I have a 6 Volts battery with 3.2 Amps, that is currently at 6.36 Volts.
I also have a DC borduino, that is connected to a bread board. Since I will be having other items on the bread board to power up, i want to power up the boarduino through its pins.
I also have a voltage regulator that outputs 5 volts.
What i have done is put a capacitor in the negative and positive strips. Then i placed the voltage regulator on the bread board and connected the middle pin to ground. I connected the batteries positive end to the input pin, and sent the output on to the positive strip. The boraduino is now connected using one 5 Volt pin and one ground pin.
What i want to know is if everything i did makes sense. The only thing that does not make sense to me is the capacitor. If the capacitor is good please explain why it should be there, i dont understand. Thanks for the help!
The capacitors recommended for voltage regulators both on their inputs and output pins are to prevent the regulator from oscillations under certain conditions and to help filter high frequency noise. For battery input an input capacitor is probably not really required and the micro board has a cap on the +5vdc to ground anyway so an additonal cap on the regulator's output is probably not absolutly required. If you can find a copy of a data sheet for your regulator it will explain better the purpose and recommended value for these input and output capacitors.
More important to get correct is the type of regulator you use. Your drawing doesn't state the type number but if it is a generic 7805 type linear regulator your battery voltage would not be high enough to insure that it remains in good regulation, typically they require a minimum of 7.5vdc on their input pin. There are low drop out type 5vdc regulators(LDO) that require less input voltage (down to 5.5vdc) on there input pin and still remain in good regulation. I don't have a part number at hand to recommend but maybe someone can jump in.
With a typical 0.6V forward voltage, that gives you a 5.4V supply. You can bridge that with a capacitor for noise.
Anyone here want to shoot down this approach? Is 5.4V still too high, or the diode too wasteful?
By the way, you may already know this, but you don't have "a 6V battery with 3.2 Amps," you have a 6V battery rated for 3.2 Amp-Hours. That means it can sustain 3.2 Amps for 1 Hour, or 1 Amp for 3.2 Hours, or any linear relationship like that, before it starts dropping voltage significantly.
The problem with a diode is that the voltage drop is dependent on the current. While a 1n40xx diode drops a nice 0.7V or so near 1A, the drop might be low enough at the low currents that the arduino draws to reach above a safe voltage...
