Reverse battery protection with two batteries.

Thread Starter

RichardO

Joined May 4, 2013
2,271
I have a circuit that uses two CR2032 coin cells for power. Using two batteries allows me to replace one battery at a time without loosing power to the circuit. I use diodes to protect against reversed battery polarity during installation.

This works fine except for the losses of the diodes. I want the circuit to run one to two years on the two coin cells. The forward voltage of the diodes significantly shortens the life of the batteries. This is because occasional high peak currents and the forward voltage drop of the diodes make the end of life voltage of the batteries effectivley about 2.7 volts (instead of the normal 2.0 volts).

I would like to replace the Schottky diodes with "ideal" diodes using MOS-FET's. The obvious solution using two MOS-FETs does not work because when one battery is replaced, the other battery in still in place. The battery that is in place turns on the MOS-FET even when the new battery is reversed.

One final complication is that the reverse battery circuit must not draw any significant current itself. If it draws more than a couple of microamps it will shorten the life of the batteries a lot.

Here are the circuits:

AAC_Ideal_diode.png


Thanks in advance for any insights into this problem.
 

#12

Joined Nov 30, 2010
18,223
.25V@ 1ma
1) Quit believing a Shottky diode always uses up 0.7 volts. You would need to draw 300 ma to get to 0.7 volts, and you can't get that out of a coin cell.
2) Don't put the batteries in backwards.
You can't make things idiot proof. There is always a better idiot being developed.
 

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crutschow

Joined Mar 14, 2008
28,547
There are some ideal diode ICs but they use more current then you can tolerate.
Some battery holders for coil cells will not make contact if the battery is incorrectly inserted.
Can't you use two of those?
 

Thread Starter

RichardO

Joined May 4, 2013
2,271
.25V@ 1ma
1) Quit believing a Shottky diode always uses up 0.7 volts. You would need to draw 300 ma to get to 0.7 volts, and you can't get that out of a coin cell.
2) Don't put the batteries in backwards.
1) I real life I expect to have momentary loads of 50ma or so. The coin cells can deliver that for a few milliseconds. They will not be delivering the open circuit voltage at that time, of course. The 0.7 volts is just my rule of thumb for a Schottky diode at full load. With the battery sagging, and the diode at lower current I am guessing that a total loss of 0.7 volts is still in the ballpark.

2) Not an option. The battery holder does not enforce correct polarity. :(
 

Thread Starter

RichardO

Joined May 4, 2013
2,271
There are some ideal diode ICs but they use more current then you can tolerate.
Some battery holders for coil cells will not make contact if the battery is incorrectly inserted.
Can't you use two of those?
The ideal diode IC's I have found don't operate at less than 2 volts either. :(

The battery holders are the cheap sheet metal ones from Keystone. I need the low profile. (And I am cheap). While looking for this picture I saw some new ones that look interesting. Thanks.
 

Thread Starter

RichardO

Joined May 4, 2013
2,271
Is it not possible to change the holders to those that do?

But, as can be seen in the LTspice simulation below, the forward drop of the Schottky is quite low at low currents:

View attachment 115458
Space is really tight so no other battery holders I have seen will work. I also need the "edge" access.

For current pulses I expect to be in the 50ma -- 250mv area of your graph. I would expect an ideal diode could do better than that.
 

crutschow

Joined Mar 14, 2008
28,547
That's better than the numbers I was getting from the datasheets.
Yup.
It appears that the LTspice model is the same for the 1N5817 through the 1N5819 and they all appear to more or less follow the 1N5717 in the Diodes Inc. data sheet I looked at.

It would thus appear that the op would want to use the 1N5817.
 

Thread Starter

RichardO

Joined May 4, 2013
2,271
It would thus appear that the op would want to use the 1N5817.
I wish I had thought of that. Seriously, it has been so long that I designed in the 1N5819 that I forgot that it could be a 1N5817. (At the time the 1N5819 was more readily available in SMD than the 1N5817).

I am still curious if there is a MOS-FET circuit that would work. It seems to me that you or maybe @MrChips has posted something similar in the past. More than a single transistor was used. Unfortunately I could not find it on the forum.
 

#12

Joined Nov 30, 2010
18,223
So, I waiting on pins and noodles to see it. :D
I said I know the requirements, not the circuit.:D
I don't think either of us is going to see it.
Mosfet gates don't go that low and bipolar transistors don't have a DC gain of 50,000:(

I did prove you can operate a bipolar transistor in the ua range, but that's a long way from 50 ma at VceSat of 0.1V:eek:
 

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Thread Starter

RichardO

Joined May 4, 2013
2,271
Don't know of any for your requirements. :(
I think I do. RichardO wants a diode-like switch that runs continuously on less than 1 ua, saturates with 2.0V or less on it's gate, and passes 50 ma pulses through an impedance of less than 2 ohms. Is that about right, @RichardO ?
Pretty much so. The 2 ohms is quit a bit higher than I would like since it would add 100 mV in voltage drop from the batteries at 50 ma.

As stated in my original post I am wanting to replace the Schottky diodes with something more efficient. Of course it's the details that make all of the difference...

A summary:

1) Replace the existing Schottky diodes with a more efficient circuit.
2) The circuit runs from two CR2032 lithium coin cells.
3) The two batteries must be isolated from each other when both are installed.
4) The circuit must be protected from one or both of the batteries being inserted in the wrong polarity.
5) The circuit must be protected from one of the batteries being inserted in the wrong polarity while the other battery is in place.
6) The batteries must not be dicharged or damaged if one or both are inserted in the wrong polarity.
7) The circuit must run on nearly dead batteries -- this is about 2 volts.
8) The circuit must not draw much more than about 1 ua per "diode".
9) There is an occasional momentary load, a few milliseconds long, of about 50 ma. The "diodes" must not drop more than a few millivolts when this happens.
 
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