Amps...?
Dun-dee
Posts: 6
Hypothetically, I want to power an LED rated at 1.7 volts, 10 milliamps. Say for example I had a 1.7 volt dc power supply rated at 10 amps. The LED is wired in directly to the power supply with no additional circuit.
Question1: Will the 10 amp supply fry the LED, or, will the LED only draw its required 10 milliamps and be fine?
Question2: Can there be such a power supply? (1.7 volts 10 amps)
Question3: What is the limiting factor in both power supplies and batteries that only disallows them from pushing more current than their rating?
Question4: What type of circuit would require high volts low current? What type of circuit would require low volts high current?
As you can see I am a little confused at how current is drawn and the practical relationship between volts and current. I am familiar with ohms law and can follow the math enough to build working circuits but the concepts are not yet completely clear.
-thanks
Post Edited (Dun-dee) : 4/19/2008 7:37:02 PM GMT
Question1: Will the 10 amp supply fry the LED, or, will the LED only draw its required 10 milliamps and be fine?
Question2: Can there be such a power supply? (1.7 volts 10 amps)
Question3: What is the limiting factor in both power supplies and batteries that only disallows them from pushing more current than their rating?
Question4: What type of circuit would require high volts low current? What type of circuit would require low volts high current?
As you can see I am a little confused at how current is drawn and the practical relationship between volts and current. I am familiar with ohms law and can follow the math enough to build working circuits but the concepts are not yet completely clear.
-thanks
Post Edited (Dun-dee) : 4/19/2008 7:37:02 PM GMT
Comments
Question 1: try it and see. LEDs are cheap and you will learn something.
Question 2: Since you have such a supply then there can be such a supply.
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- Stephen
A real LED is not ideal. It will leak a little bit of current below the threshold voltage. It also has some resistance which will limit the maximum current it will conduct (based on Ohm's Law). Practically, an LED contains small, fragile wires which will overheat and melt or explode long before the diode itself or the leads or package is damaged. The datasheet for the specific diode will give the Forward Voltage and the typical forward current for a given light output as well as the absolute maximum current that the diode can conduct before it's damaged.
2) You can buy pretty much any power supply you are willing to pay for. A 1.7V 10A supply is not difficult to build. You may be more likely to find a 1.8V supply since that's a voltage that some newer ICs are built to work at.
3) Many power supplies are designed to shut themselves off when they're asked to supply more current than they're designed for. This may be done with a fuse or a current limiter that senses the amount of current drawn and turns off the supply when that rises above the set limit. There may also be a temperature sensor to shut off the power supply when its temperature gets above a set limit.
Batteries usually do not have a current or temperature sensor, so they may be fused to limit the maximum current supplied. This is particularly important for rechargable Lithium Polymer or Lithium Ion batteries that, depending on their chemistry and construction, can violently catch fire if they get too hot or are short circuited. These often have sophisticated current and temperature limiting circuits that are part of their built-in chargers. Many other batteries have an inherent series resistance that (using Ohm's Law) limits the maximum current that can be drawn from them.
4) High voltage, low current: Vacuum tube circuitry, Neon / Argon / Krypton gas displays
Low voltage, high current: Arc welders
I guess an LED was a bad choice for the question. The purpose of my question was to find out if a "load" draws the only current it "needs" or if the current is "pushed" through thereby toasting the load if the available amps are greatly higher than the required.
thanks again for the reply
-Thanks