Can I power a Board of Education from capacitors?

winchman

In my winch project, I'm using a BoE with a BS2. See:
http://forums.parallax.com/forums/default.aspx?f=5&m=96533&g=96698#m96698

The only input is from a Hall-effect sensor, which will get its power from the BoE.

There are two outputs. Both go into a 7407 buffer which leads to two MOSFETS to control some large relays. Very little power is required on the outputs. The circuit is shown on page 157 of the Microcontroller Cookbook.

The BoE will need to be on for only four seconds each time it is used, sometimes less than that.

How much power will the BoE require to operate for this set-up? I could find the power requirements for the Stamp itself, but not for the BoE.

Would it be feasible to power the BoE from capacitors which have been charged to 12 volts just before the BoE is used?

If so, will I need to limit the input voltage to 9 volts, or can I apply the full 12 volts from the capacitors to the battery terminals or power input connector on the BoE?

What's the minimum voltage required to keep the BoE working properly?

Any suggestions for a circuit design would be appreciated, if I need some way to limit the voltage to 9 volts going into the BoE.

Thanks,
Roger

BTW, the project is coming along nicely. I decided to build a winch simulator to make it easier to test the software and circuitry. It's not quite finished, but it's already in use testing some parts of the programming.

Post Edited (winchman) : 12/22/2005 7:35:38 AM GMT

bobledoux

Try this:

http://www.bairesrobotics.com.ar/elektor/10-2003 SuperCap Battery.pdf

PJAllen

http://www.foreverflashlights.com/forever_flashlights_3.htm

Tracy Allen

The main power consumption on the BOE will be the quiescent current of the LM2940 voltage regulator and the indicator led. Those consume around 18 or 20 milliamps. Assume that the Stamp when running will consume 5 milliamps. That makes a total current of 25 milliamps. The LM2940 can operate with its input as low as 5.5 volts (and 12 volts input is no problem).

The equation for voltage change on a capacitor as it delivers current is,
I = C * dV/dt

Suppose the current is constant (which is a close approximation), you need to find the value of C that will hold the voltage above 5.5 volts for at least 4 seconds. Start with V=12 volts. That makes dV=6.5 volts. dt=4 seconds. I=0.025 amp.

C = I * dt / dV = 0.025 * 4 / 6.5 = 0.015 farad.

15000 microfarads. That is big, unless you can find a supercap that you can charge to 12 volts. If it were only the Stamp, at 5 milliamps, you could get by with 3400 microfarads.

Or possibly the circuit Bobledoux pointed to, URL. That used a 10 farad capactitor charged to 2.3 volts and a switching booster to 9 volts. It delivered a current of 8.5 milliamps at the output, and that took 50 milliamps at the input, for a current ratio of 5.88. Suppose it could be used at that efficiency to produce 25 ma at the output? That would be 147 milliamps at the input. From an initial charge of 2.3 volts down to its minimum operating voltage of 1.0 volts, that makes dV = 1.3 volts. How long could it power your BOE?
dt = C * dV / I = 10 * 1.3 / 0.147 = 88 seconds.
It might work. But the efficiency of the switcher probably drops off drastically at higher currents. It should be redesigned to deliver 6 volts instead of 9 volts (at higher efficiency) and again you'd be better off with the lower current requirement of the Stamp alone, apart from the BOE.

Why can't you use a rechargeable battery?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Tracy Allen
www.emesystems.com

Orion

You could pull the led off the BOE and drop ~15ma off the total.

winchman

Thanks, Tracy, for the info and doing the calculations. That's a real help.

The capacitors are no problem. I've got about a dozen 16V 4700mfd caps right here. I've been using them on my winches to hold a relay closed for several seconds at the end of a launch.

There two reasons I'd like to use the caps to run the BoE:

First is to get the Stamp off the winch battery when the motor is being switched on and off. It'll be pulling about 600 amps for the half second it's on, so there's bound to be some nasty transients.

The second reason is that I don't want to leave the BoE on while I'm away from the winch flying the plane. I know from experience that I'll forget to turn it off after a launch, or on before the next launch. The cap can just bleed down, and it'll be charged up automatically the next time I use the winch.

I'll probably just use a 9-volt battery for testing the system, and use the cap arrangement for a permanent installation.

Thanks again,
Roger