I connected a single cell lipo to J2 on mine just today. Started up just fine for me.
There are 3 methods to power up the board.
- 3.7 LiPo,
- 7-15 external power
- 5v via USB
The mini-breakout has all the power circuitry needed to convert each of these inputs to suitable voltage levels to power the Edison.
There are a few gochas, though, the OTG functionality won't work unless your providing power via the external input, probably because most OTG peripherals expect 5V on the USB, which is probably not generated when a LiPo is connected. I had this issue and got around it by adding a tiny DC-DC converter that converyted my LiPo's 3.7v up to a level suitable for external, then the USB-OTG peripherals worked fine.
I've solved my problem with it not powering up using just a 3.7V LiPo battery. It's a good news - bad news story. According to this post on the discussion "Edison Mini-BreakOut Board Documentation Woes...", the square pad on J2 is ground. Not so!
"Correct Answer by intel_dan on Sep 21, 2014 2:13 PM J21 is is for 7-15v (actual valid input range might lower/higher, but 7-15v will work) input into the TPS62133 which provides 5v to the rest of the board. The pin closest to the edge of the board (with a square shaped solder pad on the bottom) is Ground. J1 is for an optional themistor (just shorted together if no thermistor is present) for use when charging a 1 cell lipo battery that is hooked up to J2 (again the pin with the square shaped solder pad on the bottom/underside of the board is Ground). Hopefully the schematic for the board will be made available soon."
I should have used my meter to double check ground before connecting a battery to J2. The square pad is NOT ground on J2, at least on my board. Connecting battery positive to the square pad powered the Edison right up and I was able to communicate with it using Putty SSH wirelessly. In other words, J2 pin 2 is ground NOT the square pad (pin 1). BTW, the schematic is correct, which I should have also double-checked.
The bad news is apparently hooking the LiPo up incorrectly earlier has destroyed the TPS62133 regulator chip on the Mini breakout board. The board no longer works with power connected to J21 the 7-15V input or USB OTG. It looks like the Edison module is OK, since it works OK with the LiPo. I wonder if Intel will make the Mini breakout boards available without an Edison module?
Thanks for the tip Dave about adding a DC-DC converter to the 3.7V LiPo input. That might be a way for me to keep using this board, even though it now has a bad TPS62133 regulator.
I was lucky that I soldered wires into the GPIO's rather than using the pin header when preparing for Maker Faire Rome. I had limited time to prepare, and dealing with a blown board would have made it doubly painful.
Although in the end the DC-DC worked for me.
Probably should allow use of OTG without 5V out for powered peripherals... Maybe the board is hackable to allow this.
I was taking a closer look at the chips on my mini breakout board to see how I could bypass the regulator with +5V It doesn't seem to have the same chips as Intel's posted schematic. Some other things don't seem to match either. Have there been different revisions of the mini board? Also, is there a board layout available somewhere?
Hey, not to worry - now I've really got a good reason to dig into this thing. After doing a little more metering, I can see it is really the BQ24074 battery management chip that is bad. It has +5V coming in from the regulator chip to pin 13 (5V_SYS), but it doesn't put 3.3V out pins 10/11 (V_SYS). That's with either a power source connected to 7-15V J21 or USB OTG connected. If I connect a 3.7V LiPo battery to J2, then I do get 3.3V out on V_SYS. So, the BQ24074 is bad. Being such a tiny 16 pin surface mount chip, I could never install a new one. So, I can either use it as is with a LiPo battery all the time or maybe install a small DC-DC 5V to 3.3V Buck converter. I have a feeling the LiPo charge circuitry is also bad and would have to charge it elsewhere.
The BQ24074 should be outputting 4.4 volts to V_SYS when it's powered. I suspect the internal regulator in the BQ24074 is blown. Like you said, I would just hookup an external 3.3V regulator to the battery connector.
I must say that I have not ever seen such bad documentation as what we have for the mini breakout board. The connectors on the board itself have no polarity marking. The schematic contains errors and does not match the real board. It should take about ten minutes for the responsible engineer to edit the schematic and correct the errors. Why is it not being done? Is nobody in charge of this Edison project?
Here's my fix for the blown BQ24074 battery management chip. I planned to supply 3.3V to the board using a Buck DC-DC converter, but I knew I had to disconnect or remove the bad BQ24074 first. I also knew that would be difficult. Well, it was much harder than I expected to remove the tiny surface mounted chip. It would have been even harder to cut the offending trace. In any case, I managed to cut it out as you can see from this photo.
The orange wire is the voltage source going to the DC-DC converter. I connected it at the cathode junction of CR6 and CR7. That way it will convert either 5V coming in from the USB OTG or 7-15V coming in on J21.
The next photo shows the converter on the underside. The green wire is the converter's output and goes to J20-1 or V_SYS. I also have it connected to J2-1 so that a 3.7V LiPo battery can also provide power to the Edison module. The only downside to this fix is I no longer have a way to charge the LiPo with the breakout board. But that's OK. I have other ways to charge it back up. An advantage is I can now supply the board 5V up to 36V on J21 instead of just 7-15V.
I can't recommend this fix to anyone unless they are willing to take a high risk of ruining their board. It is extremely difficult to remove the chip and its 17 miniscule connections. BTW, Intel did offer to replace my board, but it was going to be 6-8 weeks before they had warranty stock.
Glad to hear you got it working again. Removing a QFN package with a soldering iron is not easy, but it should just pop off with a heat gun.
I just got my breakout board today. It's a pretty nice little board. Thicker than a normal PC board, but I guess that helps prevent the pins from shorting on the Edison shield.
thanks for showing how to repair the board. it may come in handy when I start working on my own mini breakout boards. right now I am impatiently trying to get familiar with the Edison mini breakout board and trying to figure out how to control the GPIO. I decided to have a closer look at the schematics I was complaining about and came to the conclusion that it most likely matches the actual board. so I was looking at the power control circuits to figure out how it works. this is what I learned so far (perhaps some Intel engineer can correct me where I misunderstood the schematic):
There appears to be four different ways to power the Edison module and the mini breakout board. As far as I can see, it is possible to use any number of these power inputs at the same time.
1. apply 5V to the micro USB connector J16 (this is the connector closest to J21
2. apply 7V-17V to J21. the polarity must be such that J21 pin 1 is ground and pin 2 is positive. there is a series diode, so if you plug in with wrong polarity the circuit will survive.
3. apply positive 7V-17V to J17 pin 4 and use the same ground as J21 pin 1. there is a series diode, so if you plug in with wrong polarity the circuit will survive.
4. plug in a Li-Ion single cell battery (3.6V nominal voltage) to J2. you must be very careful and make sure you connect the battery with correct polarity. J2 pin 2 is ground and pin 1 is VBAT (positive 3.6V). Notice that the polarity is reversed compared to J21! if the battery has a third wire for temperature sensing you should remove the jumper on J1 and connect the temperature sensor wire to J1 pin 2. if you do not use temperature sensing for the battery there is no way the charger will know if the battery is overheating and you could have a fire.
When looking at the TPS62133 power converter I can see some issues that may be worth looking closer at:
the 7-17V voltage is converted to regulated 5V that is applied to the BQ24074 and also to the current limiter circuit MIC2039 that feeds the micro USB connector J16. the steady state input current limit for the BQ24074 is just below 1A (set by R5) and the steady state current limit for the MIC2039 is 500mA, so the total load on the TPS62133 should be limited to about 1.5A. However, the MIC2069 can deliver several Amperes (typ 3.2A, max 6A) to the USB connector for short periods of time. this is to make the circuit less sensitive for current transients.
This means that the load on the TPS62133 can be much higher than 1.5A. Expect situations with 4-5A transient loads. this is where I think the TPS62133 circuit has some design mistakes. it does not appear to be designed to handle the transients:
- the inductor L1 is a very tiny device and it is likely that is saturates
- on the input side there is a 0402 size 0.10 ohm resistor R115. such a small resistor is usually rated for maximum 0.063W power dissipation, but here it is possible to have an order of magnitude higher power dissipation. R115 is likely to have a very short lifetime. the current limit of the TPS62133 can allow currents up to about 5A or more in the inductor during load transients. there is only 0.1uF at the input, so this current will flow in R115.
- I also wonder about CR2 which is in series with the output of TPS62133, what current rating it has (part number unknown). is it really able to handle 1.5A steady state plus transients?
- the output voltage sensing is connected to the cathode of CR2 instead of to the anode, do we not get in trouble if there is a power source connected to J16 and the voltage happens to be on the high side? for example, a USB charger with more than 5.2V is used, at the same time as the 7V-17V input is used? the TPS62133 could sense too high voltage and start discharging and charging its output capacitor C1? one should perhaps avoid to connect power to J16 if the high voltage input is used.
Good analysis and you have made some valid points about the breakout board's power circuitry. I have both boards, Arduino and Mini breakout, and was surprised to discover that part of the power circuitry is pretty much identical between the two boards, some of which seems redundant on the mini breakout. It seems like it could have been handled a little simpler. After all, the Edison module itself only needs a single voltage between 3.15V and 4.4V to do everything it does. Either a standard 3.3V source or a 3.7V battery level works nicely.
The Arduino breakout does need some of this complexity due to handling both 3.3V and 5V Arduino shields. Maybe Intel developed the Mini breakout after the Arduino breakout and just reused the same circuitry?
BTW, the Arduino breakout does have the battery pins marked plus and minus. Too bad the Mini doesn't. In fact, the Arduino board has all the components labeled. A diode could also avoid the problem of reversing the battery leads.