Alternatives for club meetings

So would you pay 20 cents for something valued at j^j dollars? (A. P. French, vibrations and waves)

I neglected to mention that while it's true that a 12V incandescent bulb has a cold resistance of 1.5 Ohms, a hot one has a much higher value... which is why the bulb would work in the first place! Of course you used P=IV. I used P=I²R... yes, same thing.

I still advocate testing when you're done building one since you can mess up math no matter how simple!

guyw wrote:

As John correctly points out, my numbers and formulas are OK, but my description of the bulb lighting process is almost entirely backwards. I blame early morning no caffeine syndrome. A cold light bulb has very little resistance. As it warms and lights, the resistance goes up, reducing the current flow. If not, the process would go into thermal runaway, and the filament would burn up. This means however, that initially a large current will flow, decreasing as the bulb warms up. It is this "inrush" current that risks the lighting of an e-match. An LED does not act this way, and 10ma or so of current is what it draws when it is on, making it initiator safe.

BTW... I seem to recall reading somewhere that if you replace a bulb with an LED, your also need to add a resistor of the appropriate "ohmage"... Is that correct?

Pacifico, nada mas

DeltaVee wrote:

(snip)

BTW... I seem to recall reading somewhere that if you replace a bulb with an LED, your also need to add a resistor of the appropriate "ohmage"... Is that correct?

Yes, and if you're planning on using a simple circuit where current is limited with a resistor you need to perform the calculation based on the battery voltage. The resistor is wired in series with the LED and is sized to limit the current to the LED's rated forward current. Using this LED as an example, see the Vf (2nd line) in the Electrical Characteristics table. You get both the Forward Voltage (2v typical for Red) and Forward Current (20ma).

Assuming a 12 volt battery, and Vf=2, we need to drop 12-2=10v across the resistor. To limit the current to 20ma, we need R = E / I = 10v / 0.02a = 500 ohms. The power dissipated by the resistor is P=I²R = (0.02)²* 500 = 0.2 watts. Since there's no sense in spending money for exactly this value in this application, you go with a common 510 ohms at 1/4 watt.

Note that it's perfectly save to run this "12v launcher" on 6v--the LED will just be dimmer, and in fact if you happen to run it on 24v you won't let the magic smoke out of the LED since it's maximum forward current is 50ma. However the resistor is going to get hot and could smoke as it's dissipating close to 1 watt. This assumes you're ok with 44ma of continuity current.

Given that the LED is $0.21 and the resistor is $0.17 (ok, multiply by 10 or 20 if you buy at You Do It) I can't imagine playing with incandescent bulbs...

Doug