Rich_Insane
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Voltaic cell question
So say you've built a "battery" that discharges current and exerts an electrical potential. The current is modeled by a curve I(t), and is not very
large (less than 4 mA at its peak). The potential is modeled by a curve V(t) and is also rather small (peaking at about 1-2 V). Because these two
values are measured as functions, how can I calculate the net energy output of the cell?
What I've done in the past is that I've integrated the I(t) curve to get the discharge in Coulombs, then multiplied by an average potential -- but
there has to be a better way to do this. I've taken measurements by recording the time and switching between current and potential measurement on my
multimeter. I have tried reading them both at once, but I get interference if I do that (because both the current and potential are not very large).
Is there any way I could calculate the total energy output of this cell using mathematics via manipulation of data from my I(t)/V(t) curves?
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Nicodem
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Quote: Originally posted by Rich_Insane  | | What I've done in the past is that I've integrated the I(t) curve to get the discharge in Coulombs, then multiplied by an average potential -- but
there has to be a better way to do this. I've taken measurements by recording the time and switching between current and potential measurement on my
multimeter. I have tried reading them both at once, but I get interference if I do that (because both the current and potential are not very large).
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If you measured the current in a closed circuit and the potential in an open circuit, then you can just as well throw away these measurements because
they are of no good.
In order to obtain the energy released by the cell, you need to measure the power over time continuously until you get zero current. You then
integrate this over time to obtain the released energy. The simplest way to measure electrical power is to measure current trough a load and the the
potential at the cell which is identical to the drop in the potential of the entire circuit (the drop on the wires, the ampermeter and the load).
You could get an estimation by doing this with a single multimeter, if you ignore the potential drop caused by it when it is in the ampermeter mode.
Ampermeters must have minimal resistance and thus you can, to some degree, approximate it to 0 ohm. But you need to do measurements on the closed
circuit for obvious reasons: you are measuring the drop in the potential and there is none unless the current runs. This is a bit unpractical though,
as you need to disconnect the multimeter in its ampermeter mode, reconect the circuit, measure the potential, and then reconnect it again into the
circuit. To repeat this every half a minute and ignore the time of disconnection, would give you a pretty miserable measurement. So why don't you just
buy another cheap multimeter? You can get them for a dozen of euros, if you don't mind the quality.
If your cell would give more power, you could measure the energy calorimetrically, but a few mA is a bit too low for a homemade calorimeter, so this
is out of question. And you would still need at least one multimeter, a precise thermometer and a reliable power source for the calibration of the
calorimeter.
[Edited on 24/2/2013 by Nicodem]
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watson.fawkes
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| Quote: Originally posted by Rich_Insane | I have tried reading them both at once, but I get interference if I do that (because both the current and potential are not very large).
Is there any way I could calculate the total energy output of this cell using mathematics via manipulation of data from my I(t)/V(t) curves?
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Mathematically, you want to integrate I(t) * V(t), which is best done with simultaneous measurements. If
you're having trouble reading both at once, the problem is with your test equipment. Cheap test equipment has not-high-enough input impedance to make
good measurements. You can see this yourself by drawing a schematic with the meters present in the circuit as resistors. Lower the values of these
meter-resistors to, say, 10 KΩ and you can see it distort the signal.
To solve this, you can spend a bunch of money on new test equipment. Or you can do some electronics. Personally, I think amateur scientists ought to
know some basic electronics. What you have here is a classic case of input signal conditioning. You want to measure an input voltage and an input
current. Current is measured as a voltage through a sense resistor, which should be of as low a value as practical to minimize error. There's a part
called a "four terminal current sense resistor" (or some variation thereof) which you want to convert the current to a voltage. Here are some
application notes from vendors that Digi-Key carries: (*) Ohmite, A Guide to Current Sense Resistors (*) Vishay, High Precision Current Sensing. Discrete through-hole sense resistors start in the USD 5 - 10 range. Surface mount parts, suitable for you
because of low power involved are less that USD 1.
Now that you've got a sense resistor, you need two voltage measurements. To do this, use a pair of buffer-followers and whatever voltage measuring
gear you want. A buffer-follower is an amplifier with gain 1 (the 'follower') with adequate output current to drive less-than-ideal loads (the
'buffer'). These are off-the-shelf parts these days. There are JFET input op-amps with input bias currents measured in pA (or lower). For example,
compare the bias currents of an earlier chip like the 741, at 80 nA, with that of the OPA 140 at 10 pA. That's almost four orders of magnitude better,
and will outperform most modestly-priced test equipment.
So you make your input conditioning stage, and then you run it into an even modestly-price data acquisition system. The Arduino, for example, has
analogue inputs that would be perfectly adequate to get started. It's only 8 bits, but it's also only USD 30. Spend more money on data acquisition
if/when you need better measurements.
To learn electronics of this sort, I suggest The Art of Electronics, which was written by folks supporting physics research, so they pay
attention to instrumentation issues.
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ElectroWin
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if you measure the voltage drop while you discharge the cell across a known resistance, then you can calculate the current; so now you have both
voltage and current curves, without the noise.
with 2V and 4 mA peak, this implies about a 500 ohm internal resistance in the cell (V/I= R). i suggest using a 1 k ohm resistance to discharge it.
total series R will then be around 1.5 k, and you can run it until the voltage is, say, about 0.1 V
power is then calculated as P(t) = V(t)I(t) watts and energy in joules is found by integrating power wrt time
[Edited on 2013-2-24 by ElectroWin]
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Rich_Insane
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I am using relatively high-end equipment -- Though I still think they were intended for higher potentials and currents. The meter is called a
"DM-501A" multimeter built by Tektronix. Theoretically, yes, I am measuring the potential in an open circuit (because theoretically the voltmeter has
an infinite resistance, if I recall correctly), but I believe that there is some finite internal resistance within the voltmeter. Same goes with the
ammeter setting.
Here is what I gather from this thread -- please tell me if I am correct:
I need to attach a load to the circuit and measure the potential drop across the resistor. With watson's suggestion, I gather that this resistor
should be of a higher Ohm value than the cell.
This is the circuit I designed:
Where the resistance R1 is greater than the internal resistance of the cell. That's about as complicated as I can really comprehend. I've only covered
DC circuits in physics, and am in the process of learning AC circuits (RLC circuits) via differential equations and the like. Never in my life have I
seen so many imaginary numbers in one place 
I can use Kirchoff's Loop Law such that 0 = V(source)-IR(internal)-IR1. This will allow me to calculate the current or the EMF from the cell.
Or should I attach a load and measure directly across the two terminals of the cell?
| Quote: |
You could get an estimation by doing this with a single multimeter, if you ignore the potential drop caused by it when it is in the ampermeter mode.
Ampermeters must have minimal resistance and thus you can, to some degree, approximate it to 0 ohm. But you need to do measurements on the closed
circuit for obvious reasons: you are measuring the drop in the potential and there is none unless the current runs. This is a bit unpractical though,
as you need to disconnect the multimeter in its ampermeter mode, reconect the circuit, measure the potential, and then reconnect it again into the
circuit. To repeat this every half a minute and ignore the time of disconnection, would give you a pretty miserable measurement. So why don't you just
buy another cheap multimeter? You can get them for a dozen of euros, if you don't mind the quality.
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That's exactly what I've been doing: I've used a single multimeter and switched between modes. I get three measurements (data points per set). The
multimeter is connected directly to the cell. There is no load attached.
When I used two multimeters (no load), I got ridiculously low measurements, somewhere in the order of 10 mV and 10 uA initially, then reaching a
maximum around 63 uA and 63 mV (I noted that the current and the potential had exactly same values but different magnitudes, which I assume is an
effect of the internal resistance in the meters). When the current dropped to zero, the potential shot up to about 200 mV.
After trying several times to smooth out this behavior, I gave up and switched to the single-multimeter method. This worked and gave me the readings
that I get today, but now I cannot accurately integrate the two curves.
If I do a polynomial fit and get an acceptable R^2 value (>0.95), then multiply the two obtained functions for V(t) and I(t), will I get an
accurate P(t) curve that I can integrate? I was under the impression that I needed an IV curve to do this, and in order to formulate an IV curve, I
need simultaneous measurements of potential and current -- a problem I have wrestled with for a while.
Edit: I just realized that because my I(t) curves are transient, I may face difficulty applying DC circuit rules to this situation. A friend of mine
suggested that I use root mean square values for current and potential and then utilize the i^2R1 relationship to determine the practical power
output.
I also wonder if my R(L) could be something like a low-resistance LED. This will make it easier to measure the current via ammeter... Then somehow I
can calculate the potential in the circuit, across the cell this time. The main advantage I have is that I know that the cell has a very high internal
resistance (several kilo Ohms at the least).
[Edited on 24-2-2013 by Rich_Insane]
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watson.fawkes
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Good quality test equipment will easily measure µV and µA. Once you get into nV and nA, well then you need to
start really paying attention. | Quote: Originally posted by Rich_Insane | Theoretically, yes, I am measuring the potential in an open circuit (because theoretically the voltmeter has an infinite resistance, if I recall
correctly)
[...]
When I used two multimeters (no load), I got ridiculously low measurements, somewhere in the order of 10 mV and 10 uA initially
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If you're measuring under no load, then you're going to get unrealistic answers. In particular, with no
load, you're not measuring power delivered to a load.
Now when I first replied, I thought you were measuring under load, and a variable load at that, and needed some assistance with that. You
have a much easier problem. Put in a simple resistive load and measure the voltage drop across the load. (This what @ElectroWin suggests.) Calculate
the current through the resistor with Ohm's load. Calculate your power accordingly.
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Rich_Insane
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Ah, yea, the reason I didn't want to use a load was that I wanted to get an idea of current discharge on my initial prototypes, but now that I'm in my
second phase, I think adding a load would be much more suitable and realistic.
So I should simply do what I did before using 2 multimeters, and then map the I-V curve (use a parametric function?), then integrate the IV curve? Or
should I just integrate P(t) = I(t)*V(t)?
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watson.fawkes
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This one.
And learn what Ohm's law really states which is that I(t) and V(t) are directly proportional for a constant resistance load.
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Rich_Insane
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| Quote: |
This one. And learn what Ohm's law really states which is that I(t) and V(t) are directly proportional for a constant resistance load.
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Ah, that makes sense. This must be why the I(t) and V(t) signals for AC power sources are essentially the same function, perhaps with a slight
lead/lag. And in an AC circuit, Ohm's law applies only if the reactance from inductors and capacitors are zero, otherwise we need to make a
modification and substitute R with Z, the impedance.
Am I right?
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