October 24, 2013 at 11:50 am #20273JarvisParticipant
I used a Diavolino* in a project with an input voltage of 12v but I have an Ethernet shield. Everything runs fine when I use the usb/ttl cable but the system goes bonkers when I run it off the 12v line. By bonkers I mean that the power leds blink and it restarts itself. I suspect it’s because the usb port can provide 500ma but the 75L05 can only provide 150ma?
I have a 7805 but it would have to be ‘convinced’ to fit. Is there a replacement for the 7505 that can provide around 500ma?
*I think this board is brilliant. It’s simple and the price is right.
October 24, 2013 at 2:27 pm #21520Windell OskayKeymasterYes, that looks like the likely explanation; the standard Arduino ethernet shield can draw 150 mA on its own, so you probably need around 200 mA total capacity.
- This topic was modified 3 years ago by Windell Oskay.
I don’t know of any device that can burn that much power and can fit in the TO-92 location provided. If you pull a full 500 mA, then you would be asking the regulator would be to “burn up” (12 V – 5 V) * 500 mA = 3.5 W of power as heat– more than actually goes to power the device, and much more than a TO-92 package can normally handle. In situations like this, one normally wants to use a switching regulator that will convert the input power, rather than just burning up the majority of it.As far as similarly-sized linear regulators go, the L4931CZ50-AP will fit in the location provided, and handle 250 mA, but will be thermally limited and likely will not work. At 200 mA, you’d be dissipating 1.4 W as heat. The thermal resistance of that device is 200 degrees C/W (the *worst* I’ve ever seen in a TO-92!), so at 1.4 degrees, it would operate internally at 280 degrees C above ambient– well above its rating of 125 C.If you wanted to use the L4931CZ50-AP at 200 mA, one solution would be to provide it with a lower input voltage, allowing it to dissipate much less heat. Assuming 25 C ambient, you’d need to keep the thermal dissipation to 0.5 W, for a 100 degree rise. 0.5 W = 0.2 A * 2.5 V, so you’d want no more than 7.5 V input. To get there, you could drop from 12V to 7.5 V– a drop of at least 4.5 V –with a zener diode. For example, the 1N5340BG (a 6 V zener) might be a good choice.(This is very similar to the last example in our article about zener diodes, here: http://www.evilmadscientist.com/2012/basics-introduction-to-zener-diodes/ )So, there’s no easy, simple solution. It might be worth mounting an off-board 7805 (possibly with a heat sink), mounting a 7805 with extension leads, or using the zener trick, but none of these is particularly appealing. You could also make things a lot easier by reducing the input voltage, for example by using an external regulator to step the voltage down.October 28, 2013 at 1:22 pm #21521JarvisParticipant
I was wondering why no one had replied yet so I popped in to take a look. Sure enough, I wasn’t getting notified of replies. Oops.
Thank you for taking the time for a such a thoughtful reply. It was very informative.
I had a bunch of 7805s laying around so I rigged one up off board. It’s far from ideal but it seems to be working well.
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