How to calculate the discharge current of a Farad capacitor?

   Farad capacitor discharge calculation method

　　Simple calculation method of Farad capacitor discharge Characteristics of super capacitors

　　Small size, large capacity, and energy density far greater than that of electrolytic capacitors .

　　Can be used as a backup power supply. The ESR is small, the power characteristics are good, and the power density is much larger than that of the battery. It can be used as the power compensation of the main power supply to ensure the needs of short time and high current. More than 100,000 times of charge and discharge.

　　Overcharge and overdischarge will not affect its electrical performance. Simple to use, no special charge and discharge control circuit is required. Environmentally friendly, pollution-free, maintenance-free.

　　Large working temperature range, working temperature, minus 40 degrees Celsius.

　　Simple calculation method of Farad capacitor discharge

　　T = (C & TI mes;ΔU) / IT: discharge time, in s C: capacitance, in F

　　ΔU: voltage drop, is the difference between the highest working voltage and the lowest working voltage, unit VI: discharge current, unit A

　　Supercapacitor charging and discharging time calculation method

　　When generally used in solar indicator lights, LEDs use flashing light.

　　Light, for example, using an LED and controlling the flicker discharge duration to be 0.05 seconds per second, the charging current of the supercapacitor is 100mA, and the LED discharge current is 15mA.

　　The following takes the application of 2.5V50F on solar traffic lights as an example, the super capacitor charging time is calculated as follows: C& TI mes;dv=I& TI mes;t

　　C: Capacitor rated capacity; V: Capacitor working voltage; I: Capacitor charging; t: Capacitor charging time

　　So the 2.5V50F supercapacitor charging time is: t = (C & TI mes; dv) / I = (50 × 2.5) / 0.1 = 1250s

　　The supercapacitor discharge time is: C×dv-I×C×R=I×t C: Capacitor rated capacity; V: Capacitor working voltage; I: Capacitor discharge current; t: Capacitor discharge time; R: Capacitor internal resistance

　　Then the discharge time of 2.5V50F supercapacitor from 2.5V to 0.9V is: t =C×(dv/IR)

　　=50×[(2.5-0.9)/0.015-0.02] =5332s

　　The working time of the application on the LED is 5332/0.05=106640s=29.62 hours