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how can i drop 1A dc to 500ma dc? Answered

i have an AC-DC power supply with voltages of DC 3 - 4.5 - 6 - 7.5 - 9 - 12V and 1000ma and i want to drop this down to 500ma!

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There's a common misconception that a power supply's rated current is what it supplies at all times. What it actually means is that it will supply UP TO that current. So, if you've only got one LED connected, it will be supplying around 25mA. You are quite safe to use your 1000mA supply and you will be running at half it's possible maximum capacity.

Thanks for your great answer. I'm wanting to use a 12v 1.5A ACDC adapter for some LED strip lighting that are 12v 1A rated. Will this adapter suffice?

i have 25 white leds connected in parallel which battery should I use

I want to produce DC 12V 5A from AC 12V 1.5A.

How can i do this?

Same problem too
Anyone witj solution

You can't. You need to buy a new power supply.

And 1more...., I bought a new 12v 1ah lead acid battery. Now it's giving me 12.6 output voltage. If I use this battery without charging initially will this battery become damaged? Or will it turn into a primary cell? So that it won't recharge later?

hello frns, same problem here!! I have a 12v 1amp ac-dc adapter, and i badly want to charge my 12v 1ah(three 4v 1ah batteries connected in series) using the adapter, so i have to charge my batteries with 300mA, so what do i do now???? Pls help asap

are you sure? cause it isnt a psu it is just a plug in adapter and it says on it "will supply 1000ma over all voltages" so i was looking for somthing like a current regulator? or somthing like that! thanks!

. AndyG is correct. The current is determined by the load.

yes but i think he is talking about a psu but the power supplu is a fixed current one.here is the Technical Specification Specifications: Input Voltage: AC 230V, 50Hz Input Power Plug: UK 3 Pin Mains Plug Output Voltage: DC 3 - 4.5 - 6 - 7.5 - 9 - 12V Output Current: 1A (1000mA) Output Voltage Type: Variable Output Power (Max): 12W Regulated Output: No

Still absolutely fine.
You can draw UP TO 1 Amp at each of those voltages.
The maximum rating is reached if you draw 1 Amp at 12V, and power is Volts X Amps, so 12 x 1 = 12, which is shown as the maximum output of the supply - It all ties up!
No doubts at all - This is correct.

The only thing you have to watch out for is that it says 'unregulated' so the voltage it's set to will drop a bit as the load increases. What are you planning to drive from it?

ok thanks i believe you!!! but if i want it to charge a recharable sealed lead acid battery and the manufactures charging rate says it has to be at 7.5v and 500ma but the battery is 6v and 4.5ah will the battery draw the full 1a?

Aaaargghhh! You've just opened a whole new box of frogs! What I said about the supply is entirely correct, but your battery will drag down the output to 6V when charging starts, and may draw the full 1A (and possibly a bit more, stressing the power supply), whereas, as you sort of said at the beginning, the battery wants this limited to 500mA. This will need a bit more circuitry - I'll come back to you in a bit.

Easy!
Just put a 2.7 Ohm 1 Watt resistor in series with the charging lead. This will limit the current to 500mA when the battery is fully discharged and trying to draw maximum current.

The difference between what I said earlier is that I was assuming you were planning to drive a simple resistive circuit such as a circuit board. A battery is not a simple load as it has it's own voltage and a very low internal resistance.
(Go on. After all this typing I deserve Best Answer ;¬)

Ta muchly ;¬) I'll go and answer you question about PicAxe projects too.

. Wouldn't he need at least a 6 W resistor? Can't he possibly end up wasting half of the 12 W max of the PS? Or should I have a cup of coffee after waking up and before posting? :)

I worked it as the 7.5V of the supply minus the 6V that the battery will drag it down to when charging at maximum rate = 1.5V.
To limit this to 500mA the resistor would be V / I = 1.5 / .5 = 3 Ohm (Nearest preferred value 2.7 Ohm).
Power dissipation would be I2R which is 0.5 * 0.5 * 2.7 = 0.675 W.

As the battery charges the terminal voltage rises so the resistor has less and less effect.

You're looking at a discharged battery at 0V, aren't you. From what I know of these batteries (I'm no expert) If there's any charge at all in a lead-acid, the terminal voltage will be around 6V. Here's a graph for a 12V battery. If the battery is totally and absolutely flat then it may draw over-current for a very short period but will very soon get to 6V.

> You're looking at a discharged battery at 0V, aren't you. . Yes, I was. Thanks for edifying me.

. Once again, AndyG hit the nail on the head. . You will need a charging circuit between the battery and PS. Most charging circuits will accept an inconstant voltage input (they have their own regulation built-in), so lack of regulation shouldn't be a problem.