bmachine Posted June 7, 2006 Report Share Posted June 7, 2006 Hello, dumb question here.I need to use one of those little wall mounted ac/dc converter for a little 555 based project I am doing. I found an old one that is labeled to have a 9V DC output. But I put my voltmeter to it, it reads 14volts. I then tried a couple that were rated at 24v output and they read more like 28 or 29v.Is this normal? Is it because they need to be measured under some kind of load or something? Is my transformer really going to give 9V to my project or is it going to fry everything by delivering 14 volts?Thank you very much.Bmachine. Quote Link to comment Share on other sites More sharing options...
trigger Posted June 8, 2006 Report Share Posted June 8, 2006 It is normal, you are measuring the off load voltage and it will be higher than the rated voltage. Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 8, 2006 Report Share Posted June 8, 2006 Cheap wall AC-DC adapters do not have voltage regulators. They deliver their rated output voltage only when they have their rated load current.An adapter with a higher output current rating has an output voltage that doesn't rise much without a load. Quote Link to comment Share on other sites More sharing options...
allvol Posted June 8, 2006 Report Share Posted June 8, 2006 Hi, bmachineUsing an LM7809 voltage regulator and two common capacitors will produce a steady 9 volts Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 8, 2006 Report Share Posted June 8, 2006 Using an LM7809 voltage regulator and two common capacitors will produce a steady 9 volts Quote Link to comment Share on other sites More sharing options...
ante Posted June 8, 2006 Report Share Posted June 8, 2006 Hi AG,I have to agree but, there is some hope though! If the project consumes a lot less power like one tenth of what the wall wart was originally intended for it might work! ;) Quote Link to comment Share on other sites More sharing options...
allvol Posted June 8, 2006 Report Share Posted June 8, 2006 Well, phooy. I guess I'll have to take these that are working on my bench and scrap them.AllVol Quote Link to comment Share on other sites More sharing options...
bmachine Posted June 8, 2006 Author Report Share Posted June 8, 2006 Wow! Excellent information, all. Thank you very much. I guess I'll go ahead and give this a shot.BTW, Is there a simple way to simulate a load for testing purpose?Bo. Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 8, 2006 Report Share Posted June 8, 2006 Well, phooy. Quote Link to comment Share on other sites More sharing options...
windoze killa Posted June 10, 2006 Report Share Posted June 10, 2006 Wow! Quote Link to comment Share on other sites More sharing options...
MP Posted June 11, 2006 Report Share Posted June 11, 2006 The wall wart will give you all the informatin you need. If it states [email protected], you just need to use a resistance that will cause a 500 mA load from 9 volts. Make sure your Watt rating on the resistor is high enough. You can calculate this on your own. It is easy with ohm's law, and then you learn something also. <Time to brush off the dust on the old ohm's law book.>MP Quote Link to comment Share on other sites More sharing options...
ante Posted June 11, 2006 Report Share Posted June 11, 2006 MP,You can’t use Ohms law here; Ohms law has no “P” for the wattage! ;D Quote Link to comment Share on other sites More sharing options...
MP Posted June 11, 2006 Report Share Posted June 11, 2006 This is true. Even though we have attributed the Watts calculation with ohm's law and most of the ohm's law calculators include it, power calculation was not part of the original ohm's law. It is an extended calculation. ;DMP Quote Link to comment Share on other sites More sharing options...
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