Hi all,
I need your help in designing a battery removal detection circuit.
My handheld device is powered by 2XAA batteries.
The signal should be as an interrupt to the MCU and it can't take more than 50uA.
I tried to use P-channel mosfet (S- connected to batteries and interrupt to MCU, G- GND, D- to power supply) but the problem was that the power supply has high capcaitors on the input that cause the mosfet never to turned off, than the drain and source voltage were the same all the way during the voltage decay.
I'm looking for solution that will apply fast interrupt to MCU when removing the batteries so that the MCU will be able to shout down the system correctly.
I thought using the FPF2125 load switch that contains reverse current blocking but the problem is that the input to the IC requires an 4.7uF capacitor, so still I have the same problem.

 

How about a Schottky diode between the battery and the capacitor...if you can tolerate the slight voltage drop. The MCU would sense the anode (battery) side of the diode. When the battery is removed the voltage of the capacitor is not seen by the MCU input.

Ken

 

And why 50 micro seconds, it would seem you just have to be faster than the cap. discharge time.

 

I can't tolerate a voltage drop in case of using a Schottky diode. the power efficiency is very important in this device.
russ_hensel - I ment 50uA and not uS,
The current consumption in STBY is less than 0.5mA.
The curent consumtion in operating mode could reach 250mA. (When the batteries voltage is 3v)

 

Take a look at this device and see if it suits your needs:
LTC4411.pdf

 

It looks very good IC. but the minimum voltage input is too high. (2.5v)
The device is powered by 2XAA alkaline batteries in series . (~3.2 to 1.9v)

 

How about a tamper switch or two? (Normally Open, momentary contact) If there is a cover over the batteries then use one to indicate when the cover is opened, if not, use one for each battery.

--Rich

 

Tamper switch could be great idea but impossible due to mechanical limitation.
I need an electronic solution only.

 

What do you think about the following solution:
Detecting the battery removal by sensing the current direction.
Measuring the current over low value resistor by using very low offset OPAMP like the OPA333 to be able to measure the STBY currents together with small voltage drop-out in operation mode.
The amplifier will be connected in open loop to operate as a comparator.
The OPAMP output will be in positive saturation during operation/STBY mode and in negative saturation when batteries are disconnected.

 

I think some processors have support for this perhaps thru brown out detection or adc on the power supply. The idea is that as the voltage from the caps fall below some critical voltage you still have time ( from the caps ) for an ordely shutdown. Check spec sheets for your chip and possible substitutes.

Or just add a third battery and the diode suggested earlier.

 

I think some processors have support for this perhaps thru brown out detection or adc on the power supply. The idea is that as the voltage from the caps fall below some critical voltage you still have time ( from the caps ) for an ordely shutdown. Check spec sheets for your chip and possible substitutes.

The problem is when the user replaces the batteries during STBY mode when the current is very low. Than the voltage falls down very slowlly that the batteries could be replaced and the MCU even doesn't know this.

Or just add a third battery and the diode suggested earlier.

Impossible. No room for another battery.

 

The problem is when the user replaces the batteries during STBY mode when the current is very low. Than the voltage falls down very slowlly that the batteries could be replaced and the MCU even doesn't know this.

would it be possible to reduce the filter cap so a battery removal would cause a brown-out?

the opamp solution will likely cause idle current to go up significantly.

another approach down the same "current sensing resistor" path is to use a comparator / adc in the mcu to perform the function of that opamp.

 

Use a PFET wired as a zero voltage drop "diode" this is a common setup when you need a diode effect on a DC power line but can't afford the voltage drop. Then just connect the battery side of the "diode" to a logic input (or ADC input) on your micro. Problem solved, total cost 1 PFET and maybe a couple of resistors.

 

RB,

Do you have a schematic, with approximate values? I can't seem to find anything concrete (usable) Googling.

Ken

 

you can google "perfect diode". some IC manufacturers, like Linear, have driver chips exactly for that.

 

millwood,
Thanks, but I know about those, and things like precision rectifiers. I was interested in RB's version with one PFET and a couple of resistors.

ken

 

perfect diodes are a mosfet being driven into deep saturation so they behave like a ohmic device.

the crudest version, for the positive rail, would be to ground the gate of a pmosfet to ground.

 

...the crudest version, for the positive rail, would be to ground the gate of a pmosfet to ground.

Have you tried this with in Baron's 1.9v to 3.2v supply range? Just tried it with a logic level P-MOSFET... not a diode! ???

ken

 

Have you tried this with in Baron's 1.9v to 3.2v supply range? Just tried it with a logic level P-MOSFET... not a diode! ???

ken

without a driver, it is unlikely to work because of the low supply voltage.

But if you have a driver that can push the gate below ground, it can work.

that's a fairly standard use of pmosfet (as a switch) on the positive rail.

 

There are P-channel mosfets with very low T.H point suitable to 1.9v to 3.2v range so that the gate can be tied to ground. The problem is (and I tried it) that the mosfet doesn't turn OFF when the batteries are removed because that the Vgs still exists from the power supply input capacitors. The mosfet diode has no influence because the drain and source were in the same potetial.