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I need to reduce the voltage of an 18-volt ac, wall wart, Power Supply to 9 volts ac?

How do I reduce the voltage of an 18-volt ac, wall wart, power supply to 9 volts ac? It's 900-ma which I want to remain constant.

I imagine a resister or two in the line out should do the trick.
Anyone have a moment to give me a heads up?

What you want to understand first is that if the power supply will put out 900 mA at 18 volts then you can connect a total resistance across the two leads of 20 ohms and 900 mA will flow across it. If you put two 10 ohm resistors in series across the supply wires, you get the same thing, a total of 20 ohms and 900 mA, but you can now put in a connection between the two resistors and measure across either of them and find 9 volts. The only thing is that you cannot draw any current without the voltage changing. You can fix that by using a higher value resistor. You could for example, use one 10 ohm resistor and, say, a 100 ohm pot which you can draw across and adjust for the value you want, but nevertheless, you still will have a limited current draw, never anywhere near even 450 mA for your external circuit. It is because the load presents an effective resitance in parallel with the potentiometer. It would be better to build a simple voltage regulator for your power supply and you can find those in any good circuit book. It will require some study and research in the library for you, but that is the direction to go.

Rapid-Link In-Wall Power Solution


 

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