Capacitor Voltage Drop Calculator: Formula and Examples

A capacitor voltage drop calculator estimates how far a capacitor’s voltage falls while it supplies a known current for a given time. Enter capacitance, discharge current, and duration; the calculator returns the voltage change, with the ideal constant-current formula shown below.

Formula

For a constant current drawn from an ideal capacitor, the voltage change is:

ΔV = I × t / C

• ΔV is the voltage drop, in volts (V).
• I is the discharge current, in amperes (A).
• t is the time, in seconds (s).
• C is capacitance, in farads (F).

The calculator also displays the same voltage drop in millivolts (mV). One farad is 1,000,000 microfarads (µF), and one ampere is 1,000 milliamperes (mA). Convert values before using the formula: for example, 1,000 µF is 0.001 F, while 100 mA is 0.1 A. This relationship comes from the capacitor equation i = C × dV/dt, rearranged for a steady current.

Worked Example

Using the calculator’s defaults: C = 0.001 F, I = 0.1 A, and t = 1 s.

1. Multiply current by time: 0.1 A × 1 s = 0.1 coulomb.
2. Divide charge by capacitance: 0.1 / 0.001 = 100 V.
3. Convert to millivolts: 100 V × 1,000 = 100,000 mV.

The calculated voltage drop is 100 V (100,000 mV). This is the ideal mathematical result, not a claim that a particular capacitor can safely supply that current or withstand that voltage. In a real circuit, the capacitor’s initial voltage and ratings limit the usable voltage change.

When to Use This Calculation

Use the estimate when a load draws approximately constant current and you want to understand how much a capacitor’s voltage changes over a short interval. It is useful for checking the approximate hold-up time of a supply, estimating ripple over a switching interval, or comparing capacitance values during early circuit calculations.

The result is a voltage difference, not the final capacitor voltage. If the capacitor starts at V_start, then under the stated ideal assumptions its voltage after the interval is:

V_end = V_start − ΔV

For example, a 12 V starting voltage and a calculated 2 V drop imply an ideal ending voltage of 10 V. If the calculated drop exceeds the starting voltage, the assumed current cannot continue for the full duration from that starting voltage; the calculation is signaling that the inputs or assumptions need attention.

Assumptions and Choosing Inputs

The formula assumes current stays constant, capacitance is fixed, and the capacitor behaves ideally. Real loads often do not draw constant current. A resistor across a capacitor, for example, draws less current as voltage falls, producing an exponential discharge rather than a linear drop.

• Capacitance: Use farads in the calculator. Check the component marking or datasheet, and remember that actual capacitance can differ from its nominal value.
• Current: Enter the current actually supplied by the capacitor, not automatically the load’s stated power or supply current. If you know load power and voltage, current may vary as voltage changes.
• Time: Enter seconds. Convert milliseconds by dividing by 1,000; for instance, 5 ms is 0.005 s.
• Starting and minimum voltage: These are not inputs to the drop calculation, but they determine whether the estimated drop is acceptable for the circuit’s operation.

For a changing current, the more general relationship is ΔV = (1/C) × ∫ i(t) dt. In practical terms, use the total charge drawn over the interval divided by capacitance. For a resistor-capacitor discharge, use the exponential relation V(t) = V₀e^−t/(RC), where R is resistance in ohms and t is seconds; the constant-current calculator is not the right model for that situation.

Common Mistakes and Limits

• Mixing units: Putting 1,000 for 1,000 µF while treating it as farads creates a result a million times too small. Convert µF, nF, or mF to F first.
• Confusing drop with final voltage: The answer is ΔV. Subtract it from the starting voltage to estimate the ending voltage.
• Using peak or average current without checking: A current that changes significantly can make the constant-current result misleading. Estimate charge over time or use an appropriate circuit model.
• Ignoring component behavior: Capacitance tolerance, equivalent series resistance (ESR), leakage, temperature, and frequency can affect real performance. ESR can cause an immediate voltage step when current changes, in addition to the slower capacitive drop.
• Relying on rounded figures: The display rounds to three decimal places in volts. Keep more precision in intermediate calculations when comparing a result with a narrow voltage margin.

This calculator does not choose a capacitor, verify a component’s voltage or ripple-current rating, or establish that a circuit is safe. Capacitors can retain charge after power is removed, and some circuits involve hazardous voltages. Follow the equipment documentation and appropriate electrical safety procedures; do not handle or discharge an unknown capacitor casually.

Frequently Asked Questions

How do I calculate capacitor voltage drop?
For constant current, multiply current in amperes by time in seconds, then divide by capacitance in farads: ΔV = It/C.

How much voltage does a capacitor lose per second?
At constant current, the rate is ΔV/t = I/C volts per second. A 0.1 A current from a 0.001 F capacitor corresponds to 100 V/s in the ideal model.

How long will a capacitor hold its voltage?
For a constant-current load and a permitted drop ΔV, rearrange the formula: t = CΔV/I. The actual hold-up time depends on changing load current, circuit voltage limits, capacitance variation, leakage, and other real-world effects.

Does a larger capacitor reduce voltage drop?
Yes, for the same current and duration, the ideal voltage drop is inversely proportional to capacitance. Doubling C halves ΔV, assuming the other values and operating conditions remain unchanged.

Can I use this formula for a capacitor discharging through a resistor?
Not as an exact model. Resistor current falls as capacitor voltage falls, so voltage decays exponentially. Use V(t) = V₀e^−t/(RC) for an ideal resistor-capacitor discharge.

Why is my measured drop different from the calculator?
Current may not be constant, the capacitor’s actual capacitance may differ from its label, and ESR, leakage, wiring resistance, measurement timing, or instrument accuracy can affect the reading.

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in c | Capacitance | F | 0.001 | 0
in i | Discharge current | A | 0.1 | 0
in t | Discharge time | s | 1 | 0
out dv | Voltage drop | V | i * t / c | 3
out vpc | Voltage drop | V | 100 * i * t / c | 1