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#16 CCH Concepts

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Posted 10 December 2009 - 10:47 AM

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 CCH Concepts, 10 December 2009 - 10:48 AM.


#17 PyroCreationZ

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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.


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.

Edited by PyroCreationZ, 10 December 2009 - 12:30 PM.

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#18 CCH Concepts

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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.



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.

#19 PyroCreationZ

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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.


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 ;)

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#20 CCH Concepts

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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 Posted Image



you always need resistors on LED's because they will draw more current than they can handle. hook one up without a resistor and it will just burn out. just get your self a 24v LED and to be more precise, find out the normal running current of the LED. then using ohms law calculate the resistor.

so for a 100mA current

would be

R = V/I

24/0.1A (100mA)= 240 ohm i would use a 270 ohm. but that being said its better to go higher say 1k as to be sure not to blow it as you dont need max brightness.

and if you cant find a 24v LED then you can just usage a potential divider to half the voltage. this is easily done using just 2 resistors


heres some online calculators to make things easier

led resistor

potential divider

Edited by CCH Concepts, 10 December 2009 - 02:20 PM.


#21 PyroCreationZ

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Posted 10 December 2009 - 02:36 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 Ohms (1.1k Ohms)
And the LED resistance calculator tells me as a safe pick 1.2k Ohms.
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.

Edited by PyroCreationZ, 10 December 2009 - 03:56 PM.

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#22 CCH Concepts

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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.



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.

#23 Arthur Brown

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Posted 10 December 2009 - 03:18 PM

Leds
A simple series current limit resistor will work fine. Try 100- 200 ohms per battery volt. In the dark outside on a show you want low brightness leds so that you don't kill what night vision that you have. SO even 500ohms per supply volt would work fine
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#24 PyroCreationZ

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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.


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?

Posted Image

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.)?

Edited by PyroCreationZ, 10 December 2009 - 06:04 PM.

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#25 CCH Concepts

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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?

Posted Image

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.)?


not quite.

its not saying you need 2v to the led its saying the led will use up 2 v. basically the led will need a certain voltage to run say 12v, if you were to measure the voltage after the led it would read 10v. without going into quantum mechanics its to do with the way the diode works. I'm more than happy to explained (or find a link that does). but what you need to look at is the supply voltage for the diode not the voltage it uses up, sorry the terminology has popped out of my head for the moment. but if you get a data sheet for a led you will see what i mean.

what you need is the voltage across r2 to be the voltage required at the led. ignore the 2v as that doesn't effect this type of circuit. that would only be necessary if you were trying to run something after the led that required the full supply voltage.

as for choosing a fuse, figure out how much you curiut will draw when running correctly. figure out what its faliure current is and go somewhere between the two.

#26 Arthur Brown

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Posted 10 December 2009 - 07:16 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.

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#27 PyroCreationZ

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Posted 10 December 2009 - 07:59 PM

Thanks Arthur for clearing that up :)
I do appreciate CCH helping me but I also started to think he's going a way too complicated way.
I spoke to cooperman and he also told me that a voltage devider is totally unnecessary and that for such a simple scheme I can just put a resistor in series.
I really should start reading books about electronics lol

Edited by PyroCreationZ, 10 December 2009 - 08:00 PM.

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#28 CCH Concepts

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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?


its got nothing to do with the light emitting function of a LED its because its a diode. there is a depletion lay between the P-N junction that requires a certain amount of electrons to bridge. a standard silicon diode requires 0.7 but an led requires as much as 2v because it uses different substrates and doping methods.

i would really do some reading before posting comments in such a insulting way.

diode

this is a quote from that link

" If an external voltage is placed across the diode with the same polarity as the built-in potential, the depletion zone continues to act as an insulator, preventing any significant electric current flow (unless electron/hole pairs are actively being created in the junction by, for instance, light. see photodiode). This is the reverse bias phenomenon. However, if the polarity of the external voltage opposes the built-in potential, recombination can once again proceed, resulting in substantial electric current through the p-n junction (i.e. substantial numbers of electrons and holes recombine at the junction).. For silicon diodes, the built-in potential is approximately 0.6 V. Thus, if an external current is passed through the diode, about 0.6 V will be developed across the diode such that the P-doped region is positive with respect to the N-doped region and the diode is said to be “turned on” as it has a forward bias. "

and i sited quantum mechanics not to sound clever, but to make the point that there is stuff going on that doesn't need to be understood, just expected to be able to used diodes correctly.

i will check on my understanding, but in future if you feel i am saying something in correct please just say so, state why. if i am wrong i will thank you and have learned something. but to needlessly attack me like that is pointless and not helpful to anyone.

Edited by CCH Concepts, 10 December 2009 - 09:14 PM.


#29 CCH Concepts

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Posted 10 December 2009 - 09:35 PM

just been looking at some data sheets. they have a min typical and max forward voltage. min being the min to bridge the depletion region and allow current to flow, type being the suggested voltage, max being oblivious, if any higher than the max is reached the led will burn out.



#30 portfire

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Posted 10 December 2009 - 11:26 PM

CCH- Arthur was not been insulting, in this context Quantum Therory has no place, it's just a firing system so a good understanding of electronics is all thats needed....One of my interests is the Quantum "world" and not once have I had to use it to light up an LED :blink:
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