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I have just received my first microcontroller, an Atmega8. I will soon be making a voltage regulation circuit for it to run at 5V, but I don't have all the parts yet.

Instead, I have a 5V DC 1000mA adapter, and thought there's no reason it shouldn't work just as well. Is this a good/bad idea?

Furthermore, I checked and it is actually outputting 5.48V. So I checked the Atmega8 datasheet, and it says that it will operate between 4.5 - 5.5V. I am quite close to the upper limit, so am I taking a risk with this?

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  • \$\begingroup\$ On top of the voltage, i see another risk of inverse power supply. \$\endgroup\$ Jul 19, 2012 at 1:34
  • \$\begingroup\$ What do you mean? The polarity? I checked that, and I have it right. Also, I did accidentally put it in the wrong way, but it doesn't seem like anything happened thankfully. \$\endgroup\$
    – capcom
    Jul 19, 2012 at 13:21
  • \$\begingroup\$ Yes, i meant inverse polarity. It's better if you have a rectifier in place to protect your controller from inverse polarity. Failure Proof :D \$\endgroup\$ Jul 20, 2012 at 0:40
  • \$\begingroup\$ @JeeShen - polarity is a different thing from what's asked here. And with a standard DC plug I don't see how he could reverse that. In my answer I do suggest the diode, but for the voltage drop first, the reverse polarity protection is a free extra. \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 8:55
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    \$\begingroup\$ @JeeShen - Oh yeah? Then where's the upvote :-). I mean, thanks. \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 9:10

5 Answers 5

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Yes, you're taking risks. You have to take measurement error into account, and then the 5.48 V might as well be 5.58 V, which is beyond recommended operation conditions. That may drop a bit under load, but for a good regulator that will only be a couple tens of mV.

A series diode is a good solution, but the 1N4001 is not. At 1 A it may drop as much as 1 V and then you have the same problem at the lower end of the operating range.

I would suggest to use a Schottky diode, which has a lower drop than a regular PN-diode. The 1N5818 will drop maximum 0.55 V at 1 A, somewhat less at a lower current, so you'll end up nicely around 5.0 V.

The diode will also protect against accidental polarity reversals.

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  • \$\begingroup\$ Thanks for all your insight Steven. Now, my main question is that it seems I am allowed to max out at 6V on the Atmega8. So even if it does go up to 5.58 - 5.68 V, will it still be a problem? I am waiting on those 7805s to arrive to make a regulated supply as well. Thanks. \$\endgroup\$
    – capcom
    Jul 24, 2012 at 11:54
  • \$\begingroup\$ @capcom - The 6 V is Absolute Maximum Ratings. People often think that that's the maximum they can use to operate the device but that's not rue. Datasheet page 235 says: "Exposure to absolute maximum rating conditions for extended periods may affect device reliability." So 6 V is not OK. Stay within recommended operating conditions, that's 5.5 V. If you go higher than that the controller may not immediately suffer damage, but none of the datasheet's values will be reliable. \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 12:03
  • \$\begingroup\$ @capcom - what power supply are you going to use for the 7805? \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 12:04
  • \$\begingroup\$ I see. But it seems quite unlikely that it will hit 6V. I mean, it's a 5V power supply, giving a seemingly steady 5.48V under no load. It may fluctuate because it is unregulated, but from my limited experience I don't think it will hit the max limit. Thanks for referring to the data sheet. \$\endgroup\$
    – capcom
    Jul 24, 2012 at 13:08
  • \$\begingroup\$ For the 7805, I will find a wall wart above 9V, or use my modified power supply which I took out of a PC to feed it 12.5V. Does that sound alright? The only thing is, the 100uf capacitor I want to use is rated at 10V. So I think 12.5V will be risky, right? Thanks Steven. \$\endgroup\$
    – capcom
    Jul 24, 2012 at 13:10
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Put a standard 1N4001 diode in series with the power supply. You'll drop the voltage down to 4.9V.

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    \$\begingroup\$ Running the device near it's maximum rating is risky and you may very well have some ripple on the power supply that exceeds the absolute maximum rating. This diode hack is smart, but remember to use a decoupling cap. \$\endgroup\$
    – jippie
    Jul 19, 2012 at 7:00
  • \$\begingroup\$ Sounds like a cool solution, I'll look into it. Thanks! \$\endgroup\$
    – capcom
    Jul 19, 2012 at 13:19
  • \$\begingroup\$ @capcom - diode yes, 1N4001, no. See my answer. \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 9:43
  • \$\begingroup\$ @stevenvh: A shottky makes sense, with a 0.55V drop at 1A, if an Atmel drew an amp. They draw more like 20 milliamps. What's wrong with my answer? \$\endgroup\$ Jul 24, 2012 at 15:24
  • \$\begingroup\$ @insta - I was thinking he may connect other stuff as well, maybe I focused too much on the adapter's 1 A capability. Note that a diode's voltage drop isn't a fixed 600 mV, like you calculated. That's low, even for low currents. At hundreds of mA it can go to 1 V, and then you come close to the 4.5 V. A diode is a good idea, but I think a Schottky will bring you closer to 5.0 V. (about being "wrong": note that nobody downvoted your answer!) \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 15:31
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You should be fine. The adapter will most likely reduce the voltage with load and in the datasheet under absolute maximum ratings, on page 244, the upper voltage limit for supply voltage is 6 V. The limit of 5.5 V is the highest recommended voltage and in production, voltages higher than that shouldn't be used.

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No problem.

However in general CMOS margins may improve with higher voltage and speed improves too but at the expense of heat dissipation. People will overclock their CPUs by rasing the supply voltage a 1% at a time up to 5% to check performance and temperature rise.

So if it is Hot, it means you might not be able to run at maximum ambient of 85'C but ok at room temp. More important are noise spikes on the supply, so keep it clean and within spec. with close low ESR caps.

Low power designers prefer to run at minimum voltage and see if it still works when slightly slower for prop. delays. If using an unregulated wall transformer , it will run on the high side with a light load.

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    \$\begingroup\$ Overclockers wouldn't simply raise the core voltage; they (well, "we" perhaps) raise it so that higher frequency operation becomes stable. The performance benefits of a higher vcore alone, if any (I don't see why it'd help whatsoever) are probably too tiny to measure. \$\endgroup\$
    – exscape
    Jul 24, 2012 at 8:40
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    \$\begingroup\$ Tony mentions an unregulated wall-wart, but you definitely shouldn't use that, and and don't expect you do. The "No problem" is strange advice from someone with that much experience. Even at zero measurement error (see my answer) you have 20 mV, or 0.36 % headroom. Not acceptable. \$\endgroup\$
    – stevenvh
    Jul 24, 2012 at 8:48
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What ATmega8 you are using? is Atmega8 or Atmega8L? Normally microcontroller power supply is regulated, use of Transformer based adopter is risky, as output voltages of adopter is totally depends upon input voltage. If input voltage rises than its output also rises and vise varsa.

If you have power supply of voltages upto 9 to 12V than you may use Regulator IC 7805 for regulation of microcontroller mains power supply. 7805 will give accurately 5V.

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  • \$\begingroup\$ I have the Atmega8-16PU. I am waiting on the 7805s to arrive s that I can make the regulated 5V power supply. Thanks! \$\endgroup\$
    – capcom
    Jul 24, 2012 at 11:56

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