How to Discharge 4.2V 1F Supercapacitor at 1.5A to Charge a 4V 1500mA Battery?

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#1 21673230 19 Jan 2015 15:45

Enrico Koeck Enrico Koeck

Anonymous

Post #1
21673230 19 Jan 2015 15:45

At first the energy stored in the capacitor is very low compared to the battery. The energy in the battery is 4V * 1.5Ah = 6Wh. In comparison the cap has only 0.5 * U^2 * C = 11.5Ws = 0.0032 Wh energy stored. Besides that you cannot directly charge this battery from the capacitor because if the voltage drops below that of the battery it won't be any longer possible to charge it until you add circuitry to increase the voltage.
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#2 21673229 19 Jan 2015 15:46

noneya noneya noneya noneya

Anonymous

Post #2
21673229 19 Jan 2015 15:46

Im looking at a super capacitor but I’m not sure what the discharge characteristics graphs actually mean. My goal is to be able to charge this capacitor, and discharge its energy into a battery.

The specs are:
Capacitor: 4.2V (5.5V peak) with 1F
Battery I’m looking to charge: 4V with 1500mA

There are two graphs in the image attached that I have a hard time understanding. First graph, the Constant Current Discharge @ 25C Charge 5.5V. The 1Amp goes from 5.5V and down to 0 in 4.5 seconds. There are two additional line graphs 2A and 4A. Second graph, Constant Power Discharge 25C Charge 5.5V, with line graphs of 1W, 5W, and 10W. It appears 10W line after 1000 msec, voltage becomes close to 2V.

How would you discharge this capacitor at a rate of 1.5A with 4V to charge a battery? Thanks for reading and I would appreciate any help.
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#3 21673231 19 Jan 2015 15:54

noneya noneya noneya noneya

Anonymous

Post #3
21673231 19 Jan 2015 15:54

Thanks for the reply. What formula did you use to determine the wattage and how did you get the watt hours?
#4 21673232 19 Jan 2015 17:23

Enrico Koeck Enrico Koeck

Anonymous

Post #4
21673232 19 Jan 2015 17:23

Battery: E = U * I * t = 4V * 1500mAh = 1.5Ah * 4V = 6 Wh = 21.6 kJ Capacitor: E = 0.5 U^2 C = 0.5 * (5.5V)^2 * 1F = 15.1 VAs = 15.1 Ws = 15.1 J
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#5 21673233 19 Jan 2015 17:28

noneya noneya noneya noneya

Anonymous

Post #5
21673233 19 Jan 2015 17:28

Yes I think you're math is correct, I got the same using some other formulas I found online.

I think this capacitor doesn't fit my needs but I'll use another large volume capacitor.

The problem is how would I discharge the capacitor at a certain level of amperage to charge the battery? How would I go about doing this?
#6 21673234 19 Jan 2015 17:55

noneya noneya noneya noneya

Anonymous

Post #6
21673234 19 Jan 2015 17:55

Please read below.
#7 21673235 19 Jan 2015 21:25

noneya noneya noneya noneya

Anonymous

Post #7
21673235 19 Jan 2015 21:25

For example if I used two 2.7v 3000F in series to get 5.4v @ 1500F. Do I use a step up dc-dc booster or voltage regulator to charge my 4V 1500mA battery? And when I do charge the battery, wouldn’t the voltage of capacitor drop lower to a point where the booster or regulator required input voltage is not reach therefore not being able to give the desired output of 4V?

I purchased a step down regulator to drop down the 12V input voltage @ 1Amp as the power supply to reduce it to 5.4V-6V as a way to charge the capacitors. For example if the capacitors are 5.4V at 1500F, what would be the ideal charging voltage? And how would I calculate how fast it would charge until its full?
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#8 21673236 20 Jan 2015 04:45

Enrico Koeck Enrico Koeck

Anonymous

Post #8
21673236 20 Jan 2015 04:45

At first, as you see above, you would need more than 40 1F/5.5V capacitors to fully charge your battery. How you charge your battery depends on the material used (LiPo, LiFePo, NiMH, NiCd, Pb, ...). Some are charged by constant current, some by constant voltage and so on. For every type there are charging controllers as ICs available. And yes, you have to use a Boost-Converter increase your voltage. If you are charging with constant current your charging time will be 1500mAh divided by the amount of current (like 1A charging current would result in a charging time of 1.5h).
#9 21673237 20 Jan 2015 18:02

noneya noneya noneya noneya

Anonymous

Post #9
21673237 20 Jan 2015 18:02

Continue below.
#10 21673238 20 Jan 2015 18:17

noneya noneya noneya noneya

Anonymous

Post #10
21673238 20 Jan 2015 18:17

Right that makes sense. However during the use of the capacitors to charge my 4V 1500mA battery, lets say a Li-Ion, I simply use a step up boost converter since the capacitor’s voltage will decrease as time goes and this will automatically boost the voltage up but sacrificing some current during the charge, correct? I have ordered a step down but I have not yet purchased a step up yet because I am unsure.

This is an experimentation project of simply transferring energy from capacitor to battery. I do not quite understand the discharging of capacitor especially when it comes to using that discharge to charge a device or battery. This is the capacitor I am looking at, BCAP3000: http://www.mouser.com/ds/2/257/Maxwell_K2Series_DS_1015370-4-341196.pdf. If I am charging a 4V 1500mA battery with two of these units, how will I determine the current it will discharge at and the time it takes for the voltage to drop from 5.4V down below an unreliable charge voltage?
#11 21673239 21 Jan 2015 05:19

Enrico Koeck Enrico Koeck

Anonymous

Post #11
21673239 21 Jan 2015 05:19

You could use for example a LTC3558 IC which is a battery charger circuit including a Buck/Boost-DC/DC-converter. You can programm it to the charging current you want to use which is stated in the datasheet of your battery. If you draw a constant current from the cap the voltage over time is: Uc = U0 - Ic * t / C where Ic is the drawn current and t the passing time in seconds. The LTC3558 has an input range of 4.3V-5.5V so you can use only about 40% of the energy stored in the capacitor. If thats not enough you need to look for a boost converter with the lowest possible input voltage, boost it to maybe 5V and use another IC for battery charging. But the boost converter has to be able to provide the desired output current.
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Topic summary

The discussion addresses the challenge of discharging a 4.2V, 1F supercapacitor at 1.5A to charge a 4V, 1500mAh battery. The energy stored in the capacitor (approximately 15.1 joules or 0.0042 Wh at 5.5V peak) is significantly lower than the battery's energy capacity (6 Wh). Direct charging is impractical because the capacitor voltage drops below the battery voltage during discharge, halting current flow. To maintain charging, a DC-DC boost converter is necessary to elevate the capacitor voltage to the battery charging voltage. The capacitor voltage decreases linearly with constant current discharge, following Uc = U0 - Ic * t / C. Using multiple capacitors in series can increase voltage and capacitance (e.g., two 2.7V 3000F capacitors in series yield 5.4V at 1500F). Charging the capacitor bank requires a step-down regulator from a higher voltage source. Battery charging methods depend on battery chemistry (Li-Ion, LiFePO4, NiMH, etc.) and typically involve constant current or constant voltage charging controlled by dedicated ICs. The LTC3558 IC is recommended as a programmable buck/boost charger with an input voltage range of 4.3V to 5.5V, suitable for managing the limited usable energy from the capacitor. Boost converters must have low input voltage thresholds to utilize more capacitor energy and provide stable 4V output at the required current. Charging time is estimated by dividing battery capacity by charging current (e.g., 1.5Ah / 1A = 1.5 hours). The BCAP3000 supercapacitor from Maxwell is referenced for practical experimentation.
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How to Discharge 4.2V 1F Supercapacitor at 1.5A to Charge a 4V 1500mA Battery?

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