geekhack
geekhack Projects => Making Stuff Together! => Topic started by: Thumper_ on Tue, 24 March 2015, 01:51:26
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Hello GH,
i'm about to mod my Poker II with some more LEDs in a semi-transparent case. Right now i have the Poker II with LEDs in every switch, but i want to have LEDs under the PCB.
I know, that i can desolder those LEDs and put them upside down in to get the LED under the PCB, but this will remove the whole modifier+number-row LED on the upper side, so i only get backlight in the alpha area.
But can i solder 2 LEDs on one Spot? As far as i know electricity, it'll half the Voltage each LED get, or am i missing something?
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You can add LED's, but the CURRENT will be halved, since the voltage drop across the LEDs will remain the same, but there is a resistor limiting the current and you're now splitting that between 2 LEDs. You would need to change the resistors for those that you add a 2nd LED to, or simply wire up the new LED's on their own circuit.
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Can you give me an explanation how i should make a new circuit? Especially, where shall i solder it to the PCB/where shall i take the voltage from?
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Change the led resistors from stock to something like 68ohm or 100ohm and you will be able to have an led on top and bottom.
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Other than lower A, there is not much issues with it except dimmer LEDs. I've did that for my poker2 before removing it due to personal preference. PM me if you have any questions regarding my mod.