I'm using this app called iCircuit for Android (I'm asking here because the iCurcuit forum has been forsaken) and when I have a circuit with a 3V battery and an LED, it lists the current as 2.15GA. I'm new to electronics so I'm wondering why its giving me this because I have connected an LED to a 3V battery before and its been fine.
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Brian Carlton
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MichaelK
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4.. if you don't have a current limiting resistor, then you get _all the current in the universe_ :) – pjc50 Aug 02 '13 at 17:11
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4Wow that must be a *bright* LED! – JYelton Aug 02 '13 at 17:29
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6@JYelton: For a tiny fraction of a second. Then it becomes the world's brightest arc lamp. – Ignacio Vazquez-Abrams Aug 02 '13 at 17:38
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I wonder what is the current if you set the voltage to 30V. – Codism Aug 02 '13 at 19:18
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3@Codism just for fun I tried it, it stops the simulation after 6.1V, at which point the current is 464.16YA. lol. – MichaelK Aug 02 '13 at 20:30
1 Answers
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The simulator is pretending that the battery is an ideal power source, i.e. no internal resistance. As such, it is using the largest possible value it can support, 2147483647A.
True power sources do have an internal resistance, and this is what is preventing the LED you put across it from burning out. LEDs should have a resistor placed in series with them so that you don't have to depend on this internal resistance.
Ignacio Vazquez-Abrams
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1Ah, that makes sense. Thanks. (I would upvote and not comment but not enough rep...) – MichaelK Aug 02 '13 at 16:42
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5The simulator is also evidently failing to model LED bulk resistance. Even if the power source is idealized, there is no excuse for this. – Kaz Aug 02 '13 at 17:10
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more on LED resistance: http://electronics.stackexchange.com/questions/76367/accounting-for-led-resistance/76378#76378 – Phil Frost Aug 03 '13 at 01:19