Jump to content
Electronics-Lab.com Community

Reducing 6.4V 2,700mah for a pda (ext battery)


explorador
 Share

Recommended Posts

Hi, I'm new here and found a post (in this forum) about using diodes to reduce the current but couldn't understand it quite well as others suggested other ways and putting some capacitors on the system (  ???  )

I have a PDA which uses an external folding keyboard. I use it to write when I travel (much better than a laptop). I also have a video camera battery (6.4V 2,700mah) that really lasts... so, I was thinking about using it as a external power source for my pda.

The thing is the PDA maximum external power (without frying it) is 5V. So, 6.4/2,700mah would fry it right? I tested the diode thing, and 5 of them reduce the 6.4 to almost 5.1V, but I know there are more things involved than this... Would this work? I guess it could heat up the system and burn something.

As far as I remember, volts and amps do matter... as is not the same as 5V with 850mah compared to 5V with 2A.

There is an external adapter (AC to DC) for this model of PDA and it says 5.9V and 2.0 AMPS. Is it 2,700mah the same as 2.7A?

I would apreciate any help, thanks

Link to comment
Share on other sites

Hi Explorador,
Welcome to our forum. ;D
First you say its max voltage is 5.0V, then you say its adapter is 5.9V. Which is correct?

What chemistry type is the battery? Its voltage isn't Ni-Cad or Ni-MH which would be 6.0V for 5 cells. It isn't Lithium which would be 7.2V for 2 cells. Is it lead-acid?

How did you connect 5 diodes to the battery? As a short? It wastes a lot of power. Didn't the diodes get very hot and burn out?

If you connect a single diode in series with a 6.4V battery then the voltage at the output of the diode is about 5.4V at 2A which would be fine.

2700mA is 2.7A of current. 2700mA/h is how long the charge will supply one-tenth of the rated current, 270mA for 10 hours. It might last only half an hour with a huge load of 2.7A.

Link to comment
Share on other sites

Thanks. Sorry, I have found dif informations about this. Serious and reliable sources mention 6V as the maximum. Some users tested 7V with no problems (but there wasn't enough info for me to trust).

I don't have the external adaptor (I found it on the web and it reported 5.9V), it is the model the catalog mentions (I don't have it), My PDA charges when is on the cradle. The PDA can be charged with several voltages, from 3.7 to 5 (what I tested), nothing more to avoid damaging it. Today I found a topic on the web stating 7 volts when charging but I don't want to take risks...

The video camera battery is 6V NI CAD. Today I charged it (full) and the tester reported 6.9V. I don't really know about how many cells the battery have... :(

Today I connected 4 diodes in series and it reported a drop from 6.9 to 5.8. Didn´t try it with the PDA as I'm still researching.

audioguru, the mah is not like potency? at 5V, is it too much?

Im kinda confused, as I tried another battery of 3.7 850mah to charge it, and it didn't last long... I understand, but as the voltage was getting lower and lower, the status indicator reported first "charging", then "AC".

Link to comment
Share on other sites

A diode in series is 0.3V with hardly any current, just a voltmeter as its load. With a few hundred milliamps its voltage drop is about 0.7V. With a 2A load it is about 1V.

I thought you connected 5 diodes in series then connected them across the 6.4V battery to drop the voltage to 5V and to make lots of smoke. ;D

Link to comment
Share on other sites

Audioguru, Ante, Thanks guys!

I tried with 4 diodes in series and, even that the multimeter reported 5.2+V, didn´t work to charge the PDA (it has a LED indicating "charing"...), I tired with less diodes and it worked, but didn't charge while using the PDA.

So, I used only ONE diode and it works, both on and off, it charges the device with no problem. I'm still kinda confused as the multimeter reports even more volts than when using a universal charger (4V) and didn't turn on the thing.

But anyway, now it is working. The only diode I'm using gets a little bit warm, so I will do what you say, putting a fuse to protect the system and a heat disipator to solve the heat thing and test it.

Thanks a lot guys.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

×
  • Create New...