Building and Testing a Penny Battery

This version of a battery hearkens back to the original voltaic pile invented by Alessandro Volta in 1799. Here I will be showing two versions using different anode materials. In the diagrams I've only shown batteries composed of 5 cells, but the picture of the actual battery is of a 10 cell version and that is what the instructions are for.

There are quite a few instructables about building these, but they all were severely lacking graphs and data. Not to worry, there will be plenty of both in this one!

Materials List

• 10 copper pennies (pre-1982)
• Aluminum Foil (if using anode type 1)
• 10 copper plated zinc pennies (post-1982) (if using anode type 2)
• Dremel or sandpaper (if using anode type 2)
• Cardboard to be cut into small pieces
• 3D printed case (or other holder such as these, or just rubber bands)
• Wires to connect to the electrodes
• Solder
• Solder Flux
• Soldering Iron
• Multimeter
• White Vinegar
• Salt

You can find the stl file for download here: http://www.thingiverse.com/thing:1798322

The crossbeam is used to hold the pennies together though it may take a bit of sanding the edges for it to slide through the notches in the holder.

I tried two different materials for the anode electrode, aluminum and zinc. This is the electrode that will form the "negative terminal" of the battery. For the cathode I used copper in both cases, this will be the battery's "positive terminal." In my experiments I found that the electropotential difference that develops between aluminum and copper is about 0.58 volts, while between zinc and copper the potential difference is about 0.9-1.0 volts. Clearly zinc is the better choice, but the aluminum is easier to obtain since you can just use foil from your kitchen. (For voltages between other combinations of material, see appendix 2).

To prepare to use aluminum anodes, fold a strip of aluminum foil several times till you have a long piece about as wide as the diameter of a penny. Cut squares of the strip then cut each corner of each square to round out the shapes. Strip the end off a wire, twist it into a loop and twist/smash some foil around it to use as the last anode. I've tried soldering to aluminum and had no luck at all.

To prepare to use zinc anodes, get 10 pennies (USA one cent coins) MINTED 1983 OR LATER. In 1983 the US mint started making pennies as copper plated zinc. Use a dremel or just sandpaper to grind off the copper on one side (it doesn't need to be perfect). Solder a wire to the copper side of one of them, it helps to clean it first and use lots of flux as well as an extra hot iron. This coin will be the last one in (see the last image).

To prepare the copper cathodes, get 10 pennies MINTED 1981 OR EARLIER. These are made of 95% copper. (Note: In 1982 the mint made both types of penny, so avoid that year.) Solder a wire to one of them, it helps to clean it first and use lots of flux as well as an extra hot iron. This coin will be the first one in (see the fourth image).

Use a mixture of 5% white vinegar (normal vinegar you can find at walmart) and a few teaspoons of table salt for the electrolyte solution. Then cut 5 pieces of cardboard to just a bit smaller than a penny. When the cardboard is between the layers is should only touch the layer just above and just below, and it should be thick enough to prevent those layers from touching each other.

Fill part of a glass with vinegar and mix in a few teaspoons of salt. Drop the cardboard pieces into the solution to soak for a minutes or two before placing them in the battery.

If you are using aluminum anodes, you want to be extra careful that the the piece of aluminum isn't too small which would let the cardboard and the copper cathode on each side of the aluminum touch each other.

For the copper and aluminum battery, follow the first two images. Be careful to center each piece of cardboard so that it only touches one electrode on each side. The voltage per cell is about 0.58 volts. Multiply by 10 and we get 5.8 volts, almost exactly the total output measured by the voltmeter.

For the copper and zinc battery follow the second two images. The per cell voltage is between 0.9 and 1 volt which when multiplied by 10 gives us 9-10 volts. The output of 9.4 volts fall nicely in that range.

Connect up an LED and see if you get light! Due to the high internal resistance of the battery, there is no need to connect a resistor in series with the LED.

In the second pictures the LED is pulling 0.21 mA. This is after a few days of the battery sitting around and then being soaked again in electrolyte, so you will probably see a higher current with a fresh battery.

I found that the voltage would be sustained for a day or so before dropping. Just dunk the whole thing in the electrolyte solution and let it soak for awhile. In the picture you can see the zinc and copper version. When submerged in electrolyte it is the same as having 10 cells in parallel so it only generates 0.97 volts, but the internal resistance will be much lower.

After pulling it out let it drip dry for a few hours and the voltage will climb back up.

Now that we have built a battery we would like to know how useful it is. This brings us to my favorite part, making graphs!

If you built the battery using aluminum sheets with 10 copper pennies, you should measure a total series voltage of 5 - 6 volts. If you used the zinc and copper (20 total pennies) you should measure between 9 and 10 volts. But if you try to use that to power something and keep your multimeter connected, you will notice that the voltage quickly decreases. This is due to the internal resistance of each cell in the battery.

We can think of the battery as a perfect voltage source in series with a resistor, as is shown in the first picture. (A perfect voltage source is one where the voltage doesn't change when the load changes.) If we measure the "open circuit" voltage with multimeter this gives a very good approximation of this ideal voltage, since in this mode the multimeter draws an extremely small amount of current.

To find this internal resistance value we need to determine the voltage drop across that resistance for various currents. If we measure an open circuit voltage of 10 volts and then when pulling 10 mA we measure a voltage of 9 volts, that means there was a 1 volts drop across the internal resistance. From Ohm's law we can determine that R = V/I = 1/10e-3 = 100 Ohms. Now of course we don't want just one measurement, we want to take several do a linear fit to the resulting data set of voltage drop vs current. Since V = IR is a linear equation, the slope of the best fit line will be the internal resistance.

Looking at the graph you will see 5 different data sets. There are a few interesting things to notice here.

• Look at the difference between the top two lines (orange and green). These show the difference made when adding salt to the vinegar, which turns out to be about a factor of 3 reduction in the internal resistance.

• Comparing the blue and green (both using copper and aluminum) shows that adding cells in series does indeed lead to an approximately linear increase in the internal resistance. (In other words each cell has about 1 kOhm of internal resistance and 4 together gives us between 4 and 5 kOhms while 10 together give 10 - 11 kOhms.)

• Comparing green (10 cells Cu and Al) and purple (10 cells Cu and Zn) we see that the Zn electrode not only increases the overall voltage but decreases the internal resistance to about three fifths the amount when using Al.

• The yellow line shows what happens when the whole battery is dunked in the electrolyte solution, effectively connected all the cells in parallel. Or we could think of it as one cell with 10 times the electrode surface area. This leads to a 20th the internal resistance, although of course a max voltage of only 0.6 - 0.7 volts. I was honestly expecting it to be a tenth the series resistance, if anyone has an explanation let me know.

The main take away here is that in any configuration this battery has a TON of internal resistance.

For the tl;dr, see the bold sentences.

As we saw in the graphs section, even the 9 - 10 V version of this battery can't provide anywhere near that voltage when even a few mA of current are being drawn. To actually use these to power a phone we need to combine many of them in parallel. How many? I'm glad you asked!

We know for a phone charger we need 5 volts. Let's call 'a' the number of cells in series (aka how many cells make up a single "battery") and 'b' the number of those batteries which are connected in parallel.

The Zn version has lower internal resistance (although it does take twice as many coins).

Let me define a few variables

• a = number of cells in a given battery (cells in series)
• b = number of batteries connected in parallel
• Vc = single cell open circuit voltage = 9.48/10 = .948 volts
• Ic = single cell current
• Rc = single cell internal resistance = ~6400/10 = 640 Ohms
• Vp = load voltage = 5V
• Ip = load current, USB 2.0 standard is 500 mA so let's go with that

From the voltage loop rule looking at a single battery and the load we get

a*(Vc-Rc*Ic) - Vp = 0

And from the current junction rule we get

Ip = b*Ic

Solve the second equation for Ic and plug into the first equation, then solve for 'b'.

b = (Rc*Ip)/(Vc - Vp/a)

So now we have a equation for the number of batteries we would need as function of the number of cells in each battery. Let's plot this on wolframalpha.com using the values discussed above. The y-axis is b, the number of batteries in parallel, and the x-axis is 'a', the number of cells in each battery. (If it says the computation time is exceeded, just refresh and it should work.)

We can see that when there are 10 cells each battery we are at a value of b = ~710.

The total number of coins is a*b*2 so we would need about 14200 pennies to charge a phone at 5V and .5 Amp if each battery consists of 10 cells (20 coins)

Of course the next logical question is what is the optimal ratio of series to parallel? For this we want a function for the total number of coins.

Total number = a*b = a*(Rc*Ip)/(Vc - Vp/a)

If we then plot this function and look for the location of the minimum, or better yet, the derivative of this function and look for where it equals zero, we will find the most efficient number of cells per battery.

It turns out that for this case, the best number is 10.5. Obviously we can't have half a cell, so 10 or 11 would be the most efficient. (If we use 11 cells per battery it turns out we would need 650 batteries, or 14300 coins. So 10 is very slightly better in terms of number of coins, although it would probably be easier to build 650 batteries with 11 cells than 710 batteries with 10 cells.)

So for just $142 (plus cost of 710 printed cases, bits of cardboard, and an obscene number of wires) you could build yourself a phone charger that would probably work pretty okay.

Interestingly, it turns out that the optimum number of cells in a battery is pretty much only dependant on the per cell voltage. If you were to use the Al and Cu version with a per cell voltage of about 0.58 volts, the optimum number of cells in each battery would be almost exactly 17. Of course in this case each cell is one coin and one piece of aluminum, whereas the Zn and Cu version each cell is 2 coins. Now, the internal resistance will still affect the total number of batteries needed since each will be able to supply less current if the internal resistance rises. So for the Al and Cu version the optimum number of coins comes out to by (17 cells/battery)*(1820 batteries)*(1 coin/cell) = 30940 pennies, or $310 plus the cost of all the aluminum, etc....

I tested eleven materials in all combinations of electrode pairings. That would be 66 total combinations. The highest voltage (1.7 V) resulted from using graphite (pencil lead) and magnesium as the electrodes. Although magnesium reacts quite vigorously with vinegar so it isn't a very practical choice here.

A note of warning: When doing this, be sure that only the metal you are testing is touching the liquid. I've seen a bunch of people online just dunk the whole alligator clip in along with the test material which since the alligator clip is metal it will act as an electrode which will mess up the results.