Using a coin cell as an supplementary voltage source

Thread Starter

user1397215

Joined Aug 31, 2022
3
Scenario:
I have a design for a low power device which uses 3x 1.5V (so 4.5V) AA batteries to be powered. On 99% of the time that it is running, it draws around only 1mA current. For the 1%,it can draw up to 200mA of current thanks to a vibration motor. I have noticed that my battery voltage drops during this time. The way the motor is controlled is done by a microcontroller. I have also noticed that the microcontroller resets itself as the motor tries to start spinning. It seems to be caused by the voltage drop on the cells.

I have decided to add a coin cell battery as a supplementary voltage source so that the microcontroller will not reset during this time. I tried to use a MOSFET for this circuit so that the ground of the coin cell can be connected to it and prevent the coin cell powering the microcontroller by itself when the cells are not present. However, I seem to have misunderstood how a MOSFET works in this case and wired it incorrectly where no current would flow at all from the main cells to the microcontroller. In this case, it seems the coin cell is always trying to power the device all the time.

Some info:
R19 is 15ohm resistor
VMAIN is 4.5V from 3 cells
Q5 is a BSS214NWH6327XTSA1
B1 is a CR2032MFRRV
1663290763746.png

What do I need to change for this? I just want the AA batteries to be the main voltage source, and the coin cell will only power the microcontroller (to prevent a reset) when the voltage on the cells drop to some threshold (lower than the coin cell?). Is this possible?
 

Ya’akov

Joined Jan 27, 2019
6,575
Weclome to AAC.

Do you have a cap across the MCU‘s Vcc? The sag in the battery probably recovers pretty soon after it happens, you can try a 100mF electrolytic, but oyu can increase it as necessary. It would be a lot simpler if it works.
 

DickCappels

Joined Aug 21, 2008
9,319
With only 3 volts you might not have enough gate to source voltage to get any current through the MOSFET.
What is Vmain?

What MOSFET are using?
 

MrSalts

Joined Apr 2, 2020
2,376
The Coin cell battery is not an ideal battery. That is, The voltage of a coin cell will drop dramatically at a 200mA load. The datasheet doesn't show voltage drops at 200mA but let's look at a trend. The 10mA plus causes a voltage drop across the battery of about 0.25v. 50mA load triples the voltage drop to 0.75v. I would assume a load of 200mA would double to triple the drop to 1.5 to 2 volts. That means, your coin cell is only adding 1 to 1.5v to your battery with 30 to 45ohms of internal resistance.

You mention 1% of the time that the vibration motor is running. How long (how many seconds) does the vibration motor run each time?

Also, how long do you expect the battery to last?
 

Jerry-Hat-Trick

Joined Aug 31, 2022
166
Do you have a cap across the MCU‘s Vcc?
I'm thinking, the motor probably draws maximum current as it starts, as it's a very low resistive load. If you can permanently power the processor via a diode, i.e. with just a 0.6V drop from battery voltage then a smaller capacitor across the processor power rails may be able to keep the processor going whilst the battery voltage (hopefully briefly) drops below the the voltage needed to keep the processor going? - although if a 100uF capacitor on its own will suffice Ya'akov's solution is nice!
 

Thread Starter

user1397215

Joined Aug 31, 2022
3
Weclome to AAC.

Do you have a cap across the MCU‘s Vcc? The sag in the battery probably recovers pretty soon after it happens, you can try a 100mF electrolytic, but oyu can increase it as necessary. It would be a lot simpler if it works.
Yes there is 100nF for this

With only 3 volts you might not have enough gate to source voltage to get any current through the MOSFET.
What is Vmain?

What MOSFET are using?
Vmain is 4.5V from 3 AA cells

The Coin cell battery is not an ideal battery. That is, The voltage of a coin cell will drop dramatically at a 200mA load. The datasheet doesn't show voltage drops at 200mA but let's look at a trend. The 10mA plus causes a voltage drop across the battery of about 0.25v. 50mA load triples the voltage drop to 0.75v. I would assume a load of 200mA would double to triple the drop to 1.5 to 2 volts. That means, your coin cell is only adding 1 to 1.5v to your battery with 30 to 45ohms of internal resistance.
I'm thinking, the motor probably draws maximum current as it starts, as it's a very low resistive load. If you can permanently power the processor via a diode, i.e. with just a 0.6V drop from battery voltage then a smaller capacitor across the processor power rails may be able to keep the processor going whilst the battery voltage (hopefully briefly) drops below the the voltage needed to keep the processor going? - although if a 100uF capacitor on its own will suffice Ya'akov's solution is nice!
Is there a way to make the device operate from the main AA battery cells all the time, and only have the coin cell run if the motor is running? I am just trying to prevent a reset of the MCU when the motor starts to spin (currently the motor doesn't really finish spinning because the MCU resets..). I did some measurements and using the 1mA discharge curve vs this battery it can run for about a month, the vibration motor runs about 60s each time.


I did a simulation of the circuit as it is and it seems there will be no current flowing from MOSFET? Because source and drain are respectively connected to GND of coin cell and GND of batteries. Is my understanding correct?

1663331786181.png
 

MrSalts

Joined Apr 2, 2020
2,376
I see Your biggest problem is the microcontroller resetting. Are the AA batteries in your circuit relatively new? As batteries age, their voltage tends to sag more under load than a fresh battery. The voltage drop should be minimal for a 200mA load so one possible issue may be that your batteries are not new enough. Alternatively, your actual load may be much higher (just as the motors turn on) if the steady state current draw of the motor is 200mA.

Another option is to put a second set of three AA batteries in parallel with the first set. This will allow the current draw with less of a voltage dip and hopefully not low enough to cause the micro to cut out.

Lastly, there are more powerful AA batteries. Namely, Energizer Ultimate Lithium. These are 1.5v per battery like a standard alkaline cell but these Ultimate Lithium batteries can handle loads to 1A without much sag.
 

Ian0

Joined Aug 7, 2020
6,312
Don’t run the microcontroller directly from the batteries. Use a LDO regulator to produce 3V, to keep its supply separate.
You could then use the coil cell plus a Schottky diode to keep the 3V supply going in case of dips
 

DickCappels

Joined Aug 21, 2008
9,319
As drawn you circuit should be able to switch an amp or two, it looks like you should see how it works after making some of the changes suggested.
 

DickCappels

Joined Aug 21, 2008
9,319
Yes there is 100nF for this

Vmain is 4.5V from 3 AA cells

Is there a way to make the device operate from the main AA battery cells all the time, and only have the coin cell run if the motor is running? I am just trying to prevent a reset of the MCU when the motor starts to spin (currently the motor doesn't really finish spinning because the MCU resets..). I did some measurements and using the 1mA discharge curve vs this battery it can run for about a month, the vibration motor runs about 60s each time.

I did a simulation of the circuit as it is and it seems there will be no current flowing from MOSFET? Because source and drain are respectively connected to GND of coin cell and GND of batteries. Is my understanding correct?
You need to have a connection from the drain to something (like a resistor) connected to the positive power supply. Then it should work fine. AS AN EXAMPLE

1663336607482.png
 
Last edited:
I am just trying to prevent a reset of the MCU when the motor starts to spin
I probably didn't explain my suggested solution very well. Can you measure the voltage across the 4.5V supply when the motor is running continuously? It will drop instantaneously but recover when the load becomes inductive when it's spinning. If it's just a small motor with an eccentric cam it effectively reaches no load speed pretty quickly.

And what is the minimum supply voltage that the processor can live with before it shuts down? If, for example, the processor will happily run with a 3.3V supply and the battery voltage drops to 3.0V when the motor starts, then recovers to 4.0V when the motor is running the diode from the battery to the processor supply pin would stop the processor supply dropping to 3.0V, with a capacitor that holds enough charge.

Even simpler, which may just work. I believe the vibration motor may run at a lower voltage so connecting it in series with a resistor may suffice to limit the starting current and once it's spinning it'll still see most of the power.

It may just be quicker to try it! :)
 
One last thought on this before I leave it be. You can probably tell, I favour a simple solution to a problem rather than a 'work around' which I honestly consider the addition of a coin cell to be. Sorry for my persistence!

Presumably you are not driving the motor directly from a processor output which would not be happy delivering 200mA. I'd be connecting the motor to the collector of an NPN transistor, the emmitter to ground, and the base via a resistor to the processor output. Suppose you make that resistor big enough so the collector current is limited by the gain of the transistor? If you know the current of the running motor aim for a little higher starting current than this so that the motor has a "soft start" and can settle into it's happy state.

Over and out...
 
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