[Ovmsdev] MAX942

[William Petefish]

Danger Will Robinson!

I have done exactly what is being described before. My advice, don't do it.
For reliable operation the 3.3v parts need to be isolated from the 5v
parts. You run a very real risk of burning out the circuit if you power a
3.3v device from 5v or even attempt comms without a level translator.
(found this out the hard way with a rather expensive PCB.) At the very
least you'll shorten the life on the GSM unit by running over voltage.

William

On Fri, Jul 6, 2012 at 1:39 PM, HONDA S-2000 <s2000 at sounds.wa.com> wrote:

>
> On Jul 6, 2012, at 06:04, Mark Webb-Johnson wrote:
>
>> The SIM908 is powered at 3.3V, and the PIC18F2685 at 5.0V.
>>
>> That would seem to say that the SIM908 will use 2.7V as it's minimum
>> output voltage, and the PIC18F2685 will accept anything above 2.0V. So,
>> SIM908 TX -> PIC RX should be ok.
>>
>> For the other way, the PIC18F2685 will transmit at VDD-0.7 - 4.3V, and
>> the SIM908 will accept anything above 2.4V.
>>
>> From that description, the entire MAX942 circuit is not required. The RX
>> and TX pins of the SIM908 and PIC18F2685 can be directly connected to each
>> other.
>>
>> What do you think? Perhaps it is time to find my old breadboard and try
>> it...
>>
>
> I think it should work perfectly fine from SIM908 to PIC with just a trace.
>
> However, the SIM908 probably has an absolute maximum voltage specification
> that would not allow anything above 3.3 V on any pin. Thus, the other way,
> from PIC to SIM908, would not work with just a simple trace. Look around
> your data sheet for that limit. The existing diode (not shown in your
> schematic, but mentioned in email), is surely still necessary to protect
> the SIM908 from too much voltage. If there is a resistor in there as well
> as the diode, then that's probably necessary as well, but I can't remember
> what's typical. If you have the schematic for the other board, drop me a
> link and I'll take a look.
>
>
> In general, a device cannot work when one of its pins has a higher voltage
> than its own power input pin. If you connect a higher voltage, sometimes
> current will flow backwards through the chip, out the power pin, and raise
> your 3.3 V supply to 4 V or more. This will cause your regulator to heat up
> and possibly create other symptoms. It can be especially bad if other chips
> are not rated for 4 V power.
>
> I had a board where an unused chip was exposed to improper voltages, and
> it caused problems for quite a while. I never thought to check that chip.
> It was only when I removed the chip and noticed that all the random
> problems went away that I realized one of its pins was seeing a voltage
> higher than the supply. Had I checked the 3.3 V supply rail before, I would
> have seen that it sometimes went above 4 V, and that would have been a
> clue, but I didn't think of that until I had practically solved the problem.
>
>
> Brian
>
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