EXACTLY! I hate surprises...I'm guessing that by putting a 390 mA resistor inline, it drops the amperage allowed to the igniter leads until you by pass with the launch button? Electrical schematics are not my strong point. Also, i notice that resistors come in a wattage rating...what wattage is needed, or how is the wattage determined? Thanks, rsbhunter
I think you mean ohms, not mA. The latter is a unit of current, i.e. how many thousandths of a coulomb per second. The coulomb being a unit of charge, which you might think of as a certain number of electrons. Or a shortage of that many, depending on the polarity. You COULD have a 390 mA curent regulator, but that's a more complicated beast. Not necessarily expensive, though. I think the venerable LM317 only costs a buck, or less, even now. Maybe a tenth of that in bulk.
This is all V=IR stuff. In other words, simple algebra. So it should be relatively easy to learn. When we get into fancier electronics, I'm lost too.
Let's say that, at your local launch, there is a battery which gives 14 volts if it's just been charged, despite the fact that it's nominally rated for 12 volts. LED's are made that are nominally different voltages. Let's say yours is supposed to get 3 volts. You can't just feed it 3 volts, because it's not necessarily stable. So you want a current regulator or a resistor. We'll go with a resistor, since the power involved is very low, even if we're wasting most of it. If we're trying to keep the current under 20 mA, then it won't be more than 20 mA times 14 volts. Watts are amps times volts, or maybe I should say the voltage drop across the device. So we're looking at no more than about a quarter watt. Somewhat less, because some of it will go to the LED. Anyway, if the drop across the LED is 3 volts, that leaves 11 volts for us to manage. Voltage equals current divided by resistance. A bit of algebra tells us that resistance equals voltage divided by current. In this case, 11 volts divided by 0.02 Amps is 550 ohms. I'm neglecting the resistance of the wiring, which is probably going to be much less.
Someone will probably step in if I've screwed up the math.
I hope that was helpful.
As I recall, the publications of Forest Mims were clear and helpful at times. I don't know if they're still available. Maybe at the library. I don't know for a fact that he covers this particular information, but my guess is yes,
P.S. Current is sort of analogous to water flow, voltage is kind of like pressure. The pressure drop through a pipe is not as simply predictable as the voltage drop though a resistor, but it may still be a somewhat useful analogy.
P.P.S. If we get into something that's not just quasi-static, then I"m probably as lost as you. Circuits involving inductance and capacitance, plus varying current, would stump me.
There is one JMRC launch controller that has such a buzzer and when I am doing the launch LCO, talking on the mic, moving the key, and then turning that key back off after takeoff while talking on the mic looking up in the sky and moving the key to off while not looking at it saying "It's still going up like an energizer bunny"...
