Arduino + Sabertooth + 2x12V motors powering solution

Hello guys! I'm a new Arduino user and looking on a good way to power my new robot.

I have:

Arduino Uno USB board Sabertooth 2x5 motor controller 2x 12V motor (2A max) LCD shield (4x20 from DFRrobot) PS2 Wireless Controller Some sensors... 2 Batteries, both NiMH (7.2V and 12V)

And here is what I intend to do. Is there anything bad in this setup? I'm not sure the 5V/100ma from the sabertooth is enough to power arduino with LCD,so I use a separate battery. This will also isolate both power supplies. And the 7.2V battery should not make the regulator overheat I hope. If any advice/tip can be given, I will gladly take them :) If it seems good, I'll try it as as soon as possible! Thank you!

I'd not use 7.2 battery at all, just power up everything from 12V. 7.2 is barely enough to power arduino, and if under load voltage dips down it'd only create a "hard to find" troubles. Plus , with two battery it is impossible to keep charging cycles synchronous.

Won't 12V make the regulator overheat if I use too much current?

P(regulator) = ("unused" voltage * used current) right?

So... (12-5) * 200ma = 1.4W

200ma is an approximate of my circuit. May be more or less.

Yes, could be. To save a power I'd use on of those: +12V to +5V step down DC/DC converter. When I pick up one at convenience store for $3 and open it up, there was a surprise: 34064 switching high efficient buck regulator.

FEATURES ? Supply Voltage: 3V - 36V ? Current Limiting ? Output Switch Current to 1.5A ? Adjustable Output Voltage ? Operation frequency up to 110KHz ? Low Quiescent Current ? Precision 2% Reference

Most likely you would find same IC in yours.

Seems like a good idea.

But I'm not SO short on the power with 7.2V NiMH? It has 6 cells of 1.25V which when discharged should drop to 1V. So I still have 6V when my batery is discharged.

The regulator is a NCP117, which has a 1.10V max (0.95V typical) voltage @ 100mA and slightly more if more current (1.2V max @ 800mA). So in the worst case ( battery dead and 800mA current draw), I'd drop to 6V-1.2 = 4.8V which is within the 0.5V typical margin.

Is there something I don't understand? I just don't wanna buy stuff to buy stuff. I'd like to make it work with what I already have as much as possible.

thanks again for the info though!

You math it looks o'k to me. But still, NIMH has a memory effect, it means battery has to be fully discharged before you start charging process. What you gonna to do, when 12V is "0" and 7.2 has 20-50% capacity? Leave it uncharged till next time 12V would require charging? What if it dyed earlier? One car - one battery. If power dissipation concern with supplying +12 to +5 you can put a resistors in series, to drop some 3-5 volts before it enters arduino regulator. It's not efficient, nevertheless it's small losses compare main motors consuption

That seems like a good idea. But how do I choose the resistor value? And I guess I'll need some 1W/2W too.

My thoughts would be:

R = V/I = 12/(current draw of system)

so... say at 200ma,

R = 12/.2 = 60 ohms P = RI2 = 60 * .2 * .2 = 2.4W

I hope I'm wrong...

You need to drop 5V , leaving 7 for arduino: 12 - 7 = 5. R = 5/ 0.2 = 25 OHm, P = I^2 * R = 0.04 * 25 = 1 W.

Oh, it makes much more sense now. Thank you very much for your help! :)

Have a good day