1.5 volt UPS circuit for MP3 player

[holabr]

I have a cheap MP3 player that runs on 1 AA cell. I'd like to power it off a wall wort 3 volt AC to DC converter I have. I'd also like to have a battery in the circuit so that if house power is lost, the player will continue uninterrupted off the battery. The player is going to provide background music and music on hold through a small telephone system. The problem with just running off a converter is that after any type of power interruption, the "play" button on the MP3 player has to be pressed to get it working again. The circuit should be designed in such a way as to keep the battery charged and power the player off the converter except when the power goes off.
Any ideas?

 

[iONic]

...if house power is lost, the player will continue uninterrupted off the battery.

Maybe you should tap off the phone line for the 1.5v, then you won't need a battery at all!

Of course you have to:
1) Convert the AC voltage to DC,
2) Fuse it, and
3) Regulate it to 1.5V

 

[holabr]

What would the circuit look like to do this?

 

[eblc1388]

While Mr. ionic is conjuring up the circuit from his wonderful idea, perhaps you can also consider my not so clever idea of using a simple LM317 adjustable voltage regulator.

The LM317 is a cheap three pin device working with a wide range of input voltages and allow the user to adjust its output voltage from 1.25V upwards, using two resistors or one fixed resistor plus one variable resistor.

You don't need to use a rechargeable battery but a normal 1.5V AA battery instead. The output of the LM317 should be going to the battery (+) terminal via a series 1N4001 diode to prevent the battery discharging through the voltage setting resistors should the AC power gone off.

You just need to set the LM317 voltage regulator output voltage to 2.1V(1.5V+diode 0.6V) and the AA battery will not charge or discharge under this condition and the MP3 will work happily at 1.5V. The input to the LM317 should be about 5V or more via a common AC/DC wall adaptor.

More information is available by googling "LM317".

 

[justDIY]

why not just plug the mp3 player's wart into the same ups that powers the phone system?

 

[iONic]

While Mr. ionic is conjuring up the circuit from his wonderful idea...

While Mr. ebcl1388 may have the more practical and safer approach my theoretical but possible circuit solution is posted below.

P.S. It may need some tweaking though...

 

[audioguru]

my theoretical but possible circuit solution is posted below.

P.S. It may need some tweaking though...

The phone line will be off-hook all the time when this circuit tries to steal current from it.
The current from a phone line is much too low to power an MP3 player.

The pot connected to the LM317 does not adjust its voltage. Pin 2 is grounded so the output voltage is fixed at 1.25V but will rise to about 1.5V without a load.

 

[KMoffett]

https://www.electro-tech-online.com/images/editor/menupop.gif

iONic,

Your circuit has a lot of problems.

1. The absolute maximum input-output differential voltage for the LM317 is 40Vdc. The ON-hook line voltage is nominally 48Vdc. Ring voltage is around 96Vac

2. Phone lines are not intended to be power supplies for other devices, so attaching you bridge directly across the line and supplying the needed current for the MP3 player will load down the line. As the current increases the line voltage drops...probably to the point of creating an Off-hook condition. Phone companies don't like this. If I recall correctly, the maximum current available in an Off-hook condition is about 20mA.

3. Your variable resistor from pin 3 to pin 2 will only work if there is a fixed resistor from pin 2 to ground. And with that, Cout's negative terminal would be connected to LM317's pin 2 as opposed to ground.

Ken

 

[iONic]

The current from a phone line is much too low to power an MP3 player.

The pot connected to the LM317 does not adjust its voltage. Pin 2 is grounded so the output voltage is fixed at 1.25V but will rise to about 1.5V without a load.

You don't think that you can get 100 - 200ma from tthe phone line?

Your right about the 500 ohm pot, it should be replaced with a 240 ohm fixed resistor and place the pot from pin 2 to ground.

 

[iONic]

"To invent, you need a good imagination and a pile of junk."

In other words, KMoffett, All I have is a pile of junk and a lousy imagination!

I will conceed.

 

[KMoffett]

iONic,

Your imagination is great...and your pile of junk seems adequate. Just felt that clarification was in order in case someone might try actually hooking this up. I have seen (and built) connections to the phone line to power LEDs to detect On-hook /Off-hook conditions...but these only draw <1mA. Had to do a lot of reading (pre-internet) about phones to find out where "my" designs were going south. Keep up the "what if" attitude.

Ken

 

[mvs sarma]

Generally the phone lines feed a current limited (each limb) supply of -48V and earth.the limit is around 35mA, but as the line loop resistance increases (for the far off phones) the current would less.
added to that, the lines don't care about 3 to 4 mA of leakage current in the modern exchanges. It amounts to misuse of telephone line to tap and connect devices across the phone lines, and certain telecom companies can even object to it.

I wonder that the whole thing is around externally powering an MP3 player device that would comfortably work from a AAA cell.

 

[audioguru]

Modern telephone exchanges time-out and disconnect a phone line that is off-hook for a couple of minutes.

 

[mvs sarma]

Modern telephone exchanges time-out and disconnect a phone line that is off-hook for a couple of minutes.

Yes , but they still keep battery feed, with the said current limit , on the phone line while the exchange equipment to serve the customer is taken off(thus no more tones-let it be dialtone or busy tone which comes after 90sec of inactivity).

 

[iONic]

It amounts to misuse of telephone line to tap and connect devices across the phone lines, and certain telecom companies can even object to it.

....Yikes, then I had better unplug my TV!!

 

[audioguru]

What does a TV have to do with a phone line??

 

[iONic]

What does a TV have to do with a phone line??

Was suggesting that I was running my TV from the phone line as a source, ...like a joke.

 

[Hero999]

While Mr. ionic is conjuring up the circuit from his wonderful idea, perhaps you can also consider my not so clever idea of using a simple LM317 adjustable voltage regulator.

The LM317 is a cheap three pin device working with a wide range of input voltages and allow the user to adjust its output voltage from 1.25V upwards, using two resistors or one fixed resistor plus one variable resistor.

You don't need to use a rechargeable battery but a normal 1.5V AA battery instead. The output of the LM317 should be going to the battery (+) terminal via a series 1N4001 diode to prevent the battery discharging through the voltage setting resistors should the AC power gone off.

You just need to set the LM317 voltage regulator output voltage to 2.1V(1.5V+diode 0.6V) and the AA battery will not charge or discharge under this condition and the MP3 will work happily at 1.5V. The input to the LM317 should be about 5V or more via a common AC/DC wall adaptor.

More information is available by googling "LM317".

Good idea but the problem is a silicone diode has too higher voltage drop for a 1.5V cell to work for very long. I would use a relay with NC contacts and the coil connected to the unregulated side of the mains powered supply to disconnect the battery when the power is on.

 

[audioguru]

Good idea but the problem is a silicone diode has too higher voltage drop for a 1.5V cell to work for very long.

The diode is in series with the output of the LM317 to prevent the battery from discharging into the LM317 circuit when the mains power is gone.
The circuit is powered by the 1.5V battery or by the 2.1V - 0.6V= 1.5V from the regulator.

When the battery has discharged a little then the regulator will try to charge it and might damage it.

Silicone is used to make false breasts for skinny women. Silicon is used to make diodes, transistors and ICs. A huge difference.

 

[Hero999]

The diode is in series with the output of the LM317 to prevent the battery from discharging into the LM317 circuit when the mains power is gone.
The circuit is powered by the 1.5V battery or by the 2.1V - 0.6V= 1.5V from the regulator.

Nothing wrong with that.

When the battery has discharged a little then the regulator will try to charge it and might damage it.

This is why I assumed there was a diode in series with the battery, using a relay to switch it would solve this problem. Indeed, it's a shame you can't get depletion mode power MOSFETs for this sort of application.

Silicone is used to make false breasts for skinny women. Silicon is used to make diodes, transistors and ICs. A huge difference.

I made a typographical error, big deal.