battery capacity for an oscillatory load

[rpinne]

Hi, I've got a question about the battery-life for a circuit with an oscillating load current.

The system I've designed uses no voltage regulation, and the current drain cycles from 14mA to 24mA at 600Hz, and works between 2.2V-3V. It alternates because the system enters low-power mode during each cycle of operation.

So far I've been using a 160mAh silver-oxide 3.1V battery (made up of four seperate 1.55V 80mAh cells). From measuring the voltage rail during operation I actually get a 2.65V signal with a 150mV, 600Hz sawtooth wave superimposed onto it, and this lasts for around 2.5 hours until the battery suddenly drops to 2V and the circuit stops working. When using 2xAA batteries the battery voltage during operation stays fixed at 3V (with no superimposed sawtooth wave).

I don't understand why the battery is draining so fast, but if you've got any suggestions then that'd be much appreciated. I need ideally as small batteries as I can get for around 5 hours operation. Thanks.

 

[blueroomelectronics]

Post the circuit.

 

[rpinne]

Can't post the circuit I'm afraid, but it's basically a msp430f2013 microcontroller connected to a cc2500 transceiver. The +3V from the battery is conencted directly to the input pins of these two devices. All power-supply pins are connected to 1uF decoupling capacitors, with a 10uF decoupling capacitor connected next to the power input terminal. The MSP430 switches between powered-up and powered-down modes, whereas the cc2500 switches between 'transmit' and 'receive' for each and every cycle - which results in the variable current drain.

Is the battery draining fast because 160mAh is not enough? Or perhaps because of the oscillatory nature of the load current?

 

[blueroomelectronics]

Without a schematic it's impossible to say. Watch batteries (here's the clue, high ESR) are not designed for a high load (relative to their size) but long life at a very low current drain. Could be something in your software, could be a million things. Your 2.4GHz transceiver probably gobbles up a fair amount of current as would most RF devices.
Why can't you post a schematic?

 

[ronv]

It seems like you load change is higher than you think. A 150 mv change with a 10 ma load increase implies a battery resistance of 15 ohms. How are you measuring the current?

 

[rpinne]

Hi, thanks for the replies. I've got a 10ohm resistor in series with the battery, from which I measure the voltage change across this resistor. From looking at battery manufacturers it seems that the majority of button cells are indeed designed for typical loads of 0.1-0.2mA, meaning I'm trying to guzzle almost 100 times that current for which they are designed.

 

[audioguru]

My radio-controlled model airplanes and helicopter operate the radio at 2.4GHz and fly for about 15 minutes with a 160mAh Li-Po battery cell.
The servos and radio receiver use a very low current but the propulsion motor uses up to 2A. The "powerful" Li-Po battery cell weighs 4 grams which is almost as little weight as a useless watch battery cell. Maybe your high current demand needs a Li-Po battery cell (3.0V to 4.2V).

 

[MrAl]

Hi, I've got a question about the battery-life for a circuit with an oscillating load current.

The system I've designed uses no voltage regulation, and the current drain cycles from 14mA to 24mA at 600Hz, and works between 2.2V-3V. It alternates because the system enters low-power mode during each cycle of operation.

So far I've been using a 160mAh silver-oxide 3.1V battery (made up of four seperate 1.55V 80mAh cells). From measuring the voltage rail during operation I actually get a 2.65V signal with a 150mV, 600Hz sawtooth wave superimposed onto it, and this lasts for around 2.5 hours until the battery suddenly drops to 2V and the circuit stops working. When using 2xAA batteries the battery voltage during operation stays fixed at 3V (with no superimposed sawtooth wave).

I don't understand why the battery is draining so fast, but if you've got any suggestions then that'd be much appreciated. I need ideally as small batteries as I can get for around 5 hours operation. Thanks.

Hello there,


When you have a circuit (any circuit really) that draws two or more different current levels and you want to estimate the total run time with a given battery capacity you simply add all the current draws and multiply by the duty cycle time for each current draw, then divide into the battery capacity.

For example, if we draw 10ma for 0.5 seconds and 20ma for 0.5 seconds then we have:
total=10*0.5+20*0.5
which of course is 5+10 which is 15ma total draw. With a 150mAh battery that would mean a run time of approximately 10 hours.
This means you really have to know how long the current draw is for for each interval.

Another example:
10ma draw for 0.003 seconds and 20ma draw for 0.0015 seconds and 0ma draw for 0.0015 seconds...
The total time period is:
0.003+0.0015+0.0015=0.006 seconds
The sum of currents times their 'on' times is:
10ma*0.003+20ma*0.0015+0*0.0015=0.030+0.030+0.000=0.060
and divided by the total time period of 0.006 we get:
0.060/0.006=10ma average current draw over time.
Now with a battery with capacity 150mAh that would mean approximately 15 hours run time.

So the idea is to measure all the currents for all the periods, measure the time periods for each different current level, multiply the 'on' periods times the respective current levels and sum them, sum all the periods (including the 'off' periods), then divide the sum of current levels by the total sum of all the periods. Then, look at the manufacturers data sheet to find out if you have to derate the battery due to the actual current level (skip this step for rough estimates or very low current draws). Lastly, divide the ampere hour capacity by the average current to get the estimated run time.

The formulas would look like this:
Tp=t1+t2+t3+...+tn
Iavg=(I1*t1+I2*t2+I3*t3+...In*tn)/Tp
RunTime=BatteryCapacity/Iavg
where
t1 though tn are the time periods for the currents I1 through In respectively.

Note that these estimates assume a square wave current wave shape. If you have a square plus sawtooth then you have to estimate the current level by taking the current for that period to be the very center instantaneous current level (center of the sawtooth wave).

Note also that as mentioned above this is just an estimate, and to get a better idea you have to look at the manufacturers data sheet for the battery you intend to use and see how the apparent capacity varies with different constant current levels. Most batteries show a much reduced capacity when the current draw gets higher and higher. For example, a 7.5Ah battery could end up looking like a 4Ah battery if the current level is high enough.