I am working on a project where 24v DC will be readily available. I'd like to power the arduino from this, but I don't know of a simple way. None of the linear regs I'm familiar with (i.e. 7805 & similar) can handle such a big drop, and most of the switching regs I'm familiar with are a touch more complicated than I'd like (lots of caps, an inductor, etc.)
Is there an easy solution I'm missing? If it involves more than 3 or 4 parts and $3 or so, I'll probably just use a wall wart, but I'd like to keep this clean and run everything from the 24v supply, if possible.
Well first you need to define how much total +5vdc current your application will require. The Arduino board itself will probably draw less then 40ma, so there are lots of simple cheap ways at that current level to use. How much total current is required (I/O pin loads, external components and other loads) will determine the best solution.
By the way a simple 7805 will work with 24vdc input, it's just that how much current is being drawn from it that will determine how much heat sinking would be required. Dropping down the 24vdc via a series power resistor before feeding the 7805 input is another method. Again the total current requirement is the key unknown.
I'll need to power the arduino, plus 3 PWM pins that will have pretty low load (few mA at best), and a few pins switching relays on and off. I would guess it's a pretty insignificant load, overall - the Arduino basically exists to turn on and off (or dim) things powered directly from the 24v source.
I'll have a look at a 7805 datasheet, but how does one determine power dissipation and heatsinking requirements? I know some datasheets have curves showing the max wattage a given package can dissipate, would I just assume I need heatsinking if above the curve?
but how does one determine power dissipation and heat-sinking requirements?
I use my finger If I can keep my finger on it for 5+ seconds or so, it's good to go.
Seriously, it can get pretty mathematical (more then I like to tackle) to try and calculate the heat sink size and difficult to get heat sink specifications, etc.
It sounds like you won't be drawing much 5vdc current so just find a standard TO-220 size heat-sink for the 7805 and see how hot it gets. There will be no damage as the regulator will just shutdown on over temp or over current due to self protection circuits.
You can spread out the voltage drop by pre regulating down to 12 or 9V before you put it into the final 5V regulator. This way any power dissipation is spread through two or more regulators.
I much prefer this to a series resistor because it keeps the voltage into the regulator more stable and less prone to noise.
Thanks, Mike. I was figuring there was some way to use multiple regulators to spread the dissipation, and it makes sense that essentially running two in series would help when you had a big voltage drop.
A long time ago in a galaxy far away, I saw someone mention on a forum that if you had more current than a single reg could handle, you could just put multiples in parallel. Is that equally valid?
you could just put multiples in parallel. Is that equally valid?
If you look at the application data in some manufactures 7805 type regulators data sheets, they show a couple of ways to do increased current above the nominal 1 amp regulator range. I think using parallel regulators required a series diode on each regulator output to help share the load, but it does degrade the regulation specs (and voltage due to a diode drop) somewhat, so is not the preferred way. More often recommended is to add a NPN power transistor that uses the regulator's output to control the NPN output voltage but with greater current passing.
There are 3 amp and even 5 amp, TO-3 cased, 5vdc linear regulators, fixed and adjustable. LM-350 seems to ring a bell.
If I can keep my finger on it for 5+ seconds or so, it's good to go.