* Question

How to choose super capacitor? |

* Answer

Two main applications for supercapacitors: high power pulse applications and instantaneous power retention.Features of high power pulse applications: instantaneous flow to load high current; characteristics of instantaneous power retention applications: requires continuous power supply to the load, typically for a few seconds or minutes.Two calculation formulas and application examples are provided below: C(F): nominal capacity of supercapacitor; R(Ohms): nominal internal resistance of supercapacitor; ESR(Ohms): equivalent series resistance at 1KZ; Uwork(V): normal operating voltage in the circuit; Umin(V): minimum voltage required to operate the device; t(s): required hold time in the circuitEnergy required during holding period = 1/2I (Uwork Umin) t; supercapacitor reduces energy = 1/2C (Uwork2-Umin2), thus, its capacity can be obtained (ignoring the voltage drop caused by IR) C=(Uwork Umin)t/(Uwork2-Umin2) Example: Assume that the tape drive operates at 5V and the safe operating voltage is 3V.If we choose a capacitor with a nominal capacity of 1F, the two strings are 0.5F.This choice does not provide sufficient margin given the -20% capacity deviation of the capacitor.Capacitors with a nominal capacity of 1.5F can be selected to provide 1.5F/2=0.75F.Consider a -20% capacity deviation with a minimum of 1.2F/2 = 0.6F.Pulse Power Applications Pulse power applications are characterized by a small continuous current as opposed to an instantaneous large current.Pulse power applications last from 1 ms to a few seconds.
Design analysis assumes that supercapacitance is the only energy provider during the pulse.Take 2.5V1.5F as an example.Its internal resistance R can be estimated by direct current ESR, nominally 0.075 Ohms (DCESR = ACESR * 1.5 = 0.60 Ohms * 1.5 = 0.090 Ohms).The rated capacity is 1.5F.For a 0.001 s pulse, t/C is less than 0.001 Ohms.The notebook’s limited output voltage to the PCMCIA is 3.3V / -0.3V and the notebook provides 1A of current.Many power amplifiers (PAs) require a minimum voltage of 3.0V.It is possible to output a 3.0V voltage to the notebook.The voltage to the power amplifier must first rise to 3.6V.According to the above formula, the pressure drop caused by the internal resistance is: 1A × 0.25 Ohms = 0.25V.I(t/C) = 0.04V which is small compared to the voltage drop caused by internal resistance.The series should be balanced to ensure an even distribution of voltage.The voltage drop caused by the internal resistance of the supercapacitor in pulsed power applications is often a secondary factor.The ultra-low internal resistance of the capacitor provides a completely new solution to overcome the large impedance of conventional battery systems. |

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