / 0.0022 F = 272V.
3. The Reality Check: The formula demands 272V of droop, but the rail only has 12V available. This means the 2,200 μF capacitor is completely drained in roughly 4.4 milliseconds, not the 100 ms the motor needs to spin up. The rail collapses entirely.

The Fix: To support a 6A load for 100ms with a maximum allowable droop of 4V (12V down to 8V), the required capacitance is C = (6 × 0.1) / 4 = 0.15 F (150,000 μF). Furthermore, the builder must use multiple low-ESR capacitors in parallel to drop the combined ESR below 10 mΩ, ensuring the instantaneous V_ESR drop stays under 0.1V.

Exponential Decay vs. Constant Current Draw

It is vital to recognize when to abandon the linear capacitor under load formula. As noted earlier, the linear formula (ΔV = I × Δt / C) is an approximation that holds true only when the load acts as a constant current sink.

If you are discharging a capacitor through a fixed resistor (like a bleeder resistor or a passive heating element), the current is not constant. As the capacitor voltage falls, the current falls proportionally (I = V/R). This creates an exponential decay curve. For these circuits, you must use the RC discharge equation:

V(t) = V₀ × e^(-t / RC)
Where V₀ is the initial voltage, e is Euler's number (~2.718), t is time, R is resistance in Ohms, and C is capacitance in Farads.

For example, if you are designing a defibrillator circuit or a simple RC timing delay for a 555 timer, the load is fundamentally resistive. Using the linear constant-current formula in those scenarios will yield dangerously optimistic time estimates, as the current 'tails off' rather than remaining steady. Always define your load type—active constant-current or passive resistive—before selecting your math model.