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Author Topic: 6v to DC input jack seems to work fine?  (Read 4222 times)
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I've done a lot of reading up on DC input jack voltage, and how ideally this should be 7.4v or so minimum. I have a bunch of extra unused wall warts around, 6v, 7.5v, and 9v. Took my crappy analog multimeter out to test some of their voltages with the arduino powered up. I first compared it to known reference voltages from 9v and 1.5v batteries to get an idea of the measurement error of the meter. With that in mind I found to my dismay all my 9v adapters were putting out 13v or so, and even the 7.4v was putting out 11v. As stated, these readings were with the arduino powered up. Noticed the vreg got a bit warm with all these adapters. With these results I figured the 6v one would be my best bet, but of course murphy dictated this one actually put out precisely the advertised 6v even under no load. At any rate, the arduino seems to work fine with the 6v adapter, and the vreg is running nice and cool. I suspect I will probably run into problems with this, though? I measured the 5v rail on the arduino and it's within spec no matter which adapter I use. Unfortunately I'd probably need a real multimeter to get the kind of accuracy needed to figure that out, however.
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It can work, but to absolutely ensure the regulator will output a stable 5 VDC within a reasonable tolerance the input has to be at least 1.5 Volts above 5 VDC.

Edit:  And 7.5 VDC can be easily achieved with standard alkaline batteries and is much more common voltage level for wall-warts to output than 6.5 VDC.  Technically though, a nominally 6 VDC lead acid battery with a full or nearly full charge will have an actual voltage above 6 VDC, and probably above 6.5 VDC.  So depending upon the additional current draws, a nominally 8 VDC, 10 W or more solar cell could keep a battery at a high enough voltage for a dependably stable output.
« Last Edit: September 28, 2012, 01:26:15 pm by Far-seeker » Logged

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Two points:

1) if you look at those datasheets for the devices, many of them will go down to 2-ish volt or even 1-ish volt for newer devices. So they don't have to work at 5v. The arduino happens to be designed to work with 5v. But if you were to jump that regulator, you can go down much lower than 5v. BTW, they don't need that 5v to be precise either.

2) even in a non-LDO regulator, the minimum voltage drop over the regulator is substantially lower than 1.4v at low current levels. The LM317, for example, has two p-n junction on the output path. So it needs 1.3 - 1.4v to deliver full regulation at its rated current (0.5 - 1.5a). However, your mcu will drow substantially less than that - on the tune of 20ma on a good day. So the required voltage drop will be substantially less than 1.4v.

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1) if you look at those datasheets for the devices, many of them will go down to 2-ish volt or even 1-ish volt for newer devices. So they don't have to work at 5v. The arduino happens to be designed to work with 5v. But if you were to jump that regulator, you can go down much lower than 5v. BTW, they don't need that 5v to be precise either.

If you do want to be precise, the ATmegaV328P microcontroller on the Uno has an operational range of 1.8 to 5.5 VDC.  However, the operating voltage heavily influnces the clock speed.  For example, a 16 Mhz clock is only really possible between 4.0 and 5.5 VDC (see  Figure 29-145 of the datasheet).  That's why ATmegaV328P-based boards operating on 3.3 VDC are usually only 8 Mhz (half the speed of the official Uno).

Edit: Just to be clear I am in agreement with dhenry's second point.  smiley
« Last Edit: September 28, 2012, 01:58:03 pm by Far-seeker » Logged

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@OP, it's normal for [non-regulated] wallwarts to be outputting 4-5V
above their rated voltages for "small" load currents. The voltage only
drops to near the rated voltaqe for load currents near the max for the
given device.

So-called regulated wallwarts should always produce near the rated
output voltage, however.

A 6V non-regulated wallwart will probably work fine as long as the total
Arduino load currents are small.
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Thanks for the info. Yea, even when under load the voltages from the 7.5 and 9v wallwarts are around 11-13v. Not good. The current ratings on them vary from 450ma to 1A. I would have thought the 450ma one would have dropped to the spec voltage under load, but it didn't. The 6v one is the only one that appears to work, but I guess I'll find out for sure if there are any problems once I test it under a higher load (lcd screen, ethernet shield).
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Not good.

It does not matter, as your arduino will re-regulate that anyway. So as long as the voltage coming from your little adapters doesn't go above what the arduino onboard regulator expects (18v max rating).

So anything less than that and more than 6.5v functions the same to the arduino.
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Yes, I understand it will regulate it, but the vreg chip gets hot using anything except the 6v adapter. I'd ideally like to waste less power and generate less heat. Unfortunately, I have not been able to find an adapter that puts out 7.5v, despite having two which claim to. The 6v adapters I have work great, and the vreg never gets warm, but I have read on these forums that you ideally want at least 7.5v input to the dc barrel jack due to voltage drop from the regulator. Despite this, my 6v adapter seems to work so far, but I've yet to try it under higher load.
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the vreg chip gets hot using anything except the 6v adapter.

You have then unusually high current. The mcu doesn't consume much current. Something else must be drawing that current. The onboard regulator really isn't meant to power heavy loads.
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Hm, that's odd, not sure what would be using that much current. I was only running a LED blinking sketch when testing the adapters and checking the vreg heat. The LED was wired in series with a 1k ohm resister, and was only drawing ~5ma. With the 6v adapter the vreg ran cool, but with any of the others which input 11-13v, it got warm to hot.
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With the 6v adapter the vreg ran cool, but with any of the others which input 11-13v, it got warm to hot.

This is as expected.  A linear regulator converts the excess voltage into heat, so the greater the difference between the input voltage and the output of 5 VDC the more heat it will generate.
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with any of the others which input 11-13v, it got warm to hot.

Let's say that warm means temperature increase of 30c. The package has a thermal resistance of 100c/w (less than that actually). That means a power dissipation of 0.3w. At a voltage drop of 13v -> 5v, that means a current consumption of 40ma.

With the same math, assuming hot means temperature increase of 50c, you get a current consumption of 60ma.

I would have expected the load to be 20ma max and 10ma typical for your situation.
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To give you some sense, I have a board with the following load: 3 avrs, 1 pic, 1 1602 lcd, two hc132 oscillator (14Mhz + 70k), two hc04 oscillators (4Mhz + 12Mhz), 1 4-digit 7-segment led, 1 25Mhz oscillator, 1 14.7Mhz oscillator, one rgb led, 9 bright leds, 2 7-segment leds.

Powered from a 15v Verizon phone adapter, regulated down to 3.3v, the linear regulator (to220, no heatsink) is lukewarm. I think I measured my current consumption at one time and it was around 50ma.
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Somehow those numbers don't tally with your list of equipment. 50mA sounds more like the quiescent current for the listed devices. The 2 7 segment devices alone IF displaying anything would be drawing 5mA/seg typical minimum current, assuming you are displaying 6 1's... is .005 X 2 segments X 6 digits = 60mA. A typical backlight whether EL or LED will draw 100 mA.
15V - 3V3 = 11.7 X .05A = 0.585W so the 'basic setup' doing nothing causes the regulator to dissipate 585mW, add 100 mA from the LCD backlight and the power dissipated is now 1.755W and that is well past the point where a prudent engineer would have added a heat radiator of some type. My break point is 10 deg C + ambient above that and I make available a 50mm X 2mm piece of aluminum. Everyone is different and if what you have works for you then by all means go with it... The numbers don't seem right to me so... My $0.02's worth.

Bob
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I may be able to help with your math.

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X 6 digits = 60mA.

Really? What about multiplexing?

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A typical backlight whether EL or LED will draw 100 mA.

I must have an atypical lcd: its backlighting works fine with a 330ohm resistor to the 3.3v rail.
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