Hi,
Will some one please teach me to build circuit to step down the 18DCv from my Ryobi job radio battery to 1.5DCv for my mp3 player.

Regards, Lee

 

You can use a LM317 regulator. It is not efficient, but it's the most common solution, though. You should measure how much current is sinked by your MP3 player and choose a heat-sink for the LM317 properly.
The regulator requires two resistors to set the output voltage and some capacitors.

 

Hi LeeCoughanour

consider --it is preferable touse a battery only-- at the best you may have a charger from bike bty and use a NiMh battery at 1.2 where the device still works

as regards charger you may try to make one using LM317 ( as suggested by Eng 1)
with voltage and current limits. it could perhaps be housed in a cell charger case adoptible to cigal ligher socket wired with bike battery

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Regards,
Sarma.

 

Thank you Sarma,
The information is very helpful.

Warm regards Lee

 

Thanks eng1,

Just what I needed. Warm regards Lee

 

Consider that with a linear regulator (ie - LM317) the current input and output are equal, and the entire voltage difference is across the regulator. P = V * I, and thus the power it dissipates is equal to the voltage difference, and the output current. Because of this, a linear regulator would be a TERRIBLE solution for this application, unless your mp3 player uses VERY little current and you don't care at ALL about how long the battery lasts on a charge.

18v - 1.5v = 16.5v. 1.5v/16.5v = 9%. That means that only 9% of the power coming from the battery will make it to the MP3 player, and the other 91% will be dissipated as heat by the regulator. A linear regulator like this, even with a heatsink attached, can't deal with very high power dissipation, and with a 16.5v drop across it, that's not much current to make it heat up a lot. If your mp3 player needs more than a fraction of an amp, I would steer clear of linear regulators.

A switching regulator is a much better solution; they can offer efficiencies of 75% or more. Based on your question I'm going to assume you probably don't want to try to build one just yet, because they can be rather complicated, but you might be able to find something commercially.

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EEgeek.net

 

I can't offer any chip names off the top of my head but you should be able to find a DC to DC switch mode down converter than requires very few external components and comes in a DIP package.

 

Thanks alot Evandude,
Saved me from over heat it sounds like. If you find the time could you direct my search for a switching regulator? Or maybe I should just buy rechargable batteries? I just liked the Idea of runing the two units from the same battery sence I use them all day anyway. Thanks again.

Warm regards Lee

 

Thanks for the input Sceadwian,
Every little bit helps the starving mind.

Warm regards, Lee

 

Or maybe I should just look at the top of the page at all those little ads. LOL!
Thanks again Guys.

Lee

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Lee

 

My MP3 player requires less than 100 mA. So the power dissipation would be about (18-1.5)*.1 = 1.65 W. A small TO220 heat-sink (20°C/W) is enough. As I wrote in my first post, this solution (LM317) is not efficient, but it's easy-to-do and suitable for low currents. Check with a multimeter how many mAs are required.

 

Hi
another thought
you may try out MC34063a canewll be used in stepdown mode. it is a 8 pin dip and minimum components are required. the inductor can be tried out locally itself

datasheet can be downloaded from www.alldatasheet.com


sarma

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Regards,
Sarma.

 

Interesting. I found a 25 pack of (lm317) at ebay. Might be worth messing around with this. Thanks again.

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Lee

 

Thanks Sarma,
More to think about is a good thing for me.

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Lee