Help! Universal Adaptor Question

Watts = volts * amps, so 3.1 watts / 6 volts = 0.516 and change amps =

516 mA = round up to the next hundred to give you some "wiggle room", and get a 6 volt, 600+ mA adapter (I'd expect that the closest acceptable unit you'll find will be a 750 mA, although you might be able to get your hands on a 600 mA version) and you're golden. The key is *AT LEAST* the mA rating of the device. Anything less than the device's rating, and you'll let all the magic smoke out of the adapter. More than the device's rating will do no harm, but it won't do any good, either. (beyond *MAYBE* extending the life of the adapter because it isn't working "up against the redline" constantly.)

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While you haven't said anything WRONG, that's only a small part of the issue. If you're talking REGULATED wall wart at 6V and battery charging is NOT involved, you are pretty safe using anything over the minimum required.

But most wall warts are NOT REGULATED. And the worst of the bunch are the switchable voltage ones. The voltage printed on the label is a nominal guideline voltage at some unspecified load conditions. Two different brands with the same numbers on the label can have significantly different voltage output at the same current. The voltage can be much higher at low load. So if your adapter is rated for higher current, your nominal load will be a smaller percentage of rated and the voltage is likely to be higher...sometimes a LOT higher. Especially when you turn the device off, the voltage may increase enough to damage some internal parts ahead of the switch.

Then there's the size of the barrel connector. If it's just a little too big, it can spring the contact so the thing won't run on batteries any more. It's fun to go to Radio Shack and watch the "we got answers guy" try to cram different sized connectors into the customer unit.

Your problems multiply if the adapter charges internal batteries. Battery charging is often carefully tuned to the AC adapter characteristics to save that last penny. An adapter that's too powerful can severely unbalance that equation leading to smoke.

Call me paranoid, but weigh the cost of the TV against what you'll save by not buying the vendor recommended adapter. Is the difference worth the risk. Even if it works 99% of the time, somebody has to be the 1%.

Now, if you've got the instrumentation and are willing to test the process over the range of conditions, you can make almost anything work. mike

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Also check to see if the unit requires center plus or center minus. Tom

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Hi, Patrick. The rule for voltage sources is that, as long as you don't exceed their maximum current rating, providing the voltage is a done deal. Using the power equation

P = V * I

or power equals volts times amps, you need about 0.5 amps, or 500mA. Any adapter that provides more than 500mA should be OK.

If you're in a hurry, go to any Radio Shack and get their 6V/800mA AC-to-DC Power Adapter, Catalog #: 273-1761 for the semi-outrageous price of $17.99. Bring your Casio, and have the counter person pick out the "Adaptaplug" you need.

As always at Radio Shack, you'll overpay, but that's the cost of getting the electronics component you want on a Sunday afternoon. Also, if you're an absolute newbie, there's a good chance the counter person will know more than you (but ask first anyway, that's not always the case). If you choose the wrong adapter and happen to hook it up backwards, you'll probably let the smoke out. If you want to go with one of those universal adapters, be sure to get one rated for more than

500mA. Also, if you've got kids (voice of experience), use epoxy to glue the correct adapter in place and set the voltage switch on the 6V setting. Throw the other adapters out or hide them.

Good luck Chris

For your TV type, you should get in touch with Casio, and order the original.

These sets are easily damaged, if the adaptor does not properly match for its needs! I have seen a fair number of clients have problems after. Only Casio can service the their sets, and the cost is not worth it.

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JANA _____

1000mA. The specifications on the user guide says that it runs on 6 volts, and the power consumption is approximately 3.1W.

How much mA do I need?

Regards, Patrick

Thanks everyone, I managed to purchase a brand new, original Casio adaptor at a steal(80% off!), and it turns out to be about half the price of the cheapest

800mA universal adaptor available here.

Regards, Patrick

Tom Biasi wrote in sci.electronics.misc:

Going off on a tangent here, but I have often wondered...

These devices usually have a diode in the supply line to protect against inverted voltage. At the cost of three more diodes (and another diode drop in the supply) one could put a bridge rectifier there and never have to worry about polarity again. Is this occasionally done? I've never seen it.

Anno

Yes, I have seen it in a few (very few) products.

Hi Patrick,

You would need:

I(ma):(3.1W/6)*1000=517ma minimum adaptor.

Don't forget to verify the polarity before connecting the adaptor.

Hope this helps

Fern

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