dito Posted September 10, 2005 Report Share Posted September 10, 2005 I dont understand this:Rx=(1,25+0,1)/IWhat is this 0,1? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 10, 2005 Report Share Posted September 10, 2005 i Dito,Welcome to our forum. ;DI don't know what the 0,1 is for, and I don't understand the function of the entire project.It doesn't have anything to prevent the batteries from over-charging!I just use a supply voltage a little higher than the voltage of the fully-charged battery, feed it through a simple resistor selected for 1/10th the mA/hr rating of the battery and remove the battery after an overnight charge or even after a few days. ;D Quote Link to comment Share on other sites More sharing options...
dito Posted September 13, 2005 Author Report Share Posted September 13, 2005 Oh yes! I understand now. It was a great solution. Thank's for all.I liked this is forum... Sorry my english, it's not my first language.P.S Can i only to use LM317? Didn't can i to use Lm7805?Electronics Lab 4 Ever! Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 13, 2005 Report Share Posted September 13, 2005 Hi Dito,An LM317 has a voltage drop of only 1.25V when used as a constant current source.A 7805 would have a 5V voltage drop if used as a constant current source so its current-programming resistor must be changed and the input voltage must be higher. With a higher input voltage then a 7805 would be hotter than an LM317. ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted September 13, 2005 Report Share Posted September 13, 2005 No, the LM317 has a voltage drop of 1.25 + the dropout voltage which will be between 2 to 3V so that's 3.25V to 4.25V in total, if you want very a low dropout current scource then couple of op-amps and a MOSFET are the solution. ;D Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 13, 2005 Report Share Posted September 13, 2005 Hi Alun,No problem for both regulators.The project http://www.electronics-lab.com/projects/power/037/index.html says to use a high input voltage. I don't know why. Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted September 13, 2005 Report Share Posted September 13, 2005 That's very wasteful, we could design a better one if dito wants. Quote Link to comment Share on other sites More sharing options...
ante Posted September 13, 2005 Report Share Posted September 13, 2005 May I suggest the MAX712/713 for this purpose? It can charge 1-16 cells at up to 4C! It needs only a few external components to make an excellent charger and it’s cheap or even free if you request a sample! ;D Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted September 13, 2005 Report Share Posted September 13, 2005 Good idea Ante but how easy is it to get hold of? Quote Link to comment Share on other sites More sharing options...
audioguru Posted September 14, 2005 Report Share Posted September 14, 2005 That's the way to do it!If you want to charge a battery then use a battery charger IC.It does everything correctly with its 2193 transistors. It can Fast Charge then when the battery is fully charged it automatically switches to a selectable trickle current.It can be operated linear or in switchmode, where it can take a higher voltage at low current and convert it to a lower voltage with a higher current for charging.It can even power a load while charging with its programmed current. ;D ;D Quote Link to comment Share on other sites More sharing options...
ante Posted September 14, 2005 Report Share Posted September 14, 2005 I like the chip because it’s small compared to a PCB with 2193 transistors on it! ;) Quote Link to comment Share on other sites More sharing options...
login
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.