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I am constructing a system in a solar car, and need to have capaitors large enough to power the Telemetry system for about 15s after power down.

I need to use capacitors instead of the batteries, due to race regulations.

The Telemetry system takes about 1 amp at 5v leading to a power consumption of approximately 5 watts. The minimum operating voltage is about 4.5 volts.

I am thinking of using a 5 farad capacitor, but I don't know if this will work. Could someone please shed some light on how this is calculated?

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a super capacitor is probably a better solution. I'll leave it up to someone else to suggest a brand as I don't design with them. –  placeholder Jan 10 '13 at 17:52
    
you are right, I am going to use 7 of these goo.gl/4cZnY because it fits more in my price range. (still a nice chunk of change for 15 seconds though...) –  Reid Jan 10 '13 at 19:32
    
Actually, I found a cheaper solution. I am chaining a bunch of 2.2F caps together on a board, and putting a diode in series with it, so the caps don't back power the rest of the system. I will be able to get 88F, and a descent run time. –  Reid Mar 5 '13 at 3:12
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3 Answers

up vote 5 down vote accepted

Basic capacitor equation:

\$ Q = V C \$ (stored charge = voltage * capacitance)

Current is the time derivative of charge:

\$I = \frac{dQ(t)}{dt}\$ (current = change in charge over change in time)

The time derivative of the capacitor equation is:

\$\frac{dQ(t)}{dt} = C \frac{dV(t)}{dt}\$

\$I = C \frac{dV(t)}{dt}\$

Making a coarse assumption that your discharge current is constant (at 1A), your voltage will decay linearly with a slope related to the capacitance.

That is to say, if your capacitor voltage starts at 5V and you use a 5F capacitor and you draw charge away at 1A:

\$1A = 5F \frac{dV(t)}{dt}\$

\$\frac{0.2A}{F} = \frac{dV(t)}{dt}\$

so your voltage will decay at a rate of 0.2V per second, and you will hit 4.5V in about 2.5seconds...

In order to decay 0.5V over 15 seconds plug in numbers like this:

\$1A = C \frac{0.5V}{15s}\$

\$C = 1A \cdot 15s / 0.5V = 30F\$

You can always put more capacitors in parallel to achieve an additive capacitance.

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What Vicatcu said is correct. As rawbarwb pointed out, you want a supercap for that size. For example Maxwell Technologies Inc. BMOD0083 P048 B01 digikey link is 83 F, 48 V, 10 mOhm ESR. It is expensive, ~$1,500.

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I don't know if you have the price quite right... it's listed as ~$1500 USD for me. –  helloworld922 Jan 10 '13 at 18:41
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Well, vicatcu already gave you a good answer, but here is an idea. Since energy in a capacitor scales by \$ V^2\$ there would be an advantage to storing a voltage higher than 5V. You could use a smaller cap if you started with 10 or 15 volts.

So, charge the cap up to 10 or 15 volts. You may need a boost circuit to do that. It shouldn't take much to keep it there until you need it. Then use a buck to step the cap energy down to 5V. Power out of the buck (\$ P_o\$) would be 5W for 15 seconds. It should be easy to get a buck to have 95% efficiency (\$\eta\$), so power into the buck (\$P_ {\text {in}}\$) would be \$\frac {P_o} {\eta}\$.

Lets see:

  • \$ E_c\$ = \$\frac {V^2 C_s} {2}\$ , of course you can only use down to \$ V_o\$ of 5V, so \$ E_c\$ = \$\frac{1}{2} C_s \left(V_c^2-V_o^2\right)\$
  • and \$ E_c\$ = \$T_d P_ {\text {in}}\$ where \$ T_d\$ is the time to discharge the cap.

So, put these together and solve buck input capacitance (\$ C_s\$).

  • \$ C_s\$ = \$\frac{2 T_d P_o}{\eta \left(V_c^2-V_o^2\right)}\$ = \$\frac{2 (5 W) (15 \sec )}{\left(15 V^2-5 V^2\right) \eta }\$ = 0.79 F

So, for a 15V to 5V discharge on the cap you only need a ~0.8F cap. A 20V to 5V discharge of the cap only needs a ~0.42F cap.

It is surprising how few large capacity caps are stocked in the 20V to 30V range, but here is one that could work 0.47F @ 25V. May or may not be worth the trouble.

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