Electrical engineering / Free calculator

DC hold-up capacitor

Estimate the ideal capacitance needed to support a constant-power load during a supply interruption.

01 Your inputs

Calculated on your device. Inputs are not transmitted or stored.

02 Your result

Enter your values and select Calculate.

Power during a supply interruption

Capacitor feeding a converter and constant-power loadWith the supply absent, capacitor C feeds a converter of efficiency eta, which supplies load power P. A shared return completes the circuit. Capacitor voltage falls from V high to V low during the hold-up time.
Stored energy
C supplies energy as its voltage falls from Vhigh to Vlow.
Conversion losses
Efficiency η accounts for energy lost in the converter.
Constant-power load
P is power delivered to the load, after conversion losses.
The capacitor supports the load for the entered hold-up time. Charging and source-isolation circuitry are omitted. Symbolic model; not to scale.

The method

C=2Ptη(Vhigh2−Vlow2)

A worked example

A 10 W load held for 100 ms, from 12 V down to 9 V at 90% efficiency, needs approximately 1.111 J of stored energy and 35.27 mF of ideal capacitance.

Assumptions & limits

Constant power and constant efficiency. This is an energy estimate, not a final component selection. Account separately for ESR voltage drop, capacitance tolerance, DC bias, aging, leakage, voltage rating, converter dropout, and inrush current.

Calculation source: Engineering Workbench calculation engine; capacitor energy model E = ½CV².

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