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After asking why the fuse in my ATV was blowing, I was told to insert a resistor between the switch and ground. The problem now, is determining what size resistor to use. Since fuses are ~$2.00 a piece, I'd like to avoid the trial-and-error method.

  • The power source is a 12V 12AH motorcycle style battery.
  • The voltage will likely fluctuate during operation due to the alternator (12 - 16V?).
  • The circuit is basically a direct short to ground through an LED:

  • I have no knowledge about the LED, other than it's part of a switch rated for 20A @ 12VDC.
  • The fuses are 15A.

How can I determine what size resistor I need?

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  • \$\begingroup\$ We can't possibly tell without the voltage and current ratings of the LED. I'd recommend using a 2K pot or so; start with a high resistance and decrease it slowly until the LED lights up normally. \$\endgroup\$ – user17592 May 20 '13 at 12:39
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    \$\begingroup\$ Personally I doubt you've blown a 15A auto type fuse (which are usually slow blow) with a LED, especially when you mentioned in the other question the LED is still OK. More likely the connections aren't what you think and you're shorting 12V to ground via the switch contacts. \$\endgroup\$ – PeterJ May 20 '13 at 12:50
  • \$\begingroup\$ @PeterJ All I know is when I attach the Ground pin on the switch to ground, I get two blown fuses. I don't know much about the switch, so the internal wiring is a complete guess. All I know is, pin1 is labeled Power, Pin2 is labeled ACC and pin3 is labeled Ground. The switch works fine, until I connect Ground to ground. \$\endgroup\$ – Tester101 May 20 '13 at 12:54
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    \$\begingroup\$ I would suggest that this question (and it's predecessor) are a bit hobbled by the lack of a basic multimeter, or at least no indication that one is available for some checking. A basic one can be had for as little as $5, if not borrowed from a friend. Please consider getting hold of one, so members here can walk you through some quick tests to address all the grey areas in the question. \$\endgroup\$ – Anindo Ghosh May 20 '13 at 14:44
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    \$\begingroup\$ Now, please check the poles of the switch to verify that your conception of the poles / pin-out is correct, and check which way the LED is wired, if it is still alive at all. \$\endgroup\$ – Anindo Ghosh May 20 '13 at 15:20
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I have no knowledge about the LED, other than it's part of a switch rated for 20A @ 12VDC.

There's your problem.

Since you don't know anything about the LED, you'll need a variety of resistors (or an adjustable multi-turn potentiometer) and will have to experimentally establish what looks good to you.

Assume the LED will have a forward drop of around 1V (could be less, could be more). Start with 1mA of forward current and go from there.

\$ I_{LED} = \dfrac{12V - 1V}{1mA} = 11 k \Omega \$

Then, install the resistor in series with the LED, apply power, and measure the battery voltage and the resistor voltage to actually figure out the LED voltage drop and power:

\$ V_{LED} = V_{BAT} - V_R \$

\$ I_{LED} = \dfrac{V_R}{I_R} \$

\$ P_{LED} = V_{LED} \cdot I_R \$

\$ P_R = V_R \cdot I_R \$

Repeat this iteration of increasing the LED current / reducing the resistor value until you find the minimum LED current that's useable for you (i.e. bright enough to see) and go with it. Also make sure your resistor is sized appropriately. (Using a fixed resistor once you settle on a value will be more reliable then leaving a potentiometer in the circuit.)

If you arbitrarily go with too much LED current, the LED will end up with a short lifespan and go dark prematurely.

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  • \$\begingroup\$ Wouldn't the LED blow before a 15A fuse? \$\endgroup\$ – Scott Seidman May 20 '13 at 13:06
  • \$\begingroup\$ For sure it would. That being said, an over-driven LED could die unnecessarily in days or weeks instead of lasting indefinitely. This switch is somewhat unususal in that OP was popping fuses somehow 'through' the LED. I have to wonder if it's actually some sort of lamp and not an LED per se. \$\endgroup\$ – Adam Lawrence May 20 '13 at 13:45
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I don't see an indication that the referenced part contains a LED at all, the red part just looks like a safety cover and it's not described as illuminated on the Radio Shack page. Despite your guess at the circuit I'd say it's a simple SPDT (single-pole, double-throw) switch.

In that case the center pin will be common and when the switch is pushed to the right the common and left-hand pin will be shorted. When pushed to the left the center and right-hand pin will be shorted. At the moment it's likely that one position is shorting 12V to ground.

However as Kaz pointed out in a comment you did manage to illuminate the LED in your other question and now I see it is described as SPST which makes the three connections plausible with illumination. The LED blowing the fuse while continuing to operate does seem unlikely though, and many of those switches do already have in-built current limit resistors on the LED.

Assuming you don't have much test gear available it may be worth shelling out for a cheap multimeter so you can confirm the switching contacts for sure. Either that or see if Radio Shack can provide a proper datasheet or diagram. It's been years since they operated here but sometimes the back of their blister packs had some information, but I guess you would have noticed if that was still the case.

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    \$\begingroup\$ There is in fact a LED, which is embedded into the very tip of the switch. OP has managed to light the LED before. \$\endgroup\$ – Kaz May 20 '13 at 13:28
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    \$\begingroup\$ So, the LED is probably across one side of the switch. In one position, there will be a short across the LED. OP perhaps only tested with the switch in the open position. Solution: swap the wires to the switch, put battery on left throw of switch (I suspect the diagram has it wrong up there). \$\endgroup\$ – Bobbi Bennett May 20 '13 at 14:45
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I apologize for posting a speculation as an answer.

The symptoms of this problem imply that the diagram is wrong, the switch is not as shown. Swap the red and black wires at the switch.

schematic

simulate this circuit – Schematic created using CircuitLab

On the right, Acc is switched between Battery and Ground.

On the left, Battery is switched between Acc and Ground. The circuit on the left behaves as you are describing, it is my best guess as to what is going on.

No resistor is needed for this LED, as you have shown by testing it with 12V.

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Ohm's law tells you:

V = R * I

Considering that the LED in your switch is a common one, you probably want to feed it with around 20 mA, anyways that should be enough current to make it shine bright. You got I, you got V, you need R (assuming that the voltage drop across your LED is the nominal voltage drop of a red photodiode as seen here):

R = (V - Vd) / I = (12 V - 1.8 V) / 20 mA = 510 ohm!

By the way, I''ve made the assumption that the LED is a common one, since the RadioShack page you linked to doesn't tell anything about the LED. Also, stick with Keelan's method of current limiting via a potentiometer, that should do the trick to see the actual current rating of your LED.

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  • \$\begingroup\$ You forgot about the voltage over the LED. You need R=(V-V_LED)/I instead of just R=V/I. (downvoted until fixed) \$\endgroup\$ – user17592 May 20 '13 at 12:46
  • \$\begingroup\$ I'm very familiar with Ohm's law, I just wasn't sure how much current LEDs typically require. Thanks for the help. \$\endgroup\$ – Tester101 May 20 '13 at 12:47
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    \$\begingroup\$ @Tester101 this answer is incorrect. In many cases, 300Ohm isn't enough! Please do not rely on this answer. \$\endgroup\$ – user17592 May 20 '13 at 12:48
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    \$\begingroup\$ @MickMad A quick survey of "common" LEDs from Digikey's rather vast inventory shows that 20 mA is the nominal operating current for the vast majority of LEDs out there. While it is possible that Digikey has a conspiracy to prove you wrong, frankly, 40 mA through an LED is not something I would consider a "default", only for specific LEDs that are specified for higher current in their datasheets. This answer is thus misleading, to my view. Downvoting until fixed. \$\endgroup\$ – Anindo Ghosh May 20 '13 at 14:26
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    \$\begingroup\$ You're seriously asking why one should follow the specs and not provide the highest possible current that doesn't blow up your component? Perhaps because an error is easily made, you don't want to blow up your component, and more current often isn't even useful and only consumes more energy. \$\endgroup\$ – user17592 May 20 '13 at 15:09

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