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Author Topic: Arduino chip on a breadboard - can't get it to work  (Read 1488 times)
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I'm trying to get the ATMega chip straight from my Arduino Uno to work on a breadboard.  So far there have been no signs of life.  Attached is a photo which shows how the chip is wired.

The chip has the 'blink' test program loaded on it.  I can take this chip out of the breadboard and put it back into the Arduino Uno and it blinks pin 13 perfectly.  If I put it back in the breadboard there is no response.

Things I've already tried..
- The supply voltage is correct at 5.03 volts.
- I have three 16mhz crystals and I've tried another.  I've also tried another set of capacitors from the crystal to ground.
- I completely unassembled the circuit and rebuilt it ten rows further down on the breadboard, in case there is a bad connection somewhere - still no response.
- The led and resistor definitely work as the same ones when connected between pin 13 and ground on the arduino blink.

The wiring looks correct - I've triple checked it - but maybe someone can spot something that I'm missing?


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Looks pretty good. Doesn't sound like anything is fried as it still works back in the Arduino board. Are the various crystals you tried identical? Are you sure you have the right kind of crystal? Is it a parallel resonance type (not series resonance)? What size are the crystal loading caps?
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..what is missing on the first glance are the decoupling capacitors -100nF Vcc, 100nF Vcc, 100nF Vanalog, and 10uF between the rails. And on the second glance - do you have a connection between the lefthandside (top on the photo) and righthandside (bottom on the photo) power rails (red bus with red bus and blue bus with blue bus)??
P.
« Last Edit: May 17, 2011, 07:54:16 pm by pito » Logged

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Thanks for your reply Jack.  I have it blinking on the board in front of me now!

Your suggestion to focus on the crystal and the capacitors turned out to be correct, and the answer was in the capacitors I was using.

I didn't realise the capacitor choice was so sensitive - the original capacitors (in the photo) are small blue ones with just '.1' printed on them.  I has also tried other ones with '221' on them.  After reading your post I also tried another set with '220' on them - none of these worked.

Finally the ones that work have '22' (underlined) on them.  A photo of them all, including the correct ones on the right is attached.

Again, thanks for the suggestion.


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Hey good deal! 22pF is probably in the right ballpark. Not good engineering practice, but it might also run without any capacitors. The ".1" caps are probably 0.1uF which is 100,000pF, so just a wee bit large.  Not entirely sure about the 220 and 221 ... both could be 220pF, ten times too big, and no doubt enough to stop the oscillator.
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In general, when capacitors have a decimal point as part of the value then the unit is microfarads and if there is no decimal point then the unit is picofarads.  So .1 ==> .1 microFarad or 100000 picoFarads which is way way too high.  The 221 and 220 devices are in picoFarads, about 10x too large.

Don

Edit: I only use two fingers when I type....
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That's interesting and useful thanks.

According to my reference here (in a book that's about 10 years old, describing the IEC Code) when you have three figures on a capacitor the first two figures are the value and the third is a multiplier, or number of zeros to add, to get the value in pf.

So my logic went 221 - that's 22 pf multiplied by 1 - and should be 22pf.  220 didn't make much sense as that should work out to zero.

Luckily trial and error found the solution.

And Jack, yes, I've just tried it and does work without any caps.
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Quote
...first two figures are the value and the third is a multiplier...
That works for color coded resistors.  When there are digits then the third value is the number of zeroes.

Don
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So my logic went 221 - that's 22 pf multiplied by 1 - and should be 22pf.  220 didn't make much sense as that should work out to zero.

I'm not sure if what you read it true, but for resistors:
The first two numbers are the actual number and the third is the number of zeros you add. So for 221, you have 22 plus one zero, so 220. For 220, it's 22 plus zero zeros, so just 22.

Good job on finding the problem smiley
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From reply # 8:
Quote
The first two numbers are the actual number and the third is the number of zeros you add. So for 221, you have 22 plus one zero, so 220. For 220, it's 22 plus zero zeros, so just 22.

And this differs from the answer in reply #7 by ....?

Don
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