Amps (or electrical current) is like water flowing through a pipe - the size of the pipe limits the amount of water that can flow. It doesn't matter if there is a watering can or an ocean at the other end - the same amount of water will flow. It is the device's internal resistance at a given voltage which limites the current, not the power supply itself.
Voltage (like the presure in the pipe) is the really critical issue and too much voltage will push too much power through your device and you will have a problem.
Your device will only draw what current it is designed to draw which is a function of its internal resistance. Some devices are designed to draw more when charging, but you are using a bigger power supply than you need so it will be able to draw enough for any operation that the old power supply would perform.
I have several "USB" chargers that use the standard USB plug and supply 750mA or 1000mA. When you plug in a USB device that only expects 500mA they work perfectly and in some cases they are actually designed to draw more than 500mA and will charge faster.
If you need more proof, everybody will tell you that putting a power supply rated for twice as much power / current is OK as the computer will only draw what it needs.
2007-10-03 08:49:37
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answer #1
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answered by TahoeT 6
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EEK , a lot of too technical answers.
In the real world.....
If your power supply is NON regulated and 1.2 amps (=1200mA) at 12 volt , then when you use a device that only draws 500mA (=0.5A) the load is not enough on the supply and that causes the voltage to float up too high.
The means your device could see 15 or 17 volts , which could easily be high enough to damage it.
That is IF your supply is NON regulated.
Most cheap supplies or plug packs are NON regulated.
If this is the case you have three options.
1 . Buy a regulated supply.
2. If you are electronics smart , build a regulator and use it on a non regulated supply.
3. Buy a new non regulated supply rated at 500mA.
2007-10-04 09:07:55
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answer #2
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answered by I♥U 6
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Okay, here's the deal.
You HAVE to use the 'VOLTAGE' that is required and have the positive and negative wires correct if it is D.C.
If it's A.C. the wiring doesn't matter, except it is nice to have the 'return', (white wire, green wire, return, negative, ground, neutral; are all terms used for the same), connected to the 'neutral', or 'return to ground', but not required. It's a minor safety issue.
NOW: The Amount of CURRENT, read as: AMPS supplied DOES NOT MATTER. Amps are 'TAKEN' by the Device as IT REQUIRES. The 'Device will only 'Take', (DRAW), what it needs. It will not 'draw' more than it needs. So you can have all the Amperes in the world 'AVAILABLE, and 'the device WILL ONLY USE what it needs.' Simple and sweet, huh?? VOLTS = Correct. AMPS = as long as there is enough available from the 'supply' of power.
2007-10-03 09:19:36
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answer #3
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answered by Anonymous
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It seems you are worried about having too many amps applied to your device. I understand your concern, however, you need not be. Let's explore the definition of Amperage: "The rate at which electricity flows through a conductor as regulated by the load resistance".
So, now we know that even though the power supply you want to use is capable of delivering up to 1.2 amps of current, your device will only draw 500mA (half an amp) from the supply. You can now use Ohm's Law to determine the resistance of your device by dividing 12 (volts) by 0.5 (amps) and you get 24 ohms. It is not normally important to know the resistance of common devices you wish to power, but thought it might satisfy your curiousity to know.
If you care to view the "dirty dozen" deriviatives of Ohm's Law, please visit: http://www.the12volt.com/ohm/ohmslaw.asp
At that site you will see a pie chart showing all 12 of the Ohm's Law formulas involving:
Power (Wattage)
Voltage
Amperes
Resistance
Back in tech school these formulas were known as the "dirty dozen" as there are 12 of them. Hope this helps.
2007-10-03 10:06:31
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answer #4
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answered by Anonymous
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Your load should determine the amount of current thru the circuit.
You need to provide more detail before you blow yourself up.
2007-10-03 08:55:56
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answer #5
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answered by Anonymous
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