Science Fair Project - Battery Tester - Does common ground compromise results?

[adrian1906]

I'm helping my son with his science fair project to compare the useful lives of different batteries. I helped him build a data logger using an Arduino micro controller board which as 6 Analog Input channels (the board is powered via the host computer's USB port). Originally, the circuit was designed to handle 3 batteries (A0,A1) for battery 1, (A2,A3) for battery 2, and (A4,A5) for battery 3 and log the differentials with the negative terminals connected to A0,A2,and A4. Unfortunately, I was having difficulty getting reliable values for the voltage difference computed using the pseduo command: voltage =(AnalogRead(A1)-AnalogRead(A0)) * 5volts/1023 (5 volt reference on the 10 bit A/D converter).

So, in my desire to move along, I changed the circuit. I connected all of the battery negatives to the ground of the powered board and each + terminal to A0=>A5 (This configuration allows 6 batteries to be tested at once). Flashlight bulbs are being used to drain the batteries. Each bulb had one connection to ground and another to the Analog Input channel (via a breadboard). The data logger works. My son is happy. I'm happy that he's happy and even more delighted that understood what I did AND took a liking to it (Still keeping my fingers crossed).

Nevertheless, I woke up this morning wondering if the intent of the tests was compromised. Since the bulbs and batteries are no longer independent (common grounds), is it possible that the battery capacities could be modified (ie, one battery pulling current from another)?

 

[IRstuff]

an electrical circuit requires TWO connections

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[adrian1906]

Thank you for your response IRstuff.

Are you saying that the intent of the tests is not compromised because each battery/light pairing forms its own independence circuit even though the negative terminals of each battery is connected to the computer's ground?

 

[IRstuff]

Yes

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[benta]

One thing to watch for: keep the Arduino powered all the time.
If you remove power, there is a risk that the batteries might discharge through possible protection diodes in the Arduino.
Or, remove the connections to the batteries before powering down.

Regards,

Benta.

 

[adrian1906]

Thanks for the tip Benta.

 

[OperaHouse]

One thing I found with the UNO is when you plug in a laptop the 5V voltage reference changes because Laptop power is generally higher than onboard regulator. I use a USB cable with the 5V wire cut I use a simple multiplier routine to lower A/D noise with a running average. With that multiplication and slight adjustment to voltage divider the count comes out in mv.

// Read the voltage, create a sample of 10
Volt = Volt - Volt / 10;
// Sum the readings by subtracting one average reading before conversion
Volt = Volt + AI4;

 

[adrian1906]

Thank you OperaHouse,

For a long time I was getting incorrect readings. It turned out that my reference was 5.07V. I like the idea of averaging before writing. Since you have some experience with the UNO, have you ever tried taking a voltage differential? When I did it, I got random values. (Imagine connecting your battery such that A0==>negative and A1==>positive) What should be the expected result? I do know that if I have an open pin, the values will be random. with values of 10s of mV. I wonder if this is akin to connecting the negative terminal to the A1 in which case, if A1 is random, then A1-A0 will also be random...hmmm. I think my current solution works but I am still puzzled why the differential mode did not give me the expected results.

 

[OperaHouse]

That is the same as a floating input. A floating input will ghost a prior A/D reading. Even a fairly high K resistor tied to ground will prevent this.

 

[itsmoked]

Using bulbs is a poor choice. For several reasons. "Bulbs" are not made to be very accurate. Their resistance, and hence load, changes dramatically with voltage, current, temperature, and how you even look at them, in a nasty non-linear fashion. You should be using fixed resistors as loads.



Keith Cress
kcress -

http://www.flaminsystems.com

 

[jimkirk]

Also, mind your ground loops. If you connect the loads directly across each battery and then connect the negative of each battery to ground you should be okay. If there is some distance between the loads and the batteries, there can be some I*R voltage drop adding an error, it all depends on the exact topology of your circuit.

And like Keith said, bulbs are not very precise. Some appropriately rated power resistors would be better.

You could use the PWM outputs to control some LEDs to show the battery state.

 

[OperaHouse]

Instead of differential input, ude that extra A/D input to monitor lamp current through a low ohm resistor. Then actual wattage from battery could be calculated. Lamps are a bad choice but for a science fair you need some type of visual.

 

[adrian1906]

Thank you for the suggestion to use resistors instead of flashlight bulbs. I picked up 4 10ohm power resistors. Now I have another set of questions....

When I connect the DMM across a battery carrier holding 2 AA batteries, I get the expected 3V. However, when I connect it across a 10 Ohm power resistor, the voltage across the resistor drops to 2.85V. If I remove the resistor, I get the 3V again. How is this explained?

Also, this next question may or may not have the same answer.

One of my son's friends is doing his science fair project on vegetable and fruit batteries so they wanted to compare the two types. An orange gave a voltage reading of .89 on the DMM. When we connected the 10 ohm power resistor across it, the voltage dropped to .1V! As I increased the resistance, the voltage began to approach .89 V.

My guess is that V=iR assumes that the power source has the capability to produce any current, and the problem I'm running into is a limitation on current producing capability. If this is the case, is the current (as measured using the DMM voltage and resistance), considered the maximum current the power source can produce?

 

[IRstuff]

The battery's material and electrolyte behaves like another resistor in series.. The drop will be smaller for a larger external resistor because there will be less total current, hence less I*R drop

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[adrian1906]

Thank you for clearing that up IRstuff.

So, re-arranging the voltage divider equation, the internal resistance can be determined from Ri=Re(Vin/Vout-1)

This would put the internal resistance of the battery 10*(3/2.85-1) = .52 ohms and that of the orange at 10*(.89/.1-1)=79 ohms. I guess it would be interesting to see how the internal resistance changes with the spacing of the electrodes (at least for me).

 

[IRstuff]

I was, like, "I don't remember an orange-colored battery?" But an actual orange would have pretty high resistance, since the path through the bulk of the orange involves the pulp, etc., and the orange is not that acid, and there aren't enough n*electrode*area.

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[itsmoked]

With an orange you have those sectional dividers that are sort of like insulators.

I hope you're using dissimilar metals as the two electrodes.

By the way, you can see that same 'battery internal resistance' with a car when you crank the starter. The 13.2V battery voltage will often drop several volts. To minimize this the batteries have a huge number of parallel plates in each single 'cell'. More plate area = less internal resistance.

Keith Cress
kcress -

http://www.flaminsystems.com