Home › Evil Mad Scientist Forums › Other kit and product support › Looking for replacement for a 7505
- This topic has 2 replies, 2 voices, and was last updated 11 years ago by Jarvis.
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October 24, 2013 at 11:50 am #20273JarvisParticipant
Howdy!
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?
Thanks!
*I think this board is brilliant. It’s simple and the price is right.
- This topic was modified 6 years, 5 months ago by Windell Oskay.
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.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 #21521JarvisParticipantI 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.
thanks!
Jarvis
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