Diagnose and rectify high current draw and implement loud headphone amp


Joined Sep 13, 2015
it should be 0V
that's not consistent with your chart just below.

It is inconceivable that a mcu can drain 90ma. One possibility could be that there are multiple low-impedance path from the higher voltage, through the clamping diodes, to the lower voltage that's powering the mcu.


Joined Jul 31, 2013
Im surprised the battery is damaged/faulty, it came with the development kit so it should be ok ?
You are right, it should be OK :). Parts can be faulty from the factory and that is why we have warranty.
I buy about 200 computer per year. I have about 2-3 faulty units even thought these are tested in the factory before shipping.. Just one of those things.

I might be wrong, but 200mA does not sound too high considering how many parts you have, and I will let the smarter people here help you make that lower.
You should still have ~3-4 hours of usage - what is the endurance of your aircraft?

Have fun.

Thread Starter


Joined Sep 9, 2016
Just an update - I have had success with the new battery :)

I had it playing audio with 200mA current draw for well over 30 min. without any significant drain. It could have ran for longer but it was bed time.

So after all that, it seems the problem was a faulty battery. Im not sure why it was faulty, it came with the blue tooth development kit so should have been ok.

I also managed to fix the issue where I was seeing 2.5V on the VDD_IO pin of the WT32i bluetooth module.

When the device was waking up from sleep, the dsPIC I2S module immediately began to output the I2S clock to the WT32 during the wake up process.

I have a 4 stage wake up process, so the power switch has to be pressed 4 times to wake the device up, otherwise it will go back to sleep.

When the device went back to sleep, the I2S clock would either be high or a low, depending on when exactly the dsPIC went to sleep.

Now I fully disable the dsPIC I2S module, the I2S clock remains low and the VDD_IO pin remains at 0V, as it should do.
Measuring current can also be problematic. The voltage drop of your meter could be significant.

You would probably want to look at the instantaneous product of V and I and average it and then normalize it to the battery voltage or just average the current.