The Large Capacity Reservoir Capacitance Selection

When the super capacitance replaces rechargeable batteries in products, not properly selection can cause electrolyte leakage, and it is also likely to have internal short circuit. How to choose the super capacitor the circuit needed is not so easy.

How to choose the super capacitor?

The super capacitor has two major applications: high power pulse application and instantaneous power maintaining. High power pulse application characteristics: transient flow load large current. The characteristics of the instantaneous power maintain application requires continuing to provide power for load; the duration is generally for a few seconds or minutes. A typical application of instantaneous power maintaining is the restoration of the disk drive head when cut off. Different applications on the super capacitor parameters requirements are also different. High power pulse application is using the small resistance (R) of super capacitor, and the instantaneous power maintaining is using the big electrostatic capacity (C).

C(F)：The capacitance nominal capacity

R(Ohms)：The capacitance Nominal resistance
ESR(Ohms)：1KZ below equivalent series resistance

Uwork(V)： the regular working voltage in the circuit
Umin(V)： the working minimum voltage

t(s)：the maintaining time required in a circuit or pulse duration in pulse application
I(A)： load current

Udrop(V)： the total voltage drop

Here we introduce the instantaneous power maintaining application.

Super capacitor approximate calculation formula of capacity, according to the formula, keeping the energy = super capacitor reducing energy.

Keeping the energy = 1/2I (Uwork + Umin) t;

Super capacitor reducing energy = 1/2 C (Uwork2-Umin2),

Therefore, its capacity (ignore the pressure drop caused by IR) C = (Uwork + Umin) t / (Uwork2-Umin2)
Such as microcomputer application system, the application of super capacitor as backup power, after power off it needs to use super capacitor to maintain 100 mA current, lasts for 10 s, SCM system end working voltage is 4.2 V, so how large capacity of the super capacitor can guarantee the normal work of the system?

According to above knowable formula:

The starting work voltage Vwork = 5 V; the ending work voltage Vmin = 4.2 V; Working time t = 10 s; Work power I = 0.1 A capacitance capacity required for:

C = (Vwork + Vmin) It / (Vwork2-Vmin2)

= (5 + 4.2) * 0.1 * 10 / (52-4.22)

= 1.25 F

According to the calculation results, we can choose 5.5 V, 1.5 F capacitance to meet the need.