>So, lets make an example, a simple circuit if you will. Imagine an AC power >supply, say supplying a 12v 2amp signal. On both power leads a diode (cathode >and anode pointing in opposite directions on either pole) and a LED in >parallel to the supply (hope that makes sense), anyway, if the LED is rated >at a minute 50ma, it's using that full 50ma supplied by the supply even >though the supply is capable of flowing 2A. So what will the diodes see, >exactly, 50ma across both. Ok, lets say the diode has a 1A threshold and >instead of a LED you use a 2A lamp, then the diodes will see a 2A load which >after time will burn them out. So anyway, I've still made my point. Amperage >pull will only be what is used, not what the battery or supply can produce.
No, that's just not true. Let's simplify further, and be more relevant by talking about direct current, as that is what all computers use (except for certain parts like LCD backlights). V = IR. LEDs have a minimum "breakdown" current or voltage, if you will, beneath which they will not function. But if you supply a 5V emf to a simple circuit with a 20ma LED, your current is limited by the resistance of the LED (and the source of the EMF), not how much current it "draws." Nothing "draws" current - all current is driven by voltage differences. Since you have a 5V voltage difference across the LED, the current will be I = V/R where V = 5 and R is the resistance of the component (LED). Nah, I don't buy it. If you supply an over-rated voltage to digital electronics, you have a very strong chance of burning them out. Try sticking a light bulb on a very high voltage - the bulb burns out. The current through the filament was strong enough to literally burn the filament. The same is true with an LED in series with a variable resistance resistor - turn down the resistance and the LED gets brighter. LEDs are fairly robust, though, so it takes a heck of a lot of current to burn one out. But you are wrong about them only "drawing" up to their nominal rating. Now in the case of direct current supplied by a battery, the current is limited by the sum of the battery's internal resistance and the resistance of the rest of the components on the circuit. This is why fuses are used in nearly all sensitive circuits - if there is a short, by your reasoning, the short "draws" more current, but the current is limited to whatever the lowest current "draw" upstream of the short is. Rather, the short creates a scenario where the resistance is significantly lowered. Any components upstream can be burned out since the current will be higher. (This is neglecting the problems associated with dumping too much current down too thin a pipe - i.e. FIRE). Try it yourself, I dare you. Peace, Drew -- Author of ClassicStumbler email: <alk@xxxxxxx> web: <http://homepage.mac.com/alk/> Want to know if your neighbor has Wi-Fi? Find out with ClassicStumbler! <http://homepage.mac.com/alk/classicstumbler/> -- PowerBooks is sponsored by <http://lowendmac.com/> and... Small Dog Electronics http://www.smalldog.com | Enter To Win A | -- Canon PowerShot Digital Cameras start at $299 | Free iBook! | Support Low End Mac <http://lowendmac.com/lists/support.html> PowerBooks list info: <http://lowendmac.com/lists/powerbooks.shtml> --> AOL users, remove "mailto:"; Send list messages to: <mailto:powerbooks@xxxxxxxxxxxxxxxxxx> To unsubscribe, email: <mailto:powerbooks-off@xxxxxxxxxxxxxxxxxx> For digest mode, email: <mailto:powerbooks-digest@xxxxxxxxxxxxxxxxxx> Subscription questions: <mailto:listmom@xxxxxxxxxxxx> Archive: <http://www.mail-archive.com/powerbooks%40mail.maclaunch.com/> Using a Mac? Free email & more at Applelinks! http://www.applelinks.com
