Putting 4 leds in casing?
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@hayabusafmw
If you can find out the resistance or ampage the LEDs are, I'll get all the values done for you
@Zereo_XI
1. If you mean having 2 LEDs in a circuit, and are asking if you can have several, you can, it'll just increase how much power you use - you won't need to worry unless you're using a battery or anything, then it's not much to worry unless it's 100 LEDS or something like that
2. Regulates 9v down to 5v the NES board uses - if you hooked the board directly to 9v, it would kill the board, or at least some components. It's like mains - 230V directly into almost any circuit of something which runs on mains will kill it with fire. Or at least make some smoke, or make something melt. It's explained in Ben's book
If you can find out the resistance or ampage the LEDs are, I'll get all the values done for you
@Zereo_XI
1. If you mean having 2 LEDs in a circuit, and are asking if you can have several, you can, it'll just increase how much power you use - you won't need to worry unless you're using a battery or anything, then it's not much to worry unless it's 100 LEDS or something like that
2. Regulates 9v down to 5v the NES board uses - if you hooked the board directly to 9v, it would kill the board, or at least some components. It's like mains - 230V directly into almost any circuit of something which runs on mains will kill it with fire. Or at least make some smoke, or make something melt. It's explained in Ben's book
Thanks so hooking up wires to it then putting resistor to make the 5V lower will work ok thanks I appreciate. Oh Can I just use any type of non-Copper wire? Ex:just strip a wire from a old Headset and use that.
Last edited by Zereo_XI on Sat May 19, 2007 8:35 pm, edited 1 time in total.
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@hayabusafmw
No, by resistance, I meant the resistance of the LEDs. or mA. Every component has a resistance - this defines how much voltage it gets in the circuit. Hard to explain.
@Zereo_XI
Ummm... You use a resistor, not a transistor to lower the voltage - A transistor will do nowt to the voltage, and may even make a tiny explosion.
And yes, headphone wire will work - otherwise how would you get sound through it?
No, by resistance, I meant the resistance of the LEDs. or mA. Every component has a resistance - this defines how much voltage it gets in the circuit. Hard to explain.
@Zereo_XI
Ummm... You use a resistor, not a transistor to lower the voltage - A transistor will do nowt to the voltage, and may even make a tiny explosion.
And yes, headphone wire will work - otherwise how would you get sound through it?
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AHHHH Dumb me!!
very sorry
my leds are written on a little paper
1.9V @ 30 ma
what i want to do is put 4 inside the casing. so they will be all Wired to each other
By the way Wich leg is the Contact leg? , shorter one or longer one? i always forget...
& where do you put the risistors? to the long leg or shorter one?
+ where would be the best way to Wire them too?...
Theres like 5 pins that comes out of the PSU , & like 4 other big chunky connections for the PSU to stay on the board. i need to figure which do what...
thanks!:)
very sorry
my leds are written on a little paper
1.9V @ 30 ma
what i want to do is put 4 inside the casing. so they will be all Wired to each other
By the way Wich leg is the Contact leg? , shorter one or longer one? i always forget...
& where do you put the risistors? to the long leg or shorter one?
+ where would be the best way to Wire them too?...
Theres like 5 pins that comes out of the PSU , & like 4 other big chunky connections for the PSU to stay on the board. i need to figure which do what...
thanks!:)
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- Joined:Sat Mar 31, 2007 11:47 am
You'll need 2 sets of 2 paralell wirings connected to the spots I said for it to work.
Now to calculate the resistances -
v=5v
i=0.03a
rt=v/i
rt=166 2/3
v=1.9v
i=0.03a
rl=v/i
rl=63 1/3
r1=rt-2rl
r1=166 2/3-(2x63 1/3)
r1=40
40 ohms is what you'll need. Which is yellow-black-black if you're using a single resistor.
Thus, to wire, you'll need 2 sets of 2 LEDS and a 40 ohm resistor each. As to which is the anode and cathode(positive and negative), I can't remember, sorry
Now to calculate the resistances -
v=5v
i=0.03a
rt=v/i
rt=166 2/3
v=1.9v
i=0.03a
rl=v/i
rl=63 1/3
r1=rt-2rl
r1=166 2/3-(2x63 1/3)
r1=40
40 ohms is what you'll need. Which is yellow-black-black if you're using a single resistor.
Thus, to wire, you'll need 2 sets of 2 LEDS and a 40 ohm resistor each. As to which is the anode and cathode(positive and negative), I can't remember, sorry
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Okay i connected my leds together
I decided to try with 1 resistor only
I made them all Good in the case tested with a 9volt for 7 mins
works like a charm
now i Connected them The + to the RIGHT pin (of the 5 inside the nes)
& the - to the Center one of the 5 = the 3rd (of the 5 inside the nes)
& what i get is When i Just plug the AC adapter ... the Leds SHINE GREAT
then when i push power The leds go Dim dim....
almost closed.....
I decided to try with 1 resistor only
I made them all Good in the case tested with a 9volt for 7 mins
works like a charm
now i Connected them The + to the RIGHT pin (of the 5 inside the nes)
& the - to the Center one of the 5 = the 3rd (of the 5 inside the nes)
& what i get is When i Just plug the AC adapter ... the Leds SHINE GREAT
then when i push power The leds go Dim dim....
almost closed.....
Zereo_XI wrote:Befor starting i want to know which one is going to work:
or
So is the energy divided or its stay the same for every led I add?
None of those, first check the wiring of the 7805A, Usually, you connect the + and the middle lead. Also, you have to use one resistor for each set of led you use. Because putting 2 set in parallel while use more mA. To calculate the resistor, you can see that site:
http://led.linear1.org/led.wiz
Also, to understand how electricity behave, you should check Kirchoff's Law at :
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws
Here's how you'd wire 2 led to a NES motherboard:
If you want to put more led, for example 4 led, you put two series of resistor+led+led.
Jeep
.
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- Posts:66
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