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I have come across a question that is very near and dear to my heart and I am in the hopes that somebody can answer this new years eve question of mine per se.

So I am wondering, if I were to run two 2.7V 360F super capacitors in parallel and attach them to a 5V step-up regulator, could I supply 500mA @5V (or very close to 5V) for about 2 hours?

I did some math from equations that I found online and I was wondering if the above question is plausible based off of the values stated above.

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  • \$\begingroup\$ Whatever the answer is, don't forget to match internal resistanse of source and load for maximum power transfer. \$\endgroup\$
    – GR Tech
    Jan 1, 2015 at 1:22
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    \$\begingroup\$ @GRTech What? He doesn't want maximum power transfer: he wants a power transfer of 2.5W (500mA @5V). \$\endgroup\$
    – Phil Frost
    Jan 1, 2015 at 1:40
  • \$\begingroup\$ @Phil Frost A warn to take care about supercapacitors ESR and linear discharge \$\endgroup\$
    – GR Tech
    Jan 1, 2015 at 2:00
  • \$\begingroup\$ Real world result will be more like 50 mA at 5V for 2 hours. \$\endgroup\$
    – Russell McMahon
    Jan 1, 2015 at 3:06

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The energy in a capacitor is

$$ E = \frac{1}{2}CV^2 $$

So at 2.7 V in two capacitors, you would have

$$ E = 2\:\text{capacitors} \cdot \frac{1}{2} \cdot 360\:\mathrm F \cdot (2.7\:\mathrm V)^2 = 2624.4 \:\text{joules} $$

The amount of energy you are taking per second is

$$ 0.5\:\mathrm A \cdot 5\:\mathrm V = 2.5\:\mathrm{J/s} $$

So, in a perfect capacitor with a perfect power supply, you could run this for

$$ \frac{2624.4\:\mathrm J}{2.5\:\mathrm{J/s}} = 1049.76\:\mathrm s = 17.496 \:\text{minutes} $$

Note, that this is only with a perfect capacitor. Supercaps tend to have a high series resistance that loses energy. In addition, the regulator has an efficiency that will vary according to the input, likely going down in various ranges, and certainly not really going to 0 Volts. This means you will have to have the amount of energy to maintain the minimum voltage, and add to it the amount you actually consume for the time you want.

To run a 5 V power supply with 90% efficiency at 0.5 A for 2 hours, or 7200 seconds requires a specific amount of energy to the input of the power supply: $$ 5 V \cdot 0.5 A \cdot 7200 s \cdot \frac{1}{0.90} = 20000 J$$ Note that the 90% efficiency effectively increases the amount of energy needed.

In addition, power supplies typically won't run down to 0 V. So there must be extra energy to handle the amount that is never discharged. We'll call the minimum voltage Vmin.

$$ E_{required} = E_{@Vmax} - E_{@Vmin} = \frac{CV_{max}^2}{2} - \frac{CV_{min}^2}{2}$$ $$ E_{required} = \frac{C}{2} \cdot (V_{max}^2 - V_{min}^2 )$$

So for a Vmax = 2.7 V and a Vmin = 0.5V $$ 20000 J = \frac{C}{2} \cdot (7.04 V^2)$$ $$ C = \frac {20000 J \cdot 2}{7.04} = 5681.8 F$$

This could be one huge capacitor or many smaller adding up in parallel.

However, note that I haven't considered any losses due to series resistance. That just adds on the total capacitance needed, but by using multiple parallel capacitors, you tend to reduce the effective resistance as you are putting them in parallel.

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  • \$\begingroup\$ \$Q\$ is very commonly used a a symbol for charge, so I've edited the equations to use \$E\$ instead. Sometimes also \$W\$ is used, for "work" (still measured in joules of course). Additionally, I think your calculations neglect that the question asks about two 360F capacitors, so he has twice the energy available, or an effective capacitance of 720F. Otherwise, looks good. \$\endgroup\$
    – Phil Frost
    Jan 1, 2015 at 1:36
  • \$\begingroup\$ Thanks for the fixup, but the 1/2 was cancelled by the two capacitors. But I can explicitly state it. \$\endgroup\$
    – caveman
    Jan 1, 2015 at 3:11
  • \$\begingroup\$ Ah, I totally missed that the first time. Looks great now. \$\endgroup\$
    – Phil Frost
    Jan 1, 2015 at 13:34
  • \$\begingroup\$ You said that supercaps tend to have high series resistance but I will be running my supercaps in parallel not in series, so will I not have the high resistance? @caveman \$\endgroup\$ Jan 2, 2015 at 20:53
  • \$\begingroup\$ Also I have used the formulas that you mentioned in your answer and I have come to the conclusion that if I used 10 500F 2.7V capacitors in parallel or a single 5000F 2.7V capacitor, I could provide 0.5A @ 5V for just about 2 hours. Does this seem correct to you? @caveman \$\endgroup\$ Jan 2, 2015 at 20:57
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A supercapacitor does not sound like the right device for your application. Most supercapacitors have a sizable internal resistance. This internal resistance effectively appears in series with the capacitor, and it has the same consequences:

  • As current flows through this resistor (to flow through your load), voltage is dropped across the resistor (Ohm's law). Thus, even if the capacitor is charged to 2.7V, if you pull 500mA from it, the terminal voltage will be less.
  • The energy lost in this internal resistance (according to one of Joule's laws: \$P = I^2 R\$) wastes energy making your device less efficient, and heats the capacitor, potentially damaging it.

You've not provided any link to any specific supercapacitor or datasheet, but 500mA is a lot of current for most supercapacitors.

However, 500 mA is well within the capabilities of many batteries. 500 mA for 2 hours is 1000 mAh (milliamp-hours). For comparison, Energizer's datasheet gives the capacity for their ordinary AA alkaline battery at a discharge rate of 500 mA at around 1400 mAh. Four of them in series would give you a nominal 6V and 140% of your required runtime. That's just an ordinary alkaline -- there are many other chemistries that may be even better suited, depending on what your requirements for cost, size, longevity and so on are.

Moreover, batteries maintain a more constant voltage as they discharge compared to capacitors, so if you pick the right battery, you may not even need the boost converter.

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  • \$\begingroup\$ Well the reason that I am interested in super capacitors is because i really like the idea of their 5 minute charge times versus the couple of hours that it would take to charge a decent sized Li-ion battery, but i do appreciate the comment. \$\endgroup\$ Jan 1, 2015 at 20:58
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Ok, this is what I have found from an initial google:

$$ Ah = \frac{(V_{MIN}+V_{MAX})/2 \times C}{3600} $$ Given an acceptable voltage range for your boost converter of 2.5V to 2.7V (pure assumption here), that equates to (for one capacitor): $$ Ah = \frac{(2.7 + 2.5)/2 \times 360}{3600} $$ $$ Ah = \frac{2.6 \times 360}{3600} $$ $$ = 0.26Ah $$ So two in parallel would be 0.52Ah, or 520mAh.

That could theoretically provide 520mA for 1 hour at 2.5V to 2.7V. So there's no way you can get 500mA at 5V for 2 hours.

For that you'd need 1000mAh at 5V, which pre-boost would be about double the current and half the voltage, so you'd be looking at 2000mAh at 2.7V.

For that you'd need 8 super capacitors of that type in parallel.

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