[Ovmsdev] MAX942

Sat Jul 7 08:13:27 HKT 2012

I think the BSS138 and voltage divider circuit is safest. It is cheap, simple, and gets everything at the correct voltages. I've asked the China guys to go with that.

Regards, Mark

On 7 Jul, 2012, at 7:09 AM, William Petefish <william.petefish at gmail.com> wrote:

> 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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