Low voltage when using chipKITâ„¢ WiFi Shield

Created Thu, 17 Oct 2013 21:15:02 +0000 by PICedh

PICedh

Thu, 17 Oct 2013 21:15:02 +0000

HI

When I use the chipKITâ„¢ WiFi Shield with my ChipKit MAX32, the output voltage delivered on the 5V pin by the MAX32 decreases to less than 4V So the other components (like I2C) connected to the 5V pin are disturbed.

The Max32 input voltage is 9V (Alcaline battery) as recommended (between 7v and 15V).

Is it normal ? should I increase the inpout voltage of the Max32?

regards

majenko

Thu, 17 Oct 2013 22:06:10 +0000

You shouldn't increase the voltage, but throwing away that poxy 9V battery is an absolute must.

Those batteries are worse than useless. The have a fraction of the power of other batteries, like AA batteries.

I would recommend using a cheap wall-wart, or a 6-cell AA battery pack.

PICedh

Fri, 18 Oct 2013 09:55:35 +0000

Ok Thanks a lot I didn't known that 9V batteries are so bad.

How many mAh do you suggest for Chipkit, I have Lipo 3s 350 mAh, is it enought ?

majenko

Fri, 18 Oct 2013 11:05:34 +0000

That all depends on how much power is needed. I don't know the specs of the WiFi shield, but I do know that anything with a radio transmitter in it will be quite power hungry.

The LiPo will be able to provide more current over a shorter period than the 9V battery, as the LiPo has a lower internal resistance. However, don't expect it to last very long. A 350mAh battery will provide 350mA for 1 hour, or 35mA for 10 hours, etc.

But then of course the inherent inefficiency of the linear regulators in use will drain more power from the battery than the actual electronics will.

For instance, with a 9V battery, and a target voltage of 3.3V, you have to just throw away 5.7V - that's 63% of the battery power completely wasted. For 8.4V 3-cell LiPo it's fractionally better (8.4-3.3=5.1; 5.1/8.4*100=61%), but not much.

And of course the amount of current you are drawing has an effect as well. You can think of a linear voltage regulator as basically an automatically varying resistor, which changes its resistance depending on the current to keep the voltage drop across it constant. And current flowing through a resistor generates heat. P=I²R, or I·V. So, if you are drawing 150mA through the regulator chain, then at a 5.7V voltage drop you are wasting (5.7*0.15=) 855mW of pure heat. Increase that to 300mA and you double the power wastage to 1.71W. Ouch.

When you have to run from batteries there are a few things you really should do:

1. Keep the battery voltage as close to the target voltage (plus regulator dropout voltages) as possible to reduce losses.
2. Consider replacing one or more of the linear regulators with switching "buck" regulators.

For 1) the ideal voltage would be 5V (there are two regulators - 5V followed by 3.3V) plus the dropout voltage of the 5V regulator (I think that's around 1.6V or something), so 6.6V. Using a 2-cell 7.2V LiPo / Li-Ion would be a good arrangement, but still not perfect as you have a total drop-to-target of 3.9V (54% loss)

For 2) you could use a 5V switching regulator (google for 5V UBEC) to step the battery voltage down to 5V, then bypass the on-board 5V regulator completely (feed the power into the +5V pin, or the USB port), which just leaves the 1.7V voltage drop for the 3.3V regulator, which is hardly anything on the scale of things. You're then not at the mercy of your battery voltage. You could run off, say, 3-cell 8.4V, with a 90% efficient UBEC to 5V, then you would have just (1.7/8.4*100+10=) 30% loss.

Even better would be if there are no 5V devices in your setup at all - then you could theoretically go direct to 3.3V with the buck regulator and get only 10%-15% loss (depending on buck regulator).

By the way - your typical alkaline 9V PP3 is around 565mAh. Your typical AA battery is more like 1800-2600mAh.

PICedh

Fri, 18 Oct 2013 12:16:49 +0000

Many thanks again for all these good advices. I will try an see

Eric