I need to reduce the voltage of an 18-volt ac, wall wart, power supply to 9 volts ac? - need wart
How can the voltage of 18 volts alternating current, the wall wart, 9-volt power supply? It is 900-MA, I would like to remain constant.
I guess the opponent or two in the starting line, it should do.
Has anyone a moment to give me a heads-up?
Saturday, January 30, 2010
Need Wart I Need To Reduce The Voltage Of An 18-volt Ac, Wall Wart, Power Supply To 9 Volts Ac?
6 comments:
What you first want to understand is that if, at 900 mA at 18 volt power, then you can be the total resistance through the two cables of 20 ohms and 900 mA to flow through connecting them. If you have two 10-ohm resistors are in series on the power cord, you get the same, a total of 20 ohms and 900 mA, but now measured in a connection between the two resistors, and by a set and found to 9 volts. The only thing is that you may not current consumption without the stress of change. You can fix this by using a resistance of greater value. You can, for example with a resistance of 10 ohms and that such a pot is 100 ohms, which can draw all and set the value you want, but they have limited nor a current, never anywhere, including around 450 mA for external circuitry. This is because the load has an effective resistance in parallel with the potentiometer. It would be better to build a voltage regulator for consistent performance and can not in any book of the trail are found as well. There will be some study and research opportunities to take in the Libraryr, but the direction to take.
It really is a simple and efficient to do so. When you bring two equal resistors in series, will be 9 volts obtained by one of them, but you will not get in the position, 900 mA. If you know that the device uses a constant power source that could put a resistor in series with the same value, but usually not the case. If you will be dealt with a little energy as a source of 18 volts DC probably work just as well. If you have an electronic device as a 9-volt supply are known, you'd better not be wrong.
It really is a simple and efficient to do so. When you bring two equal resistors in series, will be 9 volts obtained by one of them, but you will not get in the position, 900 mA. If you know that the device uses a constant power source that could put a resistor in series with the same value, but usually not the case. If you will be dealt with a little energy as a source of 18 volts DC probably work just as well. If you have an electronic device as a 9-volt supply are known, you'd better not be wrong.
as little as you know, you should leave him alone, back to Wal-Mart and buy a multi-voltage unit, then take over mail, no "resistance inline 'would have a fire hazard, you must create a circuit construct, not a separator does cabable,
Use two resistors of equal value in the supply of 18V.
Resistors must put together. The connection resistance is half of the total volts.
This circuit is known as a voltage divider.
Hopefully it will work.
With a transformer ratio of 2:1. This will step up to 18 V AC 9V.
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