power supply
I'm trying to convert mains AC to DC while dissipating as little heat as possible. Some ripple is tolerable if it cuts down on heat so I'm not sure if I need the regulator. I also don't know how well the values I've chosen for the various components will work together. Perhaps there's even a much easier way to go about it.
Finding parts is another concern. For instance, can a 22 volt regulator can even be bought?
Any helpful advice is greatly appreciated.
Here's the circuit diagram:
They should work, and as the info on the motor page states, you should use a 2 cell Lithium-Polymer battery.
A good calculator can be found here, it will let you know how much you can lift, run for, etc and should help you find the best propellers and best capacity and required discharge rating to use.
will it work? please feel free to point out any mistakes
The current limit potentiometer just isn't going to work - let's say you want to limit current to 1 amp; this requires a resistance of 1.25 ohms and the power rating of a 150th part of your potentiometer needs to be 1.25 watts. Given that pots have a linear power rating throughout the length of travel, the pot would need to be rated at 150 x 1.25 watts i.e. nearly 190 watts. Sure it will never see 190 watts but right at the end where you might want to limit current to 1 amp, that small fraction of pot track will see 1.25 watts thus the whole pot would need to be rated at 190 watts.
I would also recommend, that from each adjust pin to ground, you put a 10 uF capacitor to reduce the output noise of the rather ancient and noisy LM317 regulators. I'm assuming that you will also need fairly big heatsinks.
If you want a better design try this site called LEARNING ELECTRONICS. On that page they have a design for a fully adjustable power supply that uses an LM317 at the heart of it: -
Best Answer
I think it will barely work as shown.
22V is an oddball voltage. Use an adjustable regulator where the output voltage is set by the ratio of two resistors.
Peak input voltage is \$\sqrt2 * 18\$ minus diode losses (~1.5 volts), so about 24 volts. So if your regulator works with 2V or less of dropout, you will be OK. (This is kind of a tight dropout spec, not impossible, but not every regulator you find will be OK. Micrel MIC29303 appears to be OK.)
Now, how much current can you draw? Assuming 60 cycle power, let's say we can allow the supply to droop 1V in the 1/120 seconds between voltage peaks. Then the voltage regulator will have 1 volt headroom. \$I = C * {dv\over dt}\$ so if \${dv\over dt}\$ is \$1\over1/120\$, then \$I = 120 * C\$. For a current of 0.5A, then, choose C of about 4000 μF.
All of this design is cutting pretty close: If the line voltage is just a little high, you will exceed the regulator's input voltage limit. If the line voltage is just a little low, you will drop out of regulation. I suggest you find a transformer with a little higher output and use an LM317 regulator instead of a low-dropout type. Additional headroom will allow a smaller filter capacitor as well.