Alternatives for club meetings

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4 months 3 weeks ago #9922 by DeltaVee
Replied by DeltaVee on topic Alternatives for club meetings
So would you pay 20 cents for something valued at jj 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!

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4 months 3 weeks ago #9924 by guyw
Replied by guyw on topic Alternatives for club meetings
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.

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4 months 3 weeks ago - 4 months 3 weeks ago #9925 by DeltaVee
Replied by DeltaVee on topic Alternatives for club meetings

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?
Last edit: 4 months 3 weeks ago by DeltaVee.

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4 months 3 weeks ago #9926 by guyw
Replied by guyw on topic Alternatives for club meetings
It is true for a "standard" LED. However, you can obtain LED's that already have the resistor onboard. I happen to have a bunch of them. They are what I have used for the launch controllers I have built as well as the remote relay boxes. The resistor value to use for a "standard" LED depends on how much current it is rated for and what the voltage drop across the LED itself is. The data sheet for the given LED should give the particulars. If I had to guess, my favorite resistor value of 4.7K in series should be a reasonable place to start. If the LED is too dim with that value, just decrease the value until you get the brightness you want or the LED burns out ;). Also, LED;s have a polarity, so you need to make sure the voltage applied is always the correct polarity as well.

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4 months 3 weeks ago #9929 by Rild99
Replied by Rild99 on topic Alternatives for club meetings
Pacifico, nada mas

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4 months 2 weeks ago #9931 by dougs
Replied by dougs on topic Alternatives for club meetings

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

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