also some LED's might be a nice touch to show at a glance what mode your in.
Edited by CCH Concepts, 10 December 2009 - 10:48 AM.
Posted 10 December 2009 - 10:47 AM
Edited by CCH Concepts, 10 December 2009 - 10:48 AM.
Posted 10 December 2009 - 12:29 PM
that should work, i would also suggest adding a capacitor to the switch on each pole to ground. not essential, but it stops the contacts sparking when the switch contacts. which will increase the life of your switch, taking into account how much current it will be shifting when it contacts.
also some LED's might be a nice touch to show at a glance what mode your in.
Edited by PyroCreationZ, 10 December 2009 - 12:30 PM.
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Posted 10 December 2009 - 12:45 PM
Ok but where do I connect them to? And what type should I use: how much µF or F?
As for LED's then I'll need a 3 pole switch (LED seperate circuit) but would be handy idd so I'll put them in.
Posted 10 December 2009 - 01:37 PM
use some ceramic caps, not the polorised ones. something like 100nf should be fine and make sure there say about 50 volt ones, they should more than coup with anythign your doing then. they only cost pence.
as the the led, make sure you put say a 1k or 10k restistor on the postive leg. ground it to the spare switch contact you have, which will ground the LED when you go into firing mode.
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Posted 10 December 2009 - 02:08 PM
Ok I'll get some capacitors and install those.
As for the LED's I'm not quite sure what you mean?
Why do I need resistors when I'm hooking them up to a seperate AA battery pack (2x 1.2V)?
Or do you mean I should hook them up to my 24V and use resistors so the LED's won't blow out?
Really appreciate your help mate
Edited by CCH Concepts, 10 December 2009 - 02:20 PM.
Posted 10 December 2009 - 02:36 PM
Edited by PyroCreationZ, 10 December 2009 - 03:56 PM.
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Posted 10 December 2009 - 03:15 PM
Isn't it possible to just use one resistor?
So I have 24V but a normal LED wich uses about 2V & consumes 20mA.
That would be: 24V-2V=22V/0.02A=1100 Ohm (1.1k Ohm)
And the LED resistance calculator tells me as a safe pick 1.2k Ohm.
So by adding a 1.2k Ohm resistor with every (single) LED this should work or not?
I have a whole bunch of 5mm, 2V, 20mA red LED's so wouldn't want to buy 24V ones.
Besides I think those would be expensive.
Posted 10 December 2009 - 03:18 PM
Posted 10 December 2009 - 03:47 PM
i think you will find 2v is the voltage drop across the LED, what that means is the led will use 2 volts to power it. if you put it on a 12v supply it will be 10v after it. but if the led is rated for 6 volt it will blow. what the potential divider does is allow the correct current and voltage through. the ration of the resistors sets the voltage and the value sets the current.
Edited by PyroCreationZ, 10 December 2009 - 06:04 PM.
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Posted 10 December 2009 - 06:18 PM
Ok so is this correct? Considering the red diffused LED needs 2V.
And the cathode would be connected to the ground and the anode between the two resistors (V2), right?
PS: I'm also considering to put a fuse in my system...if that is necessary? It's wired so no electronic stuff except a few LED's & resistors.
Since the batteries are 7Ah, I think I'll use a 7A (slow acting) glass fuse or should I take a bit higher/lower or is a fuse not necessary at all (I think it is... just to protect from shorts, etc.)?
Posted 10 December 2009 - 07:16 PM
Posted 10 December 2009 - 07:59 PM
Edited by PyroCreationZ, 10 December 2009 - 08:00 PM.
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Posted 10 December 2009 - 09:00 PM
CCH
Invoking Quantum theory for something as simple as getting an LED to light off 24v is total POOH! And probably shows up your lack of knowledge. An LED is a current driven device and common LEDs need less than 20milliamps to light without failing, so a series resistor is ALL that is needed. In the dark a fully driven LED is too bright to look at so driving about 5mA will be plenty. (24/.005) 4800 ohms would be a good figure which by preferred value is 4K7 or 5k2.
I like the least necessary current through a LED largely because a board of LEDs needs a lot of power and hence a bigger battery.
Quantum theory IS applicable to the emission spectrum of an LED as the output light is due to energised electrons falling down to a lower energy level with the emission of a photon. Here you can invoke also the Planck equation to determine the frequency from the energy transition.
Where did you say that you graduated from?
Edited by CCH Concepts, 10 December 2009 - 09:14 PM.
Posted 10 December 2009 - 09:35 PM
Posted 10 December 2009 - 11:26 PM
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