I recently took part in a challenge called 'Hydrogen Hack' where we have a Raspberry Pi and have to make something controlled by that. The catch is that it has to be powered by a Hydrogen Fuel Cell - each fuel cell has a USB style output of 5 volts at 1 amp. This is enough to run the Raspberry Pi (we have tested) and draw 5 V off the 5 V GPIO Pin.

However, the minimum voltage our motors can run at is 8 V at 1 A but preferably we need 12 V 1.5 A as the faster the motors spin the better and the H-Bridges we are using draw some power (obviously).

Me and my friends have come up with a few ideas but all of them seem impractical, we need to draw the 8 V 1A (or 12 V 1 A) for around 20 seconds preferably but this could be shrank to 15 seconds again if necessary.

I was wondering if any of you guys could come up with some ideas as to how / if we can do this?

Brief Details:

Pi Input: 5 V 1 A

Pi GPIO 5 V Output: 5 V 200 mA

Power build up: 8 V 1 A for 15 s but preferably 12 V 1 A for 20 seconds

I have googled and cannot really find anything relevant to what we need as it is probably silly to think anyone will of done this and bothered to document it before.

The power loss in the L293D is aprox.: (1.4V + 1.2V) x 0.6A = 1.56W

The motor preferably needs: 12v * 1A = = 12w

Wattage available: 1w

So i firstly need a way of converting the 1w to 13.56 and then storing it

And by the way, You cannot use any other batteries but they did not specifically rule out capacitors.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Oct 3 '17 at 20:05

This is pretty straightforward.

You use a boost converter with a nominal 12 volt output which is also current limited to an input current which will fit your power budget from the USB. Let's say your USB power supply can spare 100 mA.

Now let's look at your energy storage capacitor. Because you want it to supply about 12 volts for 20 seconds, and the capacitor voltage will drop under load, how big a capacitor do you need? Let's say, just as a starting point, that you'll accept a 2 volt drop in 20 seconds, or 0.1 volt/sec. During this time, assume that the charging current is insignificant.

The pertinent equation relates capacitance, current and voltage by $$\frac{dV}{dt}= \frac{i}{C}$$ In this particular case, we can talk about the voltage and time discrete differences rather than the instantaneous values, so $$\frac{{\Delta}V}{{\Delta}t}= \frac{i}{C}$$ or $$C = \frac{i{\Delta}t}{{\Delta}V} $$ Plugging in your values, i = 1.5, delta V = 2, delta t = 20, we get C = 15. That is 15 farads, not microfarads.

Now, as for charging. Assuming 100% efficiency, the charger must provide the same input power as output. The input power (current times voltage) must be 5 times 0.1, or 0.5 watts. 12 volts out means the output current must be (5/12) 0.1, or about 40 mA. Since the output current is 1.5 amps the charge time must be 1.5/.04 or about 37 times longer than the discharge time. In this case, that works out to about 12 1/2 minutes of charging time for each motor activation.

You can play games with this, of course. If you let the voltage drop from 12 to 8 volts, the capacitance can be half as great, so the charge time will be approximately halved.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Oct 3 '17 at 20:03

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