Can I use a 4.5V/ .8mA wall wart with this CD player?

I have a Jensen CD 60 ortable CD player that came without a power supply. It calls for a 4.5V/500mA adapter. I have a 4.5V .8mA adpapter, and the pin polarity matches. Can I use it with the Jensen?

Thanks.

BC adapted

Reply to
BCDrums
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0.8 amps or 0.8 milliamps (ma)? If 0.8 ma. , no way that is a fraciton of 500 ma.
Reply to
hrhofmann

That depends. Do you mean 0.8 A instead of 0.8 mA?

If so, yes.

Reply to
Doug Miller

I think you slipped a decimal point. Likely 800 Milliampres =3D 0.8 AMPS!

Check that the polarity and size of the wall wart are the SAME.

You should be Ok.

Reply to
professorpaul

Whoops! It is in fact .8A. So I'm good to go? The pin polarity is the same- pin is positive, sleeve is negative.

BC

Reply to
BCDrums

On 7/19/2010 12:52 PM BCDrums spake thus:

Yes. The figure 0.8A is the amount of *current* the wall wart can supply (*please* don't use that ignorant term "amperage"!). The rule is this: as long as the power supply (wall wart) can supply *at least* as much current as the device (CD player) requires, preferably more, you're good to go. (Assuming it's the correct voltage, of course.)

If you think a power supply is big enough but aren't sure, what you can do is to run the device and carefully monitor the power supply to make sure it doesn't get too hot. When a power supply is overloaded (device wants more current than it can supply), it will overheat, and other bad things can happen then.

Reply to
David Nebenzahl

Andy comments:

It depends. You can test a dozen different "4.5 volt" wall warts and have a dozen different open circuit voltages and a dozen different internal resistances. A wall wart is designed for a particular product that has a certain load draw, and, typically, one get a LOT higher voltage than 4.5 when unloaded and approx 4.5 volts when loaded by the intended device.

It has everything to do with the internal resistance of the wall wart.

For instance, suppose the intended load is 1 amp and the wall wart, unloaded, puts out 12 volts with an internal resistance of 7.5 ohms. If this supplies a device that draws 1amp, there will be a terminal voltage of 4.5 volts. If it drives a device that draws only 100ma, the terminal voltage will be 11.25 volts.

Depending on the voltage rating of the device the wall wart is driving, something like this could break down an internal capacitor or transistor....

So you see, it just depends........

Here's how to find out, in your case. :

Connect three D cells, which provide 4.5 volts, and hook it to the device and measure the current drawn. Then use Ohm's Law to determine the load resistance the device provides.

THEN take a resistor of this value and connect it to the wall wart in question. If the loaded voltage is around 4.5 volts with the resistor, that means it will be 4.5 volts with the device you want to power.....

One caveat , tho.... the device you want to power probably draws different currents depending on whether a motor is running or stopped, or the audio is cranked up high or not. That could cause the terminal voltage to go up and down..... since the load is not constant.....

So you see, your wall wart MIGHT work OK, or it might blow out some part in the device powered. That is why wall warts are generally mated to the device being powered. It is very unusual for a wall wart to have an internal voltage regulator to supply the same voltage independent of load. If the one you want to use actually does have an internal voltage regulator, you are all set.....

Andy in Eureka, Texas Licensed EE and retired design engineer.

Reply to
Andy

Now David, don't be too harsh. I was brought up as you were with the educators who were trying to educate me insisting that there was no such word as "amperage"

Not trusting the online sources I went to my copy of Webster's International Dictionary. Not only did it define "amperage", it claimed it was dated from

1890 to 1895. I even used a little magnifier to be sure that it was not listed as slang, archic, or argot. I also verified that the usage dated back over a century an was not a modern contrivance.

I was going to check further with my good friend Tom Edison but he was busy arguing with that foreign guy Tesla.

Charlie

Reply to
Charlie

Andy is correct on all counts. You will find a slightly higher voltage from the new wall wart because of a lower current draw than it was designed for.WW

Reply to
WW

Definitely no problem. any 4.5 volt power supply capable of supplying

500ma OR MORE will do the job.
Reply to
clare

The interesting thing is, most of these units that use a 4.5 volt power supply run on 2 AA batteries - which is only 3 volts. There is a voltage regulator built in to reduce the "unregulated" 4.5 to the required 3 volts.

Reply to
clare

Hmmm, Depending on what type of circuit in the wall art.

Reply to
Tony Hwang

=A0 =A0Licensed EE and

I am grateful for all these knowledgeable responses.

The device to be driven by the wall wart is a Jensen CD player. It is supposed to be run with two AA batteries. It is a gift. My six-disc CD changer died some years ago, and I replaced it with a Panasonic portable CD player from the '90s that was in a drawer. It had a line out as well as a headphone jack. It died in '09.

My lovely bride picked up the Jensen for $22 at Target, just to have something to play CDs on through the living room stereo. I don't want a battery-driven device, and the Jensen does have an input for a DC transformer, with the specs as listed previously. The transformer for the Panasonic player is still there, thought I'd give it a try. If it works, I will be happy. If it kills the Jensen, oh well. if it burns down the house, no good! There is no compatible transformer offered for the Jensen, by the way. What is the point of a $22 CD player if the power adapter costs as much or more?

I will give it a try and keep an eye on both units for an hour and see what happens.

BC transformed

BC

Reply to
BCDrums

On 7/19/2010 5:00 PM BCDrums spake thus:

So long as the device is labeled as having a diet of 4.5 volts (you told us this, so I'm assuming this is correct), and the power supply supplies this *nominal*[1] voltage, it won't kill the CD player or burn down the house.

Good plan.

[1] "Nominal" meaning what's printed on the wall wart, even if it does deliver some higher voltage with no load, which it will undoubtedly do, since it is no doubt an unregulated power supply. There is some leeway here.
Reply to
David Nebenzahl

=A0 =A0Licensed EE and

A note...the Panasonic portable player the Jensen replaces used three AA batts, for a total of 4.5V. But yes, the Jensen says two AAs.

BC low voltage

Reply to
BCDrums

=A0 =A0Licensed EE and

My guess is that the unit has an internal voltage regulator, otherwise the motor would try to run faster on 4.5 v than on 3.0V.

Reply to
hrhofmann

[snip]

In the same way, there's no such word as "voltage". It's EMF.

Reply to
unknown

For the record, this power supply is from 1995.

BC unregulated

Reply to
BCDrums

Should work fine, if the plug is the same size. The amp rating of the plug just means the power available. Since your appliances uses less, it will, well, use less.

Example is plugging a 100 watt (0.1 amp) light bulb into a 20 amp branch circuit. Works fine, and uses 0.1 amps.

Mr. Answer is always here for you.

Reply to
Stormin Mormon

If you want to be totally correct it is electromotive force. EMF is just an abrieviation.

Reply to
clare

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