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Thread ID: 28946	2003-01-05 22:19:00	anyone who knows much about electronics...	tango (2697)	Press F1

Post ID	Timestamp	Content	User
111331	2003-01-07 08:45:00	I thought the wires which you plug onto the motherboard from the HDD LED in the front of the case had the resistance built into it.	-=JM=- (16)
111332	2003-01-07 08:51:00	> I thought the wires which you plug onto the > motherboard from the HDD LED in the front of the case > had the resistance built into it. You may be thinking of a resistor soldered into the wire and heat-shrinked up, but none of my recent ones have had this, just a connector on the mobo. This was common practice for 230v neon indicators which had a very high value soldered into the lead, for safety as these will explode if direct connected to the mains. (as will LED's) :D The HDD LED on the actual HDD itself will have on-board resistor, and the case one will have a motherboard resistor in most situations based on my observations.	godfather (25)
111333	2003-01-07 08:57:00	I'm not sure with anything to do with the actual HDD, I'm just thinking of using the connector that is on the mobo. I tested with my multimeter, and it was just below either 4 or 5 volts, I can't remember which. probably 4v. Hopefully it's about 3.6 :). The current coming out of the motherboard connector should be suitable for the superbright LED, right?	tango (2697)
111334	2003-01-07 09:27:00	Here we go again... Was the case LED connected when you measured the voltage? Is the case LED fed from 12v or 5v? If the case LED was disconnected, but trying to operate, then the voltage you measured would be the full supply voltage irrespective of the resistor. If the multimeter was the only load, the resistor would not make any real difference. A digital multimeter has an internal resistance of 10,000,000 ohms, which draws almost no current, which doesnt load the circuit. A cheap analogue meter will have about 20,000 ohms, and much the same result. If the LED was the only load, and was operating, then the voltage you measure could be the Vf of the case LED. Maths again. If the case LED has a Vf of 2.2 volts and is fed from 12 volts, then your resistor in the circuit will be (12-2.2)/.02 = 490 ohms. Say 470 or anything to 1k (you dont HAVE to run the full 20 mA) If the case LED is fed from 5v, then (5-2.2)/.02 = 140. Say 150 or so as a resistance value. Now substitute your 3.6 Vf 12v feed (12-3.6)/470 = .017 amps (17 mA) which would be fine 5v feed (5-3.6)/150 = 9mA, which would be a tad dull. The last 2 equations assume a 470 or a 150 ohm resistor. The actual value could be higher, so your currents would be lower and output lower. Ever heard of "suck it and see?" Just remember you cannot run both LEDs in parallel. Series yes, but you will run out of volts doing that. Just try the superbright in place of the case LED. If its not satisfactory, measure the current flowing in the circuit.	godfather (25)
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