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Hi,

I have opened up an old siemens A60 mobile phone and soldered wires onto the contacts which connect to the (+) and (-) of the  battery (with battery removed) so I can use an external 3.7v lipo battery. It works fine. Hower I am controlling the mobile using an Arduino uno and I want to power everything using a single power supply - a 12v, 3Ah Lead Acid battery including the phone (the arduino should be ok at 12v I think).

Does anyone have any hints or tips on how to step down the voltage on the 12v battery to get 3.7v for the phone? I believe it will need to deliver about 0.4A for the mobile to work properly - not sure - got info off mobile phone charger. Oddly (to me a novice) the phone dosn't work off it's own charger when connected direct to the power wires I soldered in, but I note it reads 5v not 3.7v

Many Thanks


PS the project is to build a GSM alarm system for my shed which has no mains power - hence the 12v lead-acid battery, which I will hook up to an off-the-shelf soral trickle charger.
« Last Edit: August 10, 2011, 08:29:11 am by trendski » Logged

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Sounds like you'll need a voltage regulator, such as the LM317T adjustable regulator. You'll need a heatsink on the regulator, too.
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Thanks Anachrocomputer,

I looked at voltage regulators but didn't know which was suitable.



I'll give that one a go.  Cheers.

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The LM317T is a commonly-available (hence cheap) 1.5Amp adjustable regulator. If you need a low dropout voltage (i.e. a low voltage differential between input and output), you could have a look at the LM2941. You won't be able to use a fixed voltage regulator such as the LM7805 because they don't make that in a 3.7V version.
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Thanks for that Anachrocomputer,

My input voltage will be 12V from a lead acid battery.

I ordered one of the LM317T. I looked at the datasheet and it gives this equation for calculating Vout from:

Quote
Vout = 1.25 V * (1 + R2 / R1 ) + Iadjj * R2

Code:
so if R1 = 1 and R2 = 2

Vout = 3.75V

ignoring Iajj because so small

But these are very small resistors so try:

Code:
R1 = 100 and R2 = 200

Vout = 375V


Is that correct? Sounds strange.


I need about 3.7 so 3.75 will do I guess, will I really have to find such small resistors?

Ideally I would like to be able to tune the Vout using a variable resistor. ie tune it between 3 and 5v to try and get the mobile phone power just right (I think) maybe because the voltage may drop under load.

Ideas most welcome

Edit:
Quote
Code:
ignoring Iajj because so small


should be

Code:
ignore Iajj*R2 'cos so small

[code]R1 = 100 and R2 = 200

Vout = 3.75V


Which makes sense now.





[/code]
« Last Edit: August 14, 2011, 01:08:37 pm by trendski » Logged

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Here is a circuit with a potmeter

http://www.electronicecircuits.com/electronic-circuits/lm-317-power-supply
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I need about 3.7 so 3.75 will do I guess, will I really have to find such small resistors?

100 Ohms and 200 Ohms are quite normal vales for resistors, nothing to worry about there. LM317 circuits are usually in this range.

But, 200 Ohms is not a common 'preferred value' for a resistor. You can find them, but more common values are 220 Ohm and 180 Ohm. In your case, it may be simpler to buy three 100 Ohm resistors and make the 200 Ohm by using two in series.  You'll find that resistors are usually sold in large quantities, sometimes in a minimum of 50 at a time!  But resistors can be had for as little as 35p for 50. I think Maplin's still sell them in one-offs, though.
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Thanks for the input guys,

Hopefully my LM317 from Maplins should come today.

I already have a 500 ohm pot and a lucky bag of resistors, so I'm going to try the circuit attached based on  http://www.electronicecircuits.com/electronic-circuits/lm-317-power-supply recommended by Erni.

I should be able to tweak it from 1.25 up to 4.1v, but if I get significant "drop-on-load" maybe I can reduce the fixed resistor to 100 ohm and get a higher upper level (7.5v)

Am I thinking along the right lines? Do I need all those diodes and capacitors [in the electronicecircuits.com circuit]  if I'm using 12v battery? (obviously not the AC step-down and rectifier bits)

Cheers


* LM317 Ajustable V-reg.gif (3.3 KB, 600x300 - viewed 18 times.)
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I don't think you'll need D2 and D3, they're to protect the regulator in various ways. Do not install C2, that's incorrect design and will reduce the regulator's ability to regulate when the output current is fluctuating. Install C1, C3 and C4 for best performance and to guarantee stability of the regulator.

The output voltage should not drop on load, as that's the purpose of the regulator! The voltage divider R1/R2 is used to sample the output voltage and feed it back to the regulator. This should make the output voltage stable even if the load draws more (or less) current -- provided we don't have C2 slowing it down!
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Got it working !

I used a  220 ohm and a 470 ohm and the LM317 is putting out 3.9v, which the mobile is working with ok, but I think I'll drop it down to 3.7 to be safe.

I'll put in those capacitors to improve stability.

I understand power regulators better now, I think.

My Arduino and the phone are running off the single 12v supply. The Arduino is sending the AT commands and the phone  is dialling out and sending SMS

Brilliant

Thanks again 

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Hi Again,

phone working with LM317 @ 3.75v. It's the TO63 package so it lies flat on a copper area of a PCB.

Should there be a large amount of heat generated by the LM317? I've got a piece of old PCB (2.5 x 4 cm) soldered to the back of it to dissipate heat.

I can only just touch the LM body - it's that hot! Although I note from the datasheet it can operate up to 150 deg C, above which it has built-in thermal shutdown.

I also noted the size of the heatsink on electroniccircuits.com's website.

Do I just need a bigger heatsink, I wonder, or could there be a more sinister problem?

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You can work out the amount of heat produced at the regulator by multiplying the current (in Amps) by the voltage drop across the regulator (in Volts). 12 - 3.75 = 8.25, yes?  Then 8.25 x 0.4 is about 3.3 Watts of heat at the regulator.  That's quite a bit, so you should expect it to get hot. You'll definitely need a heatsink, maybe a fairly big one. This is an inherent disadvantage of the standard, easy-to-use three-terminal regulators.
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Thanks again Anachrocomputer,

I think the current will also rise when transmitting so I'll up the ante with a big heatsink. Thanks to my Scots and Yorkshire forefathers  I'm a true skip-bat,  smiley-wink

and I've just scavanged 3 lovely pro heatsinks off an old pentium II motherboard, which were attached to voltage regulators. It was a dual processor system so it also has two nice chunky processor heat sinks with fans if required in future projects.

Another (perhaps nutty) idea occurred. I've got a spare 7805 5v regulator. What if I dropped 12v down in two stages 12 -> 5 then 5 -> 3.75 volts?



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Another (perhaps nutty) idea occurred. I've got a spare 7805 5v regulator. What if I dropped 12v down in two stages 12 -> 5 then 5 -> 3.75 volts?

Then you'd spread the 3.3Watts out between the two regulators. The energy has got to go somewhere, as heat!
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Yes I understood the physics. It was just one of those what if, noob moments.

It didn't work anyway 'cos I think I was right on the 1.20v minimum reference voltage and I couldn't squeeze out more than 3.3v

thanks again for all the help and insight.





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