Another set of issues are situations where current flows out of Vbat. This could happen under two quite different circumstances:
1) When the device is powered from USB, and 2) when the battery polarity is reversed. To keep USB from "charging" our primary cells, we have two options: either we cut off the battery explicitly or the boost converter takes care of this for us. Fortunately, the TPS61020 series should block reverse current just fine when properly disabling the converter. Explicitly disabling is also desirable to avoid situations where the device is powered from USB, but the 3.3 V rail is slightly below the threshold voltage of the boost converter, and the converter would thus try to "top off" the 3.3 V rail, wasting battery power. Polarity reversal is more involved. A reversed battery may damage the device and if high currents flow, also itself. AAA (and similar) cells have a mechanical anti-reversal feature: the protruding "nipple". With a suitable mechanical design, a device can thus prevent the battery from making contact when reversed. This is an example Joerg has found: http://downloads.qi-hardware.com/people/werner/anelok/tmp/mech-anti-reversal.jpg But what if this turns out to be too difficult to implement ? After all, it's already pretty hard to find battery contacts that have an even vaguely suitable shape. (I.e., that's still on my to do list.) There are again two common choices for electrical protection: 1) crowbar and fuse, and 2) diode. The crowbar consists of a diode that simply shorts the battery when reversed. This creates a large current through the battery, which then trips a fuse. The fuse can be the resettable type, which recovers once the upset is removed. While this sounds nice and easy, there are a number of problems with this approach: - at the voltages involved, it gets difficult to find a fuse that can distinguish between a normal but high current and the overload created by the crowbar diode, - worse, many such fuses have a surprisingly high resistance. In our design we should aim for not much more than 100 mOhm. - last but not least, good fuses are not cheap. E.g., the Littlefuse 0603L100SLYR that looks fairly decent costs USD 0.82 at 1000 units. So this doesn't look good for a low-voltage device like Anelok. Using a diode to only allow battery current in the right direction sounds like the most simple solution. Unfortunately, a normal diode would drop the voltage by about 0.6 V under load, which would leave scarcely anything of our 1.5 V battery voltage. But one can build an ideal diode with a FET. This approach is nicely described in http://www.ti.com/lit/an/slva139/slva139.pdf Great ! Now all we need to do is find a suitable FET ... Well, after long hours of reading dozens of data sheets, I found a small number of p-FETs that may work: Vishay SiA427ADJ, Vishay Si5499DC, A&O AON2401, and Vishay Si2329DS. They all seem to be able to admit currents of >= 1 A at 0.9 V. Then I looked at n-FETs. They seem to be a little friendlier, and On Semi MCH3484a and Vishay Si2342DS both look very promising. The overall problem is that when the battery voltage drops close to the FET's threshold voltage, the FET's resistance increases and the boost converter therefore needs to draw more current. More current means that the battery voltage drops more, the voltage loss across the FET increases, and as if all this wasn't enough, the gate voltage also decreases, thus increasing the FET resistance even more. So basically the FET could choke a perfectly healthy battery long before it would be too weak to power Anelok. Now, the FETs above may just be good enough to avoid trouble, but it's never nice to be operating so close to the envelope of what is physically possible. TI must have felt the pain, too, and created a great FET-based ideal diode that also works with very small voltages: http://www.ti.com/lit/ds/symlink/sm74611.pdf Alas, it's also quite expensive at USD 1.688 @ 1 kunits. Then Peter "whitequark" Zotov pointed me to this application note: https://www.maximintegrated.com/en/app-notes/index.mvp/id/636 This uses a boost converter to drive the FET's gate. With a high enough gate voltage, there are lots of FETs that should work great for Anelok. And we happen to have a boost converter just at the right place ! So an n-FET driven by the boost converter looks like the nicest and safest solution for this tricky problem. - Werner _______________________________________________ Qi Hardware Discussion List Mail to list (members only): [email protected] Subscribe or Unsubscribe: http://lists.en.qi-hardware.com/mailman/listinfo/discussion
