# Convert 9V "wall wart" to 4.5V? (NT/OT)



## Scottychaos (Jan 2, 2008)

Hey everyone,
I have a small "clock radio"..the AC "wall wart" adapter for it is missing..
but I have stash of other ones, but none of the right size.

The plug on the side of the radio says it wants 4.5V.
It will also run on two C batteries.

I have an AC adapter that puts out 9 volts.
It says:

Uniden AC adapter
Model AD-310
Input: AC 120V 60 Hz 4W
Output: DC 9V 210mA.
Class 2 power supply.
for use with telephones.

Could I convert this 9V output to 4.5V by wiring in a resistor in the output wiring?
If so, what size? (I tried googling it, but didnt come up with anything helpful)

And do I need to be concerned about the amps?
the output at 9V says 210mA..and the radio will run on two C-cells.
Im assuming the amps aren't a concern (are they?)..I mainly want to get the voltage down to 4.5 or 5 volts..
any suggestions?

I went to Radio Shack, $30, no thanks..

Scot


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## JPCaputo (Jul 26, 2009)

Check out allelectronics. They have one for $3 plus shipping. 

Might as well pick up anything else you need to get enough stuff to make the minimum shipping price worth it.


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## JPCaputo (Jul 26, 2009)

Check out allelectronics. They have one for $3 plus shipping. 

Might as well pick up anything else you need to get enough stuff to make the minimum shipping price worth it.


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## BigRedOne (Dec 13, 2012)

If you want to drop half the voltage using a resister, you'll need one with resistance equal to that of the clock. 

Resistance can be computed if you know the voltage and amperage; therefore measuring the current while running on batteries will let you compute the resistance. This is Ohm's law, if I remember correctly. 

The wattage rating of the resister is a function of current draw.


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## Pete Thornton (Jan 2, 2008)

Could I convert this 9V output to 4.5V by wiring in a resistor in the output wiring?If so, what size? (I tried googling it, but didnt come up with anything helpful) Yes, you need to be concerned about amps. Amps = heat, and an under-spec resistor will burn up and maybe drop burning bits on the carpet.And do I need to be concerned about the amps? the output at 9V says 210mA..and the radio will run on two C-cells.Im assuming the amps aren't a concern (are they?). 2xC-cells = 3V. The input needs 4.5 V. [Sure it runs on 2 cell ? Not 3 ?] But it isn't unusual for the batteries to supply a different voltage. The higher the voltage, the less the amps (hence the use of 11,000V power transmission lines. See answer about amps above.)

Practically speaking, I would recommend the use of 'wire wound' resistors, if you must pursue this. They can handle higher currents and won't melt.
(Incidentally, you could feed the battery compartment with a 3V wall wart just by connecting the wires to the in/out poles of the battery area.)

Or you could get your hands on a 'voltage regulator' that outputs 4.5 V (the Accucraft coach lighting comes to mind - though maybe that puts out 5v?) This device (one chip plus heat sink) outputs the same voltage whatever higher voltage (within reason) you feed it with.


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## Scottychaos (Jan 2, 2008)

thanks everyone! 
yes, its definitely only two C-cells, not three. 
and it clearly says 4.5V on the side, at the AC adapter plug. 
so I guess it can operate between 3V and 4.5V. 

I still dont know if its as simple as wiring in a resistor? 
if I can figure out what resistor is needed? 

Big Red said "If you want to drop half the voltage using a resister, you'll need one with resistance equal to that of the clock. " 

how could I determine the resistance of the clock? 

thanks, 
Scot


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## BigRedOne (Dec 13, 2012)

Amperage equals voltage divided by resistance; therefore if you can measure (or know) the voltage and amperage you can compute the resistance. 

Example, if voltage is 3.0 (2 x 1.5 v batteries), and amperage is 0.1 amps, resistance is 30 Ohms. In this case, to drop 9V to 4.5V you'd need a 30 Ohm resister. 

Wattage is amperage times voltage. Therefore, in this example, 4.5V x 0.1 amp equals 0.45 Watt. I would upsize the resister to a one or two Watt piece.


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## dbodnar (Jan 2, 2008)

Posted By Scottychaos on 02 Feb 2014 03:39 PM 
thanks everyone! 
yes, its definitely only two C-cells, not three. 
and it clearly says 4.5V on the side, at the AC adapter plug. 
so I guess it can operate between 3V and 4.5V. 

I still dont know if its as simple as wiring in a resistor? 
if I can figure out what resistor is needed? 

Big Red said "If you want to drop half the voltage using a resister, you'll need one with resistance equal to that of the clock. " 

how could I determine the resistance of the clock? 

thanks, 
Scot Scot - do you have a retired cell phone charger? Most of them put out 5 volts - that is likely to be OK with your radio.

dave


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## Dan Pierce (Jan 2, 2008)

To use the 9 volt supply, look at adding the LM317 regulator and adjust the resistors for 4 volts. 
Or use the LM7805 a 5 volt regulator and place 2 diodes in series on the output which will drop the voltage to what the clock needs. 
All solutions will cost a few $$, and a power pack of 3 to 4.5 rating is the best/easy solution. Try Savers for a used power pack.


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## Scottychaos (Jan 2, 2008)

Thanks everyone! 
I thought this might be a good idea to try, (a resistor added to the 9v adapter) because the 4.5V adapter at Radio Shack was $30! which I thought was insane.. 
(these things are essentially worthless..maybe $5 tops..) 
But I just found a few on Amazon for $3 to $10.. 

I now agree that would be a better way to go!  
I was thinking I would try the resistor, *if* it was an easy and safe job..(I didnt know if it would be or not..which is why I asked!  
thanks for the input everyone! 
much appreciated.. 
Scot


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## Scottychaos (Jan 2, 2008)

(woah! first time I have had a link added to one of my posts on MLS! not cool... 
sure, its for amazon, and I did actually write amazon..but still, I do NOT appreciate my 
words being turned into advertisements against my will..as a 1st class member, we shouldnt have to put up with that..) 
(NO ONE should have to put up with that!) 
that is the worst kind of advertising.. 
Scot


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## Gary Armitstead (Jan 2, 2008)

Posted By Scottychaos on 03 Feb 2014 06:57 AM 
(woah! first time I have had a link added to one of my posts on MLS! not cool... 
sure, its for amazon, and I did actually write amazon..but still, I do NOT appreciate my 
words being turned into advertisements against my will..as a 1st class member, we shouldnt have to put up with that..) 
(NO ONE should have to put up with that!) 
that is the worst kind of advertising.. 
Scot That's what some of us have been saying all along.


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## Dan Pierce (Jan 2, 2008)

Another idea... Get a 9 volt unregulated that is for 220 volts only. 
Wire it to 110 volts and you will get 4.5 volts!! 

Half in will give half out.


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