measure power for battery calculation - Page 1
Forum section: electronics / beginners

author: A2
author_rank: Contributor
timestamp: August 12, 2014, 11:36:42 am
content: I have a battery operated circuit.1. Circuit wake up for 100ms. Do its work & go to sleep for next 1sec.2. In 100ms, it does many different task. one after another.So current keeps on varying in that 100ms also.I have to measure power of circuit so that I can calculate the battery requirement.1. I tried to put simple ammeter in between but it couldn't measure high frequency spikes.2. Then I put a 5 ohm resistor in power path, & check the voltage across it with oscilloscope. It shows repetitive readings in the circuit.At any instant I can get current by dividing the voltage noted from CRO by 5ohm.By how to consolidate the total power across one cycle.3. Is there any low cost solution available which can measure the power drawn.

author: Simon
author_rank: Unknown
timestamp: August 12, 2014, 11:40:20 am
content: How important is it to be how accurate ? if you look at the waveform on your scope can you visually take a rough guess at the average current draw, you can then average that current draw with your sleeping state current draw by working out 9xsleep value to 1x awake value to get your average, or divide your awake current by 10 and add the sleep current to get a decent average.

author: PepeK
author_rank: Regular Contributor
timestamp: August 12, 2014, 11:51:52 am
content: If you have a scope, it is the right approach. Is it a digital scope having some build-in math ? You need sum(I * dt) = integral of I * dt, which corresponds to a charge drawn from battery. Make a screenshot for those 100ms when a circuit is active.

author: mariush
author_rank: Super Contributor
timestamp: August 12, 2014, 12:01:52 pm
content: Another solution would be to get one or two large supercapacitors, for example let's say 25 F 2.7v- they're not that expensive these days, probably about 5-10$ each.Put them in series to get 12.5 F 5.5v if your project needs more than 2.5v (you say battery but don't say what kind of battery) and charge them up to 5- 5.3v , or charge them to a bit above the level your project is designed for.Then run your project for a few seconds and disconnect the project from the supercapacitors and measure the voltage on the supercapacitors.You have original voltage, you have the voltage left , you can measure the energy consumed. Unlike batteries where the voltage will stay there for hours at low power consumption, the voltage drops pretty much linearly as energy is taken out from supercapacitors so you will notice a voltage difference right away (well, if your multimeter has a decent amount of counts, for example with my 22000 count uni-t ut61e this is a piece of cake).-You can get a 0.1-0.33 ohm resistor to act as a current shunt and measure the voltage drop on it and if you want, you can put it on a scope and log the waveform. Note that there's voltage drop on the resistor, so that screws up the estimation a bit, especially with 5ohm as you did. The more power consumption, the more voltage drop on the resistor (Ex for 5 ohm, for 100mA you have 0.5v drop, so if your battery is 3v you now suddenly have 2.5v instead of 3v)For example, if your project uses a switching regulator to convert battery voltage to some particular voltage, then just the change in inpuut voltage will screw up the efficiency of your regulator, which will make you estimate the battery life incorrectly.

author: levinite
author_rank: Newbie
timestamp: August 12, 2014, 10:44:31 pm
content: What about the internal battery resistance. Doesn't that need to be considered?