I'm not that familiar with electronics, so I'm asking here. How would I implement a capacitor so that it charges when I start the computer, discharges when the power is off (so that the LED dims quickly, instead of just turning off), and allows the LED to be lit when the power is still on?
You want to delay the turn off, but just a little, like a hot filament cooling down? What about turn on delay, should it just start to glim slowly? A capacitor alone may be difficult because at start up it will draw high current so you need extra circuitry against the surge. A big inductor and a free wheeling diode may work better. Connect inductor and LED in series. Across the input of this assembly you connect a low dropout diode, so it block the incoming signal (ring on doide housing points away from MB ground). At power up the inductor causes a slowly ramping up current. In operation the inductor does nothing, it's just a piece of wire. At power down the inductor keeps pushing current for a while, which now flows through the free wheeling diode so the MB is not affected. WARNING: I never tried this on a MB. Theory says it works. Reality may see it differently. You also need a biga## inductor, like this http://www.hammondmfg.com/153.htm .
The Inductor idea may work, but that is a really big inductor! I have tried to think of a different design, using capacitors and a few other components. I'll try to go through each component. First off the LED, you will want to use a High Bright LED, as you can get good light from it even when it is using only a few mA. If you were to supply 20mA to the LED the capacitor C1 would have to be really big (close to 10000uF). With only 2mA on the LED the capacitor C1 can be much smaller 1-3000uF. The resistor R1 determines how many mA is going through the LED. R1 however has to be adjusted to fit the type of LED you are using. In the schematic it is set to 330 ohm, this is probably a little high. Let me try with a calculation. For a High Bright Blue LED one would typically need 3.4 V to get full illumination at app. 20mA. If we were to limit the voltage to 3V we would get maybe 2mA on the LED, which would still give ample light for an ON indication. In the suggested design we loose 0.7V over the transistor and 0.7V over the Diode D1, so for the resistor R1 and LED1 we are left with 3.6V (5V - 1.4V = 3.6V) Of these 3.6V, then 0.6V is to be on R1. At 2mA then R1 should be 300 ohm (0.6V / 0.002A = 300 ohm) Should it turn out that we only get 4.9V form the 5V supply (not unusual), and that we need 3.1V on the LED then R1 should be 200 ohm (0.4V / 0.002A = 200 ohm) As you can see then maybe a 220 ohm resistor for R1 is better. But you will have to experiment. For R1 I often use a 500 ohm mini trim potentiometer for one-time designs, and then measure/adjust the mA in the final design. The adjustable resistor (470 or 500 ohm mini trim pot.) also allows me to use any color and make of LEDs. The C1 capacitor will set the turn off delay. On old motherboards you will be able to find 1000uF and sometimes 1500uF capacitors that are physically small because they are rated to max. 6.3V. D1 may not be needed, but it ensures that the voltage stored in C1 does not get used for anything other than LED1. (Actually T1 should ensure the same function, so D1 can be left out, but R1 will then need to be adjusted as well) R2 limits the power surge through the transistor T1 when the circuit is turned on. T1 and D1 can handle a max of 100mA. With R2 being larger than 3K this limit should be ensured. The combination of C2 and R2 should also help the LED to be turned on "slowly". A 1uF for C2 should be enough to ensure a slow turn on, but again you may need to experiment. 10uF capacitors are also usually available from old motherboards. For T1 I have chosen the in-expensive BC547, but any NPN transistor will do. This circuitry is very in-expensive (should be less than $3), but for more fancy and precise timing one might consider a small 8-pin microcontroller. This could probably be done for less than $10. Let me know if you would rather see such a solution. Alvin
