True RMS or DC average?

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phongvu
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Joined: Fri Apr 30, 2010 3:06 pm

True RMS or DC average?

Post by phongvu »

I made the equipment using the pico 4224 to measure the battery performance of different mobile phones under different situations such as voice call, WAP browsing, YouTube watching etc. Each phones comes with a dummy battery which has banana plugs to plug in a regulated 4.15V power supply (which is about the same as a fully charge Lithium ion battery for the mobile phone). To use the Pico 4224 for measure the current I add a 0.33Ohm resistor in series with the phone and a power supply, then capture the voltage across the 0.33Ohm and using the math functions to divide the voltage to 0.33 to get the current and plot it. I then use the Add Measurement to get the DC average and True RMS. I want to calculate how long a battey of a given phone will last under different condition. For example the LG GW620 has battery capacity 1500mA and DC Average 113mA and True RMS 136mA on a 2G voice call. Which number do I use to calculate the battery life? I have attached the picture for the
output of one capture page. I found that the number of sample and the collection time affect the value of measurements greatly. Why is that happen and which is the best setting to use? I normally have 2 phones running on 2 channels to compare the results. Is the Pico 4224 I've got good enough to get the accurate results for what I intended? Thanks
Attachments
LG GW620 voice 2G.pdf
Current versus time for LG GW620 with
True RMS and DC Average measurement.
(19.28 KiB) Downloaded 443 times

noelv
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Joined: Wed May 05, 2010 7:58 am

Re: True RMS or DC average?

Post by noelv »

Do not worry about the DC Avg or the True RMS. You should calculate the surface area under your mA vs time curves. (i.e. the Integral of the Instantaneous Current over time.) This will give you an AH (Ampere x Hour) number for each cell phone. The one with the lowest AH number is - from the power consumption point of view - the best. I am not familiar with the 4224, but, assuming you can export its measurements into an Excel spreadsheet, it should be pretty straightforward to calculate the integral in Excel.
Noel.

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