Support Note
How to Use Supercapacitors? A Brief Guide to the DesignIn Process
SN009 BY DR. RENÉ KALBITZ / FRANK PUHANE
1 EDLC – Supercapacitor
Compared to other capacitor technologies, EDLCs (Electric Double Layer Capacitor) are outstanding for their very high charge storage capacity and very low equivalent series resistance (ESR). Their high cycle life, low charging time and their large power output make them the ideal choice for many electric power applications. Possible applications are:
(Intermediate) storage devices
 To provide an application with power during battery change or poweroffline periods
 To provide power in emergency cases as uninterruptible power supplies (UPS)
Hybrid application with battery
 To relieve batteries during high power peak
 To buffer energy fluctuations in order to increase battery life time
The most important parameters for the designin process are capacitance, discharging and charging time as well as the corresponding voltages. Below we present a summary of the most important formulas and provide examples of calculations.[1,2,3]
Figure 1: General concept of charging/discharging infrastructure.
2 General Procedure of DesignIn
1^{st} Identify the mode of operation for the discharge process:
 Constant Resistance
 Constant Current
 Constant Power
2^{nd} Calculate*) the necessary capacitance depending on desired operation parameter such as operation time, output power and output current.
*) For the sake of simplicity we may neglect the losses due to ESR, leads and connections.
3^{nd} Identify the suitable charging process:
 Constant Current
 Constant Voltage
4^{nd} Calculate the charging time depending on the charging current. If necessary calculate the protective resistor.
Figure 2: Radial throughhole EDLC series WCAPSTSC
Some important formulas for the designin process are summarized in the following sections.
3 Parameter and Performance
Figure 3: Equivalent Circuit of EDLC
Basic Parameters:
V_{R} _{} 
rated voltage

C  capacitance (given in the datasheet and directly on the capacitors marking) 
R_{ESR}  equivalent series resistance (ESR) (given in the datasheet) 
R_{Leak} _{} _{} 
equivalent parallel resistance, leakage resistance

P  power output, i.e. power consumption of application 
Performance Parameters:
V1  charging voltage, usually VR = V1 
V2 
lower cutoff voltage energy storage capacity: maximum power output: 
3.1 Example
An application needs to be driven with a constant power of P = 0.4 W for t = 360 s. The lower cutoff voltage is V_{2} = 1 V. How large is the total amount of energy E and how large is the required capacitance C?
Calculation:
The required energy is E = 144 J
The required capacitance is C = 46 F, thus a capacitor with a capacitance of 50 F is recommended.
4 Constant Voltage Charging
For constant voltage charging it is recommended to use a protective resistor in series with the EDLC. It may be necessary to restrict the current with a protective resistor R_{P} to a specific value I_{max}. For a given Imax the resistance is calculated by:
The charge characteristic is calculated by (t_{0} = 0):
The corresponding charging time is calculated by:
Charging to 99.9% :
C  capacitance 
V_{1}  charging voltage 
I_{0}  current at t_{0} 
I_{max}  max. allowable current 
V_{R}  rated voltage 
V  voltage at t 
t  charging time 
t_{0}  start time 
R_{P}  protective resistance 
R_{ESR}  equivalent series resistance 
p  charging level in % 
Figure 4: Vt characteristics for constant voltage charging
4.1 Example Protective Resistance
A capacitor with capacitance C = 50 F and an equivalent series resistance R_{ESR} = 0.02 Ω shall be charged with a unprotected power source at V_{1} = V_{R} = 2.7 V. The power source has a maximal allowable current of I_{max} = 5 A. How large should the protective resistance be, to prevent overcurrent?
Calculation:
In order to prevent over current at the power source, a protective resistor with R_{P} ≥ 0.52 Ω should be used.
4.2 Example Charging Time
A capacitor with capacitance C = 50 F is charged to V = 2.16 V (80 % of V_{R}) at constant voltage V_{R} = 2.7 V with a protective resistor R_{P} = 0.5 Ω and an equivalent series resistance R_{ESR} = 0.02 Ω. How long is the charging process?
Calculation:
The charging time is ≈ 42 s.
Figure 5: It characteristics for constant voltage charging
5 Constant Resistance Discharging
The discharging characteristics of a capacitor with capacitance C over given load resistance R_{L} is calculated by (t_{0}=0):
The corresponding discharging time is calculated by:
The necessary capacitance is calculated with:
C  capacitance 
V_{0}  charging voltage at t_{0} 
I_{0}  current at t_{0} 
V  voltage at t 
t  discharging time 
t_{0}  start time 
R_{L}  load resistance 
R_{ESR}  equivalent series resistance 
Figure 6: Vt characteristics for constant resistance discharging
5.1 Example Discharging Time
A capacitor with capacitance C = 50 F is discharged from its rated voltage V_{R} = 2.7 V to V = 0.3 V with a load of R_{L} = 1 Ω. How long is the discharging process?
Calculation:
The discharge time is approximately 112 seconds.
5.2 Example Voltage Drop
A capacitor with a capacitance C = 50 F is discharged from its rated voltage V_{R} = 2.7 V with a load of R_{L} = 2 Ω for a period of time t = 280 s. What is the remaining voltage?
Calculation:
The remaining voltage is V = 0.17 V.
Figure 7: It characteristics for constant resistance discharging
6 Constant Current Charging/Discharging
If a constant current is used, the voltage V at the terminals for time t (t = 0) is calculated by:
The corresponding discharge time (t_{0} = 0) is calculated by:
The corresponding charging time (t_{0} = 0) is calculated by:
The necessary capacitance is calculated with:
I _{C,D}  constant charge / discharge current 
C  capacitance 
V_{R}  rated voltage 
V, I  voltage, current at t 
V_{0}  voltage at t_{0} (charging) 
 t  t_{0}   (dis)charge time 
t_{0}  start time 
R_{ ESR}  equivalent series resistance 
6.1 Example Charging Time
A capacitor with capacitance C = 50 F is charged from V_{0} = 0.3 V to its rated voltage V_{R} = 2.7 V with a constant current I_{C} = 2 A. How long is the charging process?
Calculation:
The charge time is 60 seconds.
6.2 Example Voltage Increase
A capacitor with capacitance C = 50 F and an initial voltage V_{0} = 0.3 V is charged with a constant current IC = 2 A for t = 5 s. How large is the capacitor voltage?
Calculation:
The capacitor voltage is V = 0.5 V.
Figure 8: Vt characteristics for constant current charging.
Figure 9: Vt characteristics for constant current discharging.
Figure 10: It characteristics for constant current charging and discharging.
7 Constant Power Discharging
If the capacitor is discharged at a constant power P_{C}, the voltage and current characteristic are calculated by (t_{0} = 0):
The corresponding discharge time (t0 = 0) is calculated by:
The necessary capacitance is calculated with:
P_{C}  constant power output 
C  capacitance 
V_{R}  rated voltage 
V, I  voltage, current at t 
I_{0}  current at t_{0} 
V_{0}  voltage at t_{0} (charging) 
t  t_{0}  discharge time 
t_{0}  start time 
Figure 11: Vt characteristics for constant power discharging
7.1 Example Discharge Time
A capacitor with capacitance C = 50 F and rated voltage V_{R} = 2.7 V is discharged at constant power P_{C} = 0.2 W. The cutoff voltage is V = 0.7 V. How long can the capacitor be operated under this condition?
Calculation:
It can be operated for t = 850 s.
7.2 Example Voltage Drop
A fully charged capacitor with capacitance C = 50 F and rated voltage V_{R} = 2.7 V has been operated for t = 180 s at constant power output of P_{C} = 0.7 W. How large is the remaining voltage?
Calculation:
The remaining voltage is V = 1.5 V
Figure 12: It characteristics for constant power discharging
A. Appendix
A.1. References
[1] N. Kularatna (2015). Energy Storage Devices for Electronic Systems – Rechargeable Batteries and Supercapacitors. Elsevier Academic Press (Print Book)
[2] F. Beguin, E. Frackowiak, G. Q. M. Lu (eds.) (2013). Supercapacitors  Materials, Systems, and Applications. WileyVCH (Print Book)
[3] B. E. Conwa (1999). Electrochemical Supercapacitors – Scientific Fundamentals and Technological Applications. Kluwer Academics / Plenum Publishers, New York (Print book)
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SN009: How to use supercapacitors
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