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kaanuki
Joined: 24 May 2006
Posts: 1
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 Posted:
Wed May 24, 2006 12:14 am Post subject:
HOW DO I --- Run 2 Devices using one AC Adapter...??? |
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Hello, All.
I am a converted computer nerd, whose last serious electronics project was in high school, 10 years ago - so, please bear with me.
I am creating a project where I am combining 2 completely seperate (off the shelf) electronics devices into one box, creating a sort of "ALL-IN-ONE" solution. As it stands, each of these devices is self powered, and comes with its own AC Adapters. I would like to create(have created) a power converting circuit board to allow me to run both devices using just 1 AC adapter. Here are the power specs(and my dilemna)...
Device #1: 12v / ~100mA
Device #2: 12v / ~500mA
What type of voltage and circuitry is called for here, and what other considerations are needed?. Any help or Guidance would be appreciated.
Thank you.
-Kaanuki-
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petemar
Joined: 01 Jun 2006
Posts: 6
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Thu Jun 01, 2006 11:55 pm Post subject:
simple |
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All you have to do is get a 12V power supply (or design one) that can handle 1Amp without more than a 5% decrease in Voltage. Then just take the output of the power supply and connect it to some sort of buss that has two or more rails. Assign one rail +VCC and another GND. Then just connect your devices to that buss. Of course, you'll need some jacks to connectorize your devices. If your working with computer-based devices, you'll probably need a 1/8" jack or something like that.
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GOD BLESS |
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neon
Joined: 25 Feb 2006
Posts: 594
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Sat Jun 03, 2006 6:20 pm Post subject:
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just add one diode to each power output that will insulate one from th other however it is a sure proof solution but try it
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petemar
Joined: 01 Jun 2006
Posts: 6
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Mon Jun 12, 2006 4:27 pm Post subject:
You don't need a diode |
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Why in the world would you need an insulating diode in this situation. You don't need a diode when you put two resistors in parrellel off of the same source. This is basically the same thing, you are putting two resistive elements in parrellel, that's it. If you put a diode in before the load, all you get is .6V less of the source voltage. That's basically the only effect.
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