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How can I calculate charging and discharging time of capacitor? The capacitance is 2100F, rated voltage 2.8V, internal resistance .0158 mohm. Let's say I have 15W of power to feed in the capacitor. how do I calculate charging and discharging time for that? The specific energy is 37 Wh/Kg and specific power 5.6 kW/Kg. based on the charging and discharging time I need to determine how many capacitor do I need to feed 15W. Can you please tell me how can I solve that? Thanks

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  • \$\begingroup\$ What value of resistor you will be using? what voltage ? You have any information on this? \$\endgroup\$ – abhishek tyagi Mar 26 '16 at 3:03
  • \$\begingroup\$ I am using constant power. I am considering the rated voltage and internal resistance. Unfortunately I do not have more information. \$\endgroup\$ – Nayeem Mar 26 '16 at 5:50
  • \$\begingroup\$ Do you have the datasheet or at least the part number? \$\endgroup\$ – jms Mar 26 '16 at 5:59
  • \$\begingroup\$ I am using yunasko li ion capacitor.yunasko.com/en/technology/competition \$\endgroup\$ – Nayeem Mar 26 '16 at 6:26
  • \$\begingroup\$ If anyone can suggest me how can I simulate the li ion capacitor in pspice or simulink would be highly appreciated \$\endgroup\$ – Nayeem Mar 26 '16 at 6:27
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If that 2100F is not a typo, so capacticance = 2.1kF, then you can do the sums as follows.

Stored energy at 2.8v = 0.5CV^2 = 8400J

However, you would not be able to charge/discharge efficiently over more than about a 2:1 voltage range, so let's put the minimum voltage at 1.4v. The available energy change by swinging between 1.4v and 2.8v is then 75% of the maximum stored energy, which is 6300J.

If you read the Yunasko data you linked to, then you will see that the specific energy is computed according to note 1, which calls for a voltage swing between rated and 50% rated voltage.

At 15 watts (assuming 100% efficiency in a switchmode converter) that would take 420 seconds = 7 minutes to charge or discharge.

You can reduce the lower voltage further, but you are running into diminishing returns, and making the converter less efficient. If you swung between 2.8 and 1v for instance, then you would be able to use 87% rather than just 75% of the maximum stored energy.

Probably the best way to model a capacitor in SPICE is to use a capacitor, in simulink a current integral.

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  • \$\begingroup\$ Thanks Neil_UK. How about the discharge?lets say I have a depth of discharge 80%. \$\endgroup\$ – Nayeem Mar 26 '16 at 7:08
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    \$\begingroup\$ \$ {1 \over 2} CV_1^2 - {1 \over 2} CV_2^2 = E \$ \$\endgroup\$ – Daniel Mar 26 '16 at 8:22
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    \$\begingroup\$ \$ {E \over t}=P \$ \$\endgroup\$ – Daniel Mar 26 '16 at 8:23
  • \$\begingroup\$ Solve for t.... \$\endgroup\$ – Daniel Mar 26 '16 at 8:24
  • \$\begingroup\$ @Daniel thanks. Let's say I have rated voltage 2.8v which is V1. How should I take the value of V2 as I do not have any indications. \$\endgroup\$ – Nayeem Mar 26 '16 at 8:28
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If R is negligible, a fully charged capacitor holds

energy(W-seconds) =0.5 * 2100 * 2.8^2 = 8232 W-sec

and takes (lower limit, not exact value) 8232/15 = 548.8 seconds to charge to 2.8V from zero.

More accurately, the power limit applies to the heat generated in the resistance, as well as to the rate at which the stored energy of the capacitor increases.

If the capacitor charge is V(t), its measured terminal potential is V(t) + I*R One can set up equations, and solve:

15 W = I^2 * (1.58E-5 Ohm) + d/dt(0.5 * 2.1E3 F * V^2(t))

I = C * dV/dt

but it's clear that when V is near zero, your 15W supply would be sourcing current at a rate only limited by that low series resistance: 970 amperes. That's not realistic, I hope.

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