Author Topic: Urgent info  (Read 9121 times)

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Offline vaualbusTopic starter

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Urgent info
« on: April 12, 2013, 06:04:50 pm »
I've a motorola MC68000 micro. For the datasheet the maximum voltage is 5V +- 10% (5.25 V). If the supply voltage is just one milli volt higher (about 5.35). The cpu could be break?
 

Offline Mr Smiley

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Re: Urgent info
« Reply #1 on: April 12, 2013, 06:10:16 pm »
Isn't that 100mV  (0.1v)    :bullshit:

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Offline c4757p

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Re: Urgent info
« Reply #2 on: April 12, 2013, 06:13:40 pm »
If the supply voltage is just one milli volt higher

If it's really just 1mV higher I'm sure it will be fine.

Quote
(about 5.35).

That is not 1mV, that's 100mV, and much more likely to break something.
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Offline Galaxyrise

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Re: Urgent info
« Reply #3 on: April 12, 2013, 06:16:29 pm »
Probably depends on how long and how often it's overvolted, too, but... voltage regulator and done, right?
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Offline KJDS

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Re: Urgent info
« Reply #4 on: April 12, 2013, 06:20:48 pm »
errrrm, 5B +10% is 5.5V, or am I missing something?

Offline c4757p

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Re: Urgent info
« Reply #5 on: April 12, 2013, 06:23:05 pm »
Good point.
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Offline vaualbusTopic starter

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Re: Urgent info
« Reply #6 on: April 12, 2013, 06:33:02 pm »
Yes I'm wrong on the +-5 percent. Thanks for the help. Best regards Alberto.
 

Offline mariush

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Re: Urgent info
« Reply #7 on: April 12, 2013, 06:41:39 pm »
Read the datasheet.

It should have a table called "Absolute maximum specifications" or something like that, which should tell you how big of a voltage will the IC handle for short durations of time.

If you mean Freescale MC68000, then the datasheet is here : http://cache.freescale.com/files/32bit/doc/ref_manual/M68000UM.pdf?pspll=1  and page 157 has such table.  It says there maximum 7 volts. 
On page 163 you have there electrical characteristics, and the subtitle says 5v +/- 5%
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #8 on: April 12, 2013, 07:22:29 pm »
Thank a lot now I'm ok. The chip is not break. Despite I don't have a MC, I have a SCN68000, The maximum table still say +7.V.
Thank you for your help.
Also notice that this chip use a lor of current, more or less 1.5A.
Best regards Alberto Vaudagna. 
 

Online Dr. Frank

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Re: Urgent info
« Reply #9 on: April 12, 2013, 07:41:49 pm »
..
Also notice that this chip use a lor of current, more or less 1.5A.
Best regards Alberto Vaudagna.

1.5A for the µP seems to be much too high, Wikipedia states 1.35W, or so.
Check datasheet, 1.5W is specified as max. Power.
Maybe there's an ESD/EOS damage inside.

If the power consumption is a problem, the 68HC000 version should still be available.

Frank
« Last Edit: April 12, 2013, 07:53:57 pm by Dr. Frank »
 

Offline mariush

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Re: Urgent info
« Reply #10 on: April 12, 2013, 08:27:50 pm »
The CMOS should use little power, about 0.35 - 1 watts, a few hundred mA.

The non-cmos version could use up to 7 watts or, at 5v that would be over 1A.
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #11 on: April 12, 2013, 09:52:05 pm »
So is normal (like I thought).
At first before to read the datasheet eletrical information I touched my little 7805 that use for the 5V power(I know is not the best solution of all, I'm working on it) and it was too hot. Immidantly I thought that the chip sould be shorted but that fortunatly I read the datasheet!
Best regards Alberto Vaudagna.
 

Online Dr. Frank

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Re: Urgent info
« Reply #12 on: April 13, 2013, 07:49:28 am »
So is normal (like I thought).
At first before to read the datasheet eletrical information I touched my little 7805 that use for the 5V power(I know is not the best solution of all, I'm working on it) and it was too hot. Immidantly I thought that the chip sould be shorted but that fortunatly I read the datasheet!
Best regards Alberto Vaudagna.

Once again, please read the datasheet carefully!

In the SCN68000 datasheet, a max. current of 1.5A is mentionend in a footnote (p51/59), but that's obviously peak current, not average supply current.
On the next page, an average Ptot of 1.3-1.8W is specified, depending on T (ambient) and clock freq.
As the package thermal resistance is around 40K/W, a P dissipation of 3W or more would destroy the chip over time by overheating. Malfunction will happen from about 2W onwards.

I don't know, where you folks have read about (average) 7.5W dissipation, but all the datasheets definitely tell a different story.
There's also the question, if the processor itself consumes the majority of the current, or the rest of the circuitry.

Therefore, Please measure, how hot the processor case gets.. if its over 60..70°C, something might be wrong probably.
Frank
« Last Edit: April 13, 2013, 07:58:03 am by Dr. Frank »
 

Offline SeanB

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Re: Urgent info
« Reply #13 on: April 13, 2013, 08:24:10 am »
I have had TTL survive quite large overvoltages, no problem. Would not recommend it, but most have a minimum breakdown voltage of 7V, as a lower level of voltage that will damage a portion of parts. Some might survive 10V, some might not. All will survive 7V though. With more complex devices though they will probably only survive 8V or less. There have been many simple TTL designs where they are operated off a 4 pack of cells with no problem, they do eventually stop running at around 3V.
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #14 on: April 13, 2013, 03:40:59 pm »
So, the case temeparture is not hot but it could  be demage?
Thanks for everybody.
Best regards Alberto Vaudagna.
« Last Edit: April 13, 2013, 03:47:27 pm by vaualbus »
 

Online Dr. Frank

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Re: Urgent info
« Reply #15 on: April 13, 2013, 06:35:29 pm »
So, the case temeparture is not hot but it could  be demage?
Thanks for everybody.
Best regards Alberto Vaudagna.

No, if the case is not too hot, the chip inside will also not be too hot, and therfore, there will also be no damage to the chip.

You stated, that the whole circuitry, which we others cannot "see",  consumes so much current, that your external 7805 gets too hot.
Therefore, there was the question, where the power goes; obviously not to the µP.

Frank
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #16 on: April 13, 2013, 08:48:40 pm »
Ok thanks al lot.
PS: On the circuir there is only the microprocessor, a 8MHz quartz and the 74HC04 and a green led. So I suppose that the power go to the micro. Isn't it right?
Best regards Alberto.
 

Online Dr. Frank

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Re: Urgent info
« Reply #17 on: April 13, 2013, 09:27:08 pm »
Ok thanks al lot.
PS: On the circuir there is only the microprocessor, a 8MHz quartz and the 74HC04 and a green led. So I suppose that the power go to the micro. Isn't it right?
Best regards Alberto.

From your description, that's right.
On 8 MHz, 1.4W is dissipated by the µP, that makes about 280mAn of current, which you could check at your PSU.

Frank
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #18 on: April 13, 2013, 10:48:48 pm »
Ok, but why the 7805 begin hot? It rated for 1.5A.
Now I'll be going to use a pc power supply that should resolve the problem.
 

Offline mariush

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Re: Urgent info
« Reply #19 on: April 13, 2013, 11:14:05 pm »
The CMOS version of the microcontroller uses up to 0.35 watts.

The older HMOS version uses up to 1.35 w according to Wikipedia :
Quote
The original HMOS MC68000 consumed around 1.35 watts at an ambient temperature of 25 °C, regardless of clock speed, while the MC68HC000 consumed only 0.13 watts at 8 MHz and 0.38 watts at 20 MHz. (Unlike CMOS circuits, HMOS still draws power when idle, so power consumption varies little with clock rate.) Apple selected the 68HC000 for use in the Macintosh Portable.

I was the one that said 7 watts by accident in one of my posts above, I apologize for that. I meant to say 7 volts absolute maximum voltage back then.

So even if your microcontroller uses 1.35 watts, that's about 1.35/5  = 0.27A

The 7805 needs at least 2v above the 5v to output a stable 5v, and the difference between the input voltage and output voltage is dissipated as heat.

I went through the posts and I couldn't see any note about the input voltage you use... if you use 7.5v, then your 7805 will dissipate approximately (7.5v - 5v ) x 0.27A =  0.675 watts, which should be easily dissipated by heatsink.

but if you power the 7805 from 12v, then you have  (12v-5v)x0.27a = 1.89 watts which would make it normal to get the heatsink a bit hot.

Now there's another note to make... these regulators can actually handle quite a bit of heat. It's normal for the heatsink and regulator to be about 40-50c hot when running, even with such little current. It might feel hot to you, but the 7805 can run up to 80-100c for long time without problems.

 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #20 on: April 13, 2013, 11:51:35 pm »
I'm use 13.3 v. So it is the reson of the hot. Bute I think that the computer psu is the best solution.
 

Offline vaualbusTopic starter

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Re: Urgent info
« Reply #21 on: April 14, 2013, 04:44:38 pm »
Now that I'm back home where I get the u I've done some mesurament. The entire circuit use 0.75A (suppose all go the u) and the temperature is between 50 - 60°.
I think the chip isn't good at all it is a quite lot. I'm right.
PS: I've arleady oreder another MC68000 to be sure.
 


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