Current stabilization circuit - Page 1
Forum section: electronics / beginners

author: Krunky
author_rank: Newbie
timestamp: May 15, 2016, 10:11:00 am
content: Hello. So I have a project going on: arduino battery capacity testerSo as we all know: battery is more empty = lower current (diagram below)Equation here is a bit more complicated because it uses integrals and its not that accurate.And to make it easier and more accurate to measure and calculate I need a stabile current like this:Equation for this is: mAh = mA * timeI would need a diagram of a current stabilization circuit. That will hold 500mA until battery reaches 1V (when it reaches 1V it will stop the test and begin charging the battery to prevent damage).Thank you for your help!*I will be testing all kinds of batteries (from 1,5V up to 5V).

author: Dave
author_rank: Super Contributor
timestamp: May 15, 2016, 11:08:10 am
content: You need a constant current sink. It has been done way too many times by people on this forum, just do a search.

author: danadak
author_rank: Super Contributor
timestamp: May 15, 2016, 11:36:18 am
content: Consider using least squares error fit, or a power curve fit of thedischarge curve. Or look at battery chemistry, there must be a f()that describes the discharge curve ?http://users.ece.gatech.edu/rincon-mora/publicat/jrnls/tec05_batt_mdl.pdfhttp://www.ti.com/lit/an/slva194/slva194.pdfhttps://pdfs.semanticscholar.org/989c/bd5ddda33516d42d50b3e251bea236c80ace.pdfRegards, Dana.

author: Kalvin
author_rank: Super Contributor
timestamp: May 15, 2016, 11:43:06 am
content: You may want to use quite small sense resistor (like 0.1 ohms) in order to be able to measure the battery terminal voltage down to hundred millivolts. At 500mA the sense voltage will be 50mV so you can drain the battery completely empty. Using a 1 ohm sense resistor the voltage drop will be 0.5V at 500mA which may be a bit high if you want to drain the 1.5V battery completely.Using either 1 ohm resistor or 0.1 ohm resistor means that the sense voltage is quite low. You may want to amplify the sense voltage by 5 - 10 for example, so that the reference voltage can be in range of volts instead of millivolts.You will save yourself from a lot of trouble if you select a good rail-to-rail op amp if you are using a single positive voltage as the sense voltage is quite small. If you are using +/- supply voltages you can get away with a jellybean op amp.You may also want to measure the battery terminal voltage.

author: Krunky
author_rank: Newbie
timestamp: May 15, 2016, 11:45:54 am
content: Quote from: Kalvin on May 15, 2016, 11:43:06 amI had that in mind. But I will also test LiPo batteries which can explode or get damaged if they have under 1V of power. This is why I cannot use this method. Arduino will be programmed that the cutoff voltage will be 1V (because some batteries dont have protection for over-discharge and that stuff).

author: Kalvin
author_rank: Super Contributor
timestamp: May 15, 2016, 11:49:00 am
content: Quote from: Krunky on May 15, 2016, 11:45:54 amIn that case I would build a simple safety mechanism using an op amp which would monitor the battery voltage and will cut off the load current when the battery terminal voltage goes below a certain limit. That would work even if the Arduino would fail for a reason or another.

author: Krunky
author_rank: Newbie
timestamp: May 15, 2016, 12:46:33 pm
content: Thanks for all of your help. Here I have one ghetto solution that might work. Can you please check it out:So there we have a Li-Ion charger that charges the battery with 1A at 5V. Lets say that here we have a 5V battery (I know it says 3.7V on the battery). But if I want to use for example a 3.7V one i could just add a voltage redulator (0,8V-5V to 5V module) and it would also work?So if I use 3.7V battery then:*Kepp in mind that voltage regulator has output current of around 200mA on 3.7V Vin. Will then battery charger make it 1A constant?Thanks for your help.

author: Aodhan145
author_rank: Frequent Contributor
timestamp: May 15, 2016, 01:24:27 pm
content: Quote from: Krunky on May 15, 2016, 12:46:33 pmThat does not look like a good idea. I think you have misunderstood what they mean by constant current. They mean constant current drain from the battery. There is a circuit everywhere on these forums where it uses a logic level mosfet and an opamp to make a constant current dummy load. Search for this first. Make sure you include a cut off voltage as well as you do not want to discharge a battery too much without any protection circuitry on it.Here is a video where Dave explains how the constant current dummy load which most people make works. Most people on the forum replace the pot with a DAC as to digitally control the circuit. Hopefully this will help.

author: sjd.aliyan
author_rank: Contributor
timestamp: May 16, 2016, 06:03:02 am
content: Quote from: Krunky on May 15, 2016, 10:11:00 amIt is not completely right Higher dischared battery have lower voltageYou need a Constant Current load for testing your battery.You could design a CC circuit with mosfet or a LM358 these are usual choices.http://www.instructables.com/id/Arduino-True-Battery-Capacity-Tester-Li-IonNiMH/Read articles in BatteryUnversity.com this site is very helpful.