Hello all,

I am doing project of cypress controller CY8C3244PVI-133

I am using 2 'AA' duracell batteries connected in series with 1500mAH each rating to power ON the board.

My active current is 20mA and sleep current is 40uA

I put cell of 1.568V and 1.547V in circuit on 11/06/2014

I didnt use the board at all. I monitored the current using Ammeter which is equal to 40uA.

Today i.e 13/06/2014 I check the voltage of batteies it is showing 1.565V and 1.545V.

why voltage is dropped so much. can U tell me how to calculate battery life.

Please reply.

regards
PAZ

 

A few mV is not much of a drop.

To calculate battery life in hours you divide the ampere-hours by the current draw. Thus for a 40μA load the life would be 1500mAH / .04mA = 37,500 hours or about 4.2 years.

 

Hi,
thanks for reply.

but battery voltage is dropping day by day.

in some other circuit I put the batteries of 1.589V and 1.561V on 22-05-2014

and today i.e 13-06-2014 I checked the voltage of batteries it came: 1.565V and 1.545V

Is this OK?

can you tell me how much voltage drop to be expected in a day if sleep current is 40uA?

regards
Paz

 

can you tell me how much voltage drop to be expected in a day if sleep current is 40uA?

Check the datasheet for your particular brand of battery.

If you're measuring battery voltage to mV accuracy then have you considered the temperature dependence of the voltage?

 

Stop looking at three decimal places.
You need not worry until the voltage drops to about 1.2V.

 

Hello all,

thanks for reply.

mV can create a problem if my circuit is used for operation.

I have LCD where indication of the batteries are shown.

please tell me how I should calculate for battery drop in a day with sleep current and active current

On google I got calculation for battery life but for a day drop how I should calculate?

How I should ensure that controller is properly in sleep mode?

regards
paz

 

You cannot reliably measure the voltage drop over one day if the expected battery life is 4 years.

If a mV drop is a serious problem for you then you have to redesign your system.

 

..............................
can you tell me how much voltage drop to be expected in a day if sleep current is 40uA?

If you look at this discharge curve for an alkaline battery, you can see the the voltage drops off rapidly at the beginning of the discharge and then the slope becomes less until near the end of the battery life. Doing a rough calculation based upon this curve (using about the first 0.05 of the AH capacity causes a drop from 1.5V to 1.4V) I estimated that the initial voltage drop with a 40μA load is 1.28mV/day, which is close to what you observed.

 

Hello all,

thanks for reply and graph explained.

with all your reference my concern is if I keep it operated i.e active mode for 10 hours i.e 25mA
and rest of hour i.e (24-10) = 14hrs OFF i.e. sleep mode i.e 40uA
that is requirement of the project

will my batteries last for atleast a year?

please tell me the calculation.

regards,
paz

 

10 hours x 25mA = 250mAh
Your battery will last 5 days.

 

thanks for your guidance.

Is there any way in software or hardware to maximize the battery life?

I am using cypress IC i.e Cy8C3244PVI-133

Please guide.

regards
paz

 

If you reduce the length of time the MCU is in run mode to 10 minutes each day your battery will run for 1 year.

Perhaps if you gave us the big picture we would be in a better position to advise you.

 

In many projects the micro can be kept in sleep mode until woken by an external event. Can't you use that method?

 

.....................
with all your reference my concern is if I keep it operated i.e active mode for 10 hours i.e 25mA
and rest of hour i.e (24-10) = 14hrs OFF i.e. sleep mode i.e 40uA
that is requirement of the project

please tell me the calculation.

The calculation is simple. You multiply the current consumed by the time to get ampere-hours. Thus for 10 hours/day @ 25ma the consumption is 250mAH/day (ignoring the very small 40μA consumption when it is in sleep mode). Thus the batteries you selected will only last 1500mAH / 250mAH = 6 days.

For the batteries to last one year requires a capacity of 250mAH/day * 365 days = 91,250 AH (91.25AH). That would require a large array of batteries. For example a D-cell alkaline Energizer battery has about a 20AH capacity (down to 1V output). Thus you would need at least ten of those in series-parallel (five in parallel, in series with another five in parallel).

To have the circuit operate for a year with a smaller size battery, you will need to reduce the time the circuit is ON and/or reduce the current consumption when it is on.