Werner Almesberger
2016-03-25 16:33:43 UTC
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:
Loading Image...
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
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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:
Loading Image...
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): ***@lists.en.qi-hardware.com
Subscribe or Unsubscribe: http://lists.en.qi-hardware.c