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Control Box Circuit Diagram - Proofreading

Unread postPosted: Sun Apr 27, 2008 11:13 pm
Author: mark.f
Well, so far, high school physics and a little googling has taken me this far, but I'd like somebody who knows a little bit more about electronics than me to proofread this. The values in the diagram are "average" values, and I can change them once I order parts. I know there aren't going to be exact matches for some resistor values, but they're there just for calculation's sake.

So far, I think I've got most everything right. The first parallel "leg" consists of a small capacitor, and the section wired parallel off of THAT is a small dirtbike ignition coil. The resistor under the capacitor is to reduce the voltage going to the LED. The LED will light up when the capacitor is charging (supposedly :wink: ) and go out when it's charged.

The second leg is a simple indicator LED for the fan. The resistor produces a voltage drop to bring the LED down to 3.6 volts and 20 mA. The last parallel leg is the fan itself, which uses a resistor to bring voltage down to 12 volts and 220 mA.

Based on all of this information, I should get a nice, low current draw on the battery (260 mA theoretically), and everything should work, but I could be wrong.

That's it. If anybody sees a problem I'd like to hear it before I go ordering parts from allelectronics (next paycheck).

Unread postPosted: Sun Apr 27, 2008 11:53 pm
Author: starman
Looks generally OK. Be sure the computer fan you use really is 220 mA. The ones I'm familiar with typically pull somewhat less than that....100-150 mA.

Unread postPosted: Mon Apr 28, 2008 5:16 am
Author: TurboSuper
Looks good to me. I don't think the resistor in series with the fan is necessary; they can run off 18V, IIRC.


...and is the internal resistance of a CPU fan really that low? They only draw mabye a hundred milliamps at 12V, which means that their resistance/impedance should be much higher....right?

Unread postPosted: Mon Apr 28, 2008 1:09 pm
Author: jimmy101
Circuit looks good. None of the resistor values will be all that critical, +/- 50% should work for all of them.

If the fan is a brushless fan then you can't really measure it's resistance or calculate it from the voltage and current ratings. Doesn't matter too much though, a few tens of ohms resistnce in series will work. Turbo is probably correct though, you probably don't need the resistor at all.

You probably want the cap to be a fairly big one. Perhaps a 120 uF (or more) photocap. Photocaps have low equivalent resitance so they discharge faster than generic caps. A 120uF cap charging through the 720 ohm resistor will take less than one second to fully charge so the LED indicator might not tell you much. If you omit the reistor and LED then you'll probably want to add a K ohm or so of resitance between the cap+ and the + power supply.

Unread postPosted: Tue Apr 29, 2008 6:10 am
Author: mark.f
Thanks jimmy, I was wondering what type of capacitor to use yesterday. I was planning to go on a millifarad (or larger) capacitor, but since it's really only the rate of change of voltage through the coil that matters, your idea sounds better. If I had the gumption to make a HV charging circuit or remove one from a camera for the capacitor, I would, but I think the 18v from the battery should provide sufficient enough dI/dt through the coil to produce a good enough spark (the coil is a dirtbike ATV coil since I finally figured out why I fried my last one).

EDIT: just one last quick question to anybody who knows. Are photoflash capacitors under "Snap-In Capacitors" on AllElectronics? Because clicking on photoflash capacitors link yields no items.

Unread postPosted: Tue Apr 29, 2008 1:01 pm
Author: jimmy101
Mark

You can get complete photoflash boards (including the photocap) from Allectronics for $1.85 each (http://www.allelectronics.com/cgi-bin/i ... MBLY_.html ) Normally, allelectronics has photocaps, looks like they don't currently have any.

There is a way to estimate whether or not you need a cap. The cap won't do anything if the resistance of the coil is about the same or more than the resistance of the battery. In this situation, the current through the coil is limited by the input voltage and the resistance of the coil. A cap, mow matter how big, won't do anything.

If the resistance of the coil is less than the resistance of the battery then the battery resistance limits the current through the coil. A cap will help since it is (usually) much lower resistance than the battery.

What's the resistance of a typical battery? Another very rough estimation ... the resistance can be derived from ohms law (V=IR) based on the rated voltage and maximum (short circuit) current. A 9V battery will source about 3A (but not for very long) so the internal reistance is about 3 Ohms. A car battery will source perhaps 300A so it's internal resistance is about 12V/300A=0.04 ohms. A rechargeable 1.5 AA battery will do about 8A, so it's internal resistance is about 0.2 ohms.

If you use a wall wart as the power supply then a cap will probably help since most wall warts will only put out about 100mA. Roughly 1/30th what a fresh 9V battery will do. (But the 9V battery will go dead pretty quick.)

I would make a wild-ass guess that the primary of your coil is less than 0.1 ohms. So, a cap would be helpful if you use a 9V battery but it won't do anything to boost the current if you are using a 12V car battery.

The other thing the cap does is it produces a power pulse in a more reproducable fashion than does does just a battery in series with a switch. So, you might decide to use a cap for that reason alone.

Unread postPosted: Tue Apr 29, 2008 7:01 pm
Author: mark.f
Okay, thanks Jimmy. I'm using two 9v's so I'll probably use the capacitor.

EDIT: alright, better yet. I went to look for my friend Nick who started working at Wal-Mart in the garden center. Didn't find nick, but I found an electronic BBQ igniter for 16 bucks. That's what I'm gonna probably wind up using.