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C (F): the nominal capacity of the supercapacitor;
R (Ohms): the nominal internal resistance of the supercapacitor;
ESR (Ohms): equivalent series resistance under 1KZ;
Vwork (V): normal working voltage
Vmin (V): cut-off working voltage;
t(s): continuous working time required in the circuit;
Vdrop (V): The total voltage drop at the end of discharge or high current pulse;
I(A): Load current; Approximate calculation formula of super capacitor capacity, The energy required to maintain = the energy reduced by the super capacitor. The required energy during the holding period=1/2I(Vwork+ Vmin)t; Supercapacitor reduced energy=1/2C(Vwork2 -Vmin2), Therefore, its capacity can be obtained (ignoring the voltage drop caused by IR) C=(Vwork+ Vmin) *I*t/( Vwork2 -Vmin2) An example is as follows: For example, in a single-chip application system, a super capacitor is used as a backup power supply. After a power failure, a super capacitor needs to be used to maintain a current of 100 mA for 10 seconds. 4.2V, then how much supercapacitor is needed to ensure the normal operation of the system? It can be known from the above formula: Working starting voltage Vwork=5V Working cut-off voltage Vmin=4.2V Working time t=10s Working power supply I=0.1A Then the required capacitance is:
C=(Vwork+ Vmin)*I*t/( Vwork2 -Vmin2) =(5+4.2)*0.1*10/(52 -4.22) =1.25F According to the calculation result, you can choose 5.5V 1.5F capacitor to meet the needs Up.
Example: Assume that the working voltage of the tape drive is 5V and the safe working voltage is 3V. If the DC motor requires 0.5A to hold for 2 seconds (it can work safely), how much super capacitor should be used? Solution: C=(Uwork+ Umin)It/(Uwork*Uwork -Umin*Umin)=(5+3)*0.2*2/(5*5-3*3=0.5F Because the voltage of 5V exceeds the single capacitor Therefore, two capacitors can be connected in series. If two identical capacitors are connected in series, the voltage of each capacitor is the nominal voltage of 2.5V.
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