Voltaic piles and multimeters

Factual Questions
1

I am tutoring some high school students, and we are building voltaic piles (an early form of battery). We’re investigating the effect of different metals and electrolytes on performance, i.e. voltage and amperage. When measuring voltage, we get very stable readings in line with what you’d expect from a copper/zinc cell. When we try to measure current, however the readout is ambiguous. The initial reading is on the order of about 100 uA, but it immediately starts dropping, and never settles to a constant value.
Yes, we have the leads plugged in correctly, and yes we have the multimeter set to read current in the right scale, and yes, we have tried other multimeters with the same result.

Is there some phenomenon that I am not aware of or taking into account? It would be nice to know how much current these cells are capable of producing. Any insights would be appreciated. Thanks in advance.

Bizz

2

Are you measuring current through some sort of load, or by connecting the ammeter leads directly to the battery terminals? The latter is essentially a short circuit across the battery, so it’s not surprising that you would see the current reading start out high and then drop off as the battery runs out.

3

That was my thought, I don’t know enough about electronics to recommend specific resistors, but I think you’re just killing the battery.

You didn’t say what kind of voltage you’re getting but I’ll bet if you attach an LED or small incandescent to it (like for a little flash light), it’ll light up but for a very short time. That should verify your meter reading.

To take this a step further, when you hooked up the ammeter, what are you trying to measure the amperage of? I’ll guess that you’re not sure. Typically you’re measure how much current is flowing though a load (resister, light bulb, motor, whatever), but since there’s not a load attached, the multimeter is just taking all the electricity the battery can give it and with a crude battery, it doesn’t have much.

If you set the meter to amps and stick it in an outlet you’ll probably blow a fuse (either in the meter or in the panel) too. Like andrewm said, you’re shorting the battery out. It’s like putting a paperclip across the terminals.

4

The measurements you are getting are absolutely normal.

I’m not a battery expert, but if I understand correctly they work somewhat like this:

-When a battery is standing idle for a while, ions (charged particles) from the electrolyte will stick near the plates or electrodes. This will create an electric charge on the plate surface that is opposite to the ion charge.

-When the battery is shorted, this stored charge will generate a big current. But as the charge is depleted the current will drop. The charge is regenerated by the chemical reactions in the electrolyte, but this is a slow process, so the current will drop to some low steady-state value.

This phenomenon is called the ionic resistance of the battery. See here some resources:

http://data.energizer.com/PDFs/BatteryIR.pdf

Also, when you are doing this kind of measurements, you have to take into account the burden voltage of your multimeter.

Multimeters measure current by making it flow through a precision resistor and measuring the voltage drop across it. At the uA range this resistor will be something in the order of 1Kohm or more. This can lead to weird readings if not accounted for.

See more about burden voltage here: Can you live with the burden? | Fluke

5

The “proper” way to evaluate each cell is to determine the open-circuit voltage and internal resistance. We sometimes call this the “Thévenin equivalent.” In doing so, you are building a first-order model of the cell.

You only need two resistors and a voltmeter to do it. It is not even necessary to measure the current with an ammeter. (Current can be calculated if you want to know what it is.)

Procedure: Stick a resistor across the cell (e.g. 100 kΩ) and measure the voltage. Stick a different-valued resistor across the cell (e.g. 50 kΩ) and measure the voltage. After some simple algebra you’ll have the Thévenin equivalent. For an even better model, include the resistance of your voltmeter, which is in parallel with the resistor during each measurement.

6

multimeters measure current this way- they have an internal “shunt” of a known, very low resistance. when you perform a measurement of current through a circuit, the meter actually measures the voltage drop across the shunt, then calculates the current via Ohm’s law.

so basically you’re not getting a meaningful reading because you’re measuring the current your battery is dumping through the shunt. As has been said, you’re more or less creating a short circuit.

7

Thanks to all. Sounds like the current is simply too low/meter is too crappy to get a meaningful reading. I suppose I could put 10 of them in parallel to up the amperage.

Or buy a decent meter.

8

No, the problem is not necessarily due to the low current or the quality of the meter. You could run into the same problems even with a top of the line 8.5 digit bench multimeter.

When measuring something with a multimeter you will inadvertently “load” it in some way. In some circuits it doesn’t matter and the loading effect will be of no concern.

But these home-brew batteries are extremely sensitive. You need a deeper understanding of how batteries AND multimeters work to make meaningful measurements.

9

To be really clear - there is no such thing as the “amperage” of a battery. You can’t measure it, as it isn’t a defined thing.

As has been described above, putting at current meter across a battery simply shorts it out. This isn’t a useful thing to do, and the readings you get are equally useless.

Crafter_Man has described the right answer. You can measure the Thévenian equivalent. A battery is definable as a perfect voltage source (ie one that can deliver an infinite current without the voltage changing) in series with a resistance. This equivalent is a fundamental way of describing any electrical device.

Measuring the two quantities - voltage and internal resistance gets you the two parameters that define the battery. The two together then define what the battery will deliver into any real life load - which is probably what you are trying to get a handle on with an attempt to measure the battery’s “amperage”.

Don’t be confused by many batteries having a rating of amp-hours. That is a measure of the total energy they hold. It isn’t a measure of the current the battery can deliver. The energy a battery holds is voltage times capacity. So a 1.5 volt battery that can deliver one amp for one hour holds 1.5 x 1 x 3600 = 4800 Joules. You can measure this - but you would need to kill the battery in order to do it.

10

I don’t think you’re understanding what’s going on here. If you make 10 of these and connect an ammeter from one end to the other you’re going to burn something out.

11

I think you’re seeing polarisation:

http://en.wikipedia.org/wiki/Primary_cell#Polarization

This is a well-known problem with simple batteries.

12

You can put another meter across the load to measure voltage.
If the current is dropping then you will see the voltage drop too.
As has been said, you do need a load. Start with a big one, like 100K and then substitute smaller and smaller ones to test. A tiny incandescent bulb will be the best load to use. You’ll be able to see the bulb brighten or dim.

13

An incandescent bulb won’t be a good load to use if you’re hoping to get measurements. The resistance of a bulb varies with temperature, and it heats up when it’s in use, so it’ll make your calculations much more complicated.

14

Yeah, I’ve always found electronics to be daunting. However, what I am really after is not absolute values as much as a relative number with which to compare different battery constructs. Voltage is a slam dunk, and I (er, I mean the students) can calculate the Thévenian equivalent and compare this number as we vary different parameters (such as electrolyte composition and metals). This will be a great way to talk about simple circuits and Ohm’s law. Thanks, everybody!

15

Good stuff: what is the “simple algebra” that I will need to do?

16

For heaven’s sake. POLARISATION

17

Thanks, Bert. I’ll keep that in mind.

18

The first-order model for the cell is an ideal voltage source in series with an ideal resistor. The former is called the cell’s Thévenin equivalent voltage, a.k.a. the cell’s open circuit voltage. We’ll call its voltage V[sub]th[/sub]. The latter is called the cell’s Thévenin equivalent resistance, a.k.a. the cell’s internal resistance. We’ll call its resistance R[sub]th[/sub]. Just like the schematic here. Both of these quantities are unknown for the cell. Measurements will make them known.

You will take two voltage measurements (V[sub]1[/sub] and V[sub]2[/sub]) with two different load resistors (R[sub]1[/sub] and R[sub]2[/sub]). In other words, connect R[sub]1[/sub] to the cell and measure the voltage (V[sub]1[/sub]), then connect R[sub]2[/sub] to the cell and measure the voltage (V[sub]2[/sub]).

During the first measurement, V[sub]th[/sub] must be equal to the voltage across R[sub]th[/sub] + the voltage across R[sub]1[/sub] according to Kirchhoff’s voltage law. The voltage across R[sub]1[/sub] is simply what you measure with the voltmeter (V[sub]1[/sub]). The voltage across R[sub]th[/sub] is determined using Ohm’s Law; the voltage across it is (I[sub]1[/sub])( R[sub]th[/sub]), where I[sub]1[/sub] is the current in the circuit. The current in the circuit can be calculated by applying Ohm’s Law to R[sub]1[/sub], i.e. I[sub]1[/sub] = V[sub]1[/sub]/ R[sub]1[/sub]. Therefore, the voltage across R[sub]th[/sub] is (I[sub]1[/sub])( R[sub]th[/sub]) = (V[sub]1[/sub]/ R[sub]1[/sub])( R[sub]th[/sub]) = V[sub]1[/sub] R[sub]th[/sub]/R[sub]1[/sub]. Thus

V[sub]th[/sub] = (V[sub]1[/sub] R[sub]th[/sub]/R[sub]1[/sub]) + V[sub]1[/sub]

Doing the same for the second measurement we get

V[sub]th[/sub] = (V[sub]2[/sub] R[sub]th[/sub]/R[sub]2[/sub]) + V[sub]2[/sub]

Now we have two equations with two unknowns (V[sub]th[/sub] and R[sub]th[/sub]).

Combining the two and solving for R[sub]th[/sub]:

R[sub]th[/sub] = (R[sub]1[/sub] R[sub]2[/sub]( V[sub]2[/sub] - V[sub]1[/sub])) / (V[sub]1[/sub] R[sub]2[/sub] - V[sub]2[/sub] R[sub]1[/sub])

Now that R[sub]th[/sub] is known, you can easily solve for V[sub]th[/sub]:

V[sub]th[/sub] = (V[sub]1[/sub] R[sub]th[/sub]/R[sub]1[/sub]) + V[sub]1[/sub]

Keep in mind, however, that V[sub]th[/sub] and R[sub]th[/sub] are dependent on everything. This includes the load resistance (current), temperature, amount of charge in the cell, phase of moon, etc. If you were to make these measurements with a set of resistors that were twice the values of R[sub]1[/sub] and R[sub]2[/sub], for example, you would get a somewhat different answer for V[sub]th[/sub] and R[sub]th[/sub]. Also note that time can be an issue; the voltage across the resistor will vary with time. Ideally, therefore, you want both measurements to occure within one or two seconds of each other. A smart way to do this is to use two resistors in parallel for R[sub]2[/sub]. Here’s an example: grab ya two, 100 kΩ resistors. During the first measurement, R[sub]1[/sub] will be one of the 100 kΩ resistors. During the second measurement, R[sub]2[/sub] will be both 100 kΩ resistors in parallel (R[sub]2[/sub] = 50 kΩ). The second resistor can be quickly connected across the first resistor using alligator clips, and thus the second measurement can be taken very quickly after the first measurement.

And for bonus points, include the input resistance of the voltmeter.