explorador
- Oct 3, 2006
- 13
- Joined
- Oct 3, 2006
- Messages
- 13
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
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