Sad thing is..thats how I use to turn on one of my old servers. I didnt have any sort of switch laying around so I ran two wires out and rubbed em together to turn it on/off.Originally Posted by Frakk
I think you only have to short the two pins for a second to turn the pc on (atx mobos). same with the reset. so you could just have two bare wires hanging and you could turn on the pc just by connecting them. therefore you could use any switch, you could even make one. tell me if im wrong
How did you come to trade the fiddle for the drum?
There you go. make your own switch and go nuts
I used to do that too lol. Hey, whatever works
I should get myself a set of those mini switches you see with the senfu techstation. During rig testing I just short out the pins with a coin or screwdriver
All you got to do is understand what a switch does. It connects metal to metal. All it does is compleats a wire. Once you know that you can do whatever it takes to let the current pass. If it does then you have a switch.
How did you come to trade the fiddle for the drum?
to make the plugs that attach to the mobo pins for leds and switches, I have heard of (but not yet tried) soldering wires into a modified jumper. not totally sure how that would work but I know that jumpers would be made with the same plastic assembly.
You want 100 mil or .100" (yes that is 1/10 of an inch)
Connector blocks, also just a FYI those "pins" are called more coomonly called headers also quite a few ppl will call them jumpers, a lot of people will call a "jumper" those little blocks you put on header to conect two of them together, those are shorting blocks or "shunts" for short.
Anyway you can get the 100mil connector houseings at http://www.jameco.com
Jameco Part numbers for commonly needed mobo front pannel connectors:
*******************
If you don't have one of these you shold buy one
Crimper for the pins below
Jameco #: 99442
*******************
Female Crimp Pin: Jameco#: 100765
2-pin houseing: Jameco#: 100811
used for Power Switch, Reset Switch, HDD LED, and Power LED for µATX systems
3-pin houseing: Jameco#: 157382
used for Power LED for standard ATX systems
There was a topic at POC.net a bit about about LEDs was a webpage that sold them cheep, I'll be danged if I didn't froget the name of it. I "normally" use http://www.superbrightleds.com but there was a cheeper webpage out there...
OK. Here we go...
First, you need a soldering tool, wire, solder, shrink tube and/or electrical tape.
For your power and reset buttons, use what's called a momentary switch. These unlit come with 2 pins. The job of the momentary switch is to send signal (electricity) through only when pressed and held. Once you release the button, the circuit is broken.
As for the end that plugs into the mobo, the end from any atx case power button will work.
If you get into more advanced switching, i.e., I'd suggest googling for electrical switch tutorials, as not all switches are equal.
LEDs are more tricky as most times they require a resistor when hooked up. This requires a little math and the aid of this tool.
Anything else needs further explanation.
Last edited by fishies; 04-02-2005 at 01:11 AM.
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hooking up led's to the mobo headers dont require resistors. most LEDs need 2V, some ultra bright ones need 5V. there is 2V coming from the header, so you dont need anything. if you hook the 5V up to it, it will be a little dimmer, but will last a lot longer. you only need resistors when you hook them up to the power supply or other higher voltages.
A Resistor in a LED circuit isn't for voltage, though it does dissipate the extra voltage from the circuit. The resistor is to limit the amount of current drawn though the LED. The majority of motherboards have a 330Ω resistor and +3.3Vdc or 5Vdc connected to the HDD and Power LED headers. The vast majority of LEDs are around 20mA max current and drawing more current will damage the LED. Say you have 25mA though a LED; well it might not blow it right away, as 100mA will but the lifetime of the LED will lower when compared to running it at 20mA or less.
The formula to calculate a resistor for a LED is
RLED = (Vsource – VLED) / ILED
The current stated for a LED is the max the LED is allowed to run though it. The voltage is the min power source needed to turn the LED on.
Remember a LED is still a semiconductor diode it will only take as much voltage as it needs and it will always get it's voltage before anything else. So no matter what resistor you use the LED will have the same voltage. Using different resistors will allow different currents and thus different "strengths" of brightness from the LED. Only way I can think of to blow a LED with voltage is to break the LEDs reverse bias voltage barrier.
If it is a 2.4V 20mA LED then when it is in a circuit 2.4V will go away never to be seen again as it is used up when pushing though the NP junction of the LED.
Example: a 2.4V LED inside a 5V circuit, you have 2.6V floating around with nothing but the trace and wire resistance so we might as well say no resistor in series with the LED. So that 2.6V will be pulling infinite current which will go past the 20mA and thus blowing it.
So you need to calculate how big of a resistor is needed to control the current
So explaining that formula stated before
The Vsorce - Vled is how much voltage you have left and Ohms law states that V = I * R so R = V / I
So in our example
R = (5V - 2.4V) / .02A
Which equals 130ohm or greater resistor value needed for the LED.
Lets say you put a 10ohm resistor with our example LED will prob die.
Reason is our remainder voltage 2.6V / 10ohms (our random resistor) = 260mA or .26A which is well over 10x our maximum allowed current.
Also note that measuring the “voltage” across the headers will give you what we call a “false meter reading.” This is due to the internal resistance of the meter. Most DMM are 10MΩ internal resistance, because of this touching your probes to the header pins will put the meter in “series” with the motherboard’s resistor and we all know you can’t read voltage in series.
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