Thread: mp3 player hack

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Hello everyone, I am in need of people far smarter than me to help with an electronic question. I have several wall-warts that I got cheaply at goodwill that I would like to use to provide continuous power to a small mp3 player. The problem is their outputs range from 3 to 9 volts dc depending which one I use and the player only uses 1.5 volts. Does anyone know of a simple way to safely reduce the voltage. I did a google search and saw some adjustable voltage circuits but they seemed beyond my understanding. Any help would be appreciated.

I've never seen a discrete 1.5VDC regulator, but you could use a voltage divider circuit to drop the 3 volt wart to 1.5 volts. They're very easy to understand and build. You just need a small perf board, a pair of resistors, and a possibly a terminal block to attach the wires. If you wanted to make it adjustable, you could use a 250 ohm pot in place of one of the fixed resistors.

Here's a link to a voltage divider calculator:
http://www.raltron.com/cust/tools/voltage_divider.asp

A pair of 150 ohm resistors wired as shown on the calculator page will give you 1.5VDC. Just make sure that the wart really is 3 volts - verify both the input and output voltages with a DMM before connecting your player.

Edit - I should clarify the resistor values. Any 10X multiple of 15 ohms will result in 1.5VDC (150, 1.5K, 15K etc).

Originally Posted by Otaku

Just make sure that the wart really is 3 volts - verify both the input and output voltages with a DMM before connecting your player.

Qft. I've seen 3 vdc rated wall warts range all the way to 4-4.6 vdc. Also, they are nearly always an unfiltered design which can play hell when combined w/ an audio device. On the upside, it is very little trouble to add a filter to them.

Otaku & rnmully,

The voltage divider idea may be problematic in this case. While it's true that using a voltage divider can be used to create a reduced voltage, the calculators [Vout = Vin * (R2/(R1+R2))] will only give you correct results if there's no load on the output voltage. Any load presented here (i.e.: mp3 player) will lower the resulting voltage from what the equation will say.

You could 'tune' the resistance, with the mp3 player connected, and that would get you the right voltage, by experimentation. But then it would be important to make sure the load didn't change, meaning that the headphone outputs would probably need to be hooked to another amp, instead of headphones or speakers. Otherwise, as soon as a it played a loud sound, the load would increase, and as a result, the input Voltage to the mp3 player would drop - possibly resetting it.

In general, electronics don't like pulling their supply through resistance. That's why supplies usually have big capacitors in them, and every chip has a capacitor right next to its power pin.

Here's a regulator you can get from Mouser, that will put out 1.5V To use this, connect your supply to a 100uF cap, and pins 5 & 4 of the regulator. Connect ground from your supply to pins 1 & 2 of the regulator. Finally, connect pin 3 of the regulator to a 100uF cap, and to your mp3 player.

- Hook

Hey, cool, Hook! That's a very useful device and I'll be ordering some for a couple of projects I have in mind. That's true that changes in the player voltage demand would cause a sag and a possible device reset. Could this be fixed by using a 100uF bypass cap across the divider output?
PS - not trying to hijack the thread. Might be better to answer via PM.