Battery operation test completed

A project log for sensor node

a learning project to gain experience with RF design and the ARM Cortex M0+ platform

Hannes HochreinerHannes Hochreiner 04/17/2017 at 09:344 Comments

I finally ran the last test I wanted to do with the current version of the board: the battery operation test. Since it was unclear to me from the documentation of the sensor breakout board whether it would operate normally below 3.3 V, I gave it a try. I powered the board from the CR2032, which my multi-meter claimed to be 3.05V, and looked at the measurement results in the debugger. Everything, seemed to work fine. The next step will be to order a new batch of boards with the correct pin assignment for the radio chip. I am really looking forward to seeing the full assembly work.

Background on the breakout board power confusion:

On the Adafruit description page of the breakout board it claims to work from 3-5 volts input and both 3.3V and 5V logic. However, I also had a look at the circuit diagram and the model of regulator chip (Microchip MIC5225-3.3) that has been used. This chip is rated to output 3.3V with a dropout voltage of typically about 300mV at full load. My confusion was mainly that I thought that below about 3.3V (actually rather 3.6V) it would not work. This did not fit with the documentation and also did not reflect what I experienced powering the breakout board from a 3.3V source. Now that I tried with a 3V source and had another read of the data sheet, I think the more accurate interpretation is that below 3.6V the regulator chip rated for 3.3V cannot provide the 3.3V anymore. Even so, it does not go to zero straight away. There is actually a nice graph in the data sheet showing the output voltage remains about 0.5V below the input voltage down to about 2.3V. Only then it drops to zero. Since the sensor chip works down to 1.9V, everything should work out just fine down to a battery voltage of about 2.4V.


Hannes Hochreiner wrote 05/11/2017 at 17:45 point

@Arsenijs thanks for the hint on how to remove small parts from a board. So far, I always thought that I would need a hot air gun. I will look into the method you described.

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Arya wrote 05/10/2017 at 09:47 point

Wouldn't it make sense to desolder the regulator from the breakout and jumper VIN to VOUT? Then, it could run on even lower voltage - and you'll get less temperature shift from the regulator heating up.

I think that you could also increase the battery life - by removing the level-shifting circuitry, too - it has all these pullups =) But then, it'd be literally just sensor + capacitors, so I'd look into making my own breakout at that point (isn't this sensor leaded, as per your requirements?)

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Hannes Hochreiner wrote 05/10/2017 at 17:59 point

Yes, your suggestions make perfect sense. However, I do not have the tools / ability yet remove parts from the break-out board. Unfortunately, the sensor only comes in a DFN package. I would have loved to use it directly. It would have been more energy efficient, cheaper, and more compact. At some point, I will have to give lead-less packages a try.

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Arya wrote 05/10/2017 at 18:26 point

You can remove small SMD parts like these with a blob of solder on the end of the soldering iron tip, just cover all part's leads with solder and it's going to detach from the board =) If you remove them, you'll need three jumpers - across VIN/VOUT, SDA and SCL (two level-shifting transistors), don't forget.

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