Best Method for Accurate Measurement of DC Current?

Google "current shunt resistor". These are high precision resistors made especially for measuring current. Choose a resistor that will develop a small (insignificant to your load) voltage. 50mV is a common value. Measure the voltage across the current shunt resistor, divide by the resistance and you have your current.

For a maximum current of 5mA, you would need a 10Ω resistor, the current reading would then be V_resistor/10. So a voltage reading of .01 V would correspond to a current of 1.0mA. The accuracy of your measurement is determined by the accuracy of your resistor (and you should be able to find one with at least 0.1% tolerance) and the accuracy of your means of reading the voltage.

You can read the voltage with a high accuracy digital volt-meter, and if you need the process to be automated, these can often be read by a computer.

Hello,

I normally use LEM current sensors for the stuff I do. You should take a look at the LA 25-NP/SP11 from LEM. You can find them at: http://www.lem.com/hq/en/component/option,com_catalog/task,displaymodel/id,90.08.19.011.0/

These are very good current sensors. You can normally find them at digikey too so it's easy to get them.

The "small load" can itself be your current-sensing element if the load is a precision resistor whose value is accurately known.

In this diagram your "small load" is shown as a resistor with a digital multimeter (DMM) connected across its terminals. When set to measure voltage, a digital multimeter has a very high input impedance (the higher the better) which minimizes its influence on the measurement (a meter having a lower input impedance diverts some of the current from the load to the meter, upsetting the measurement).

A battery can be modelled as an "ideal" voltage source (voltage stays constant regardless of the current, implying a zero source impedance) in series with the battery's internal resistance, Rb. As the battery discharges, the value of Rb changes. How Rb changes over time is a complex function of the battery chemistry. There are ways to measure a battery's instantaneous internal resistance, so let me know if you're interested in making such measurements.

Two resistors labelled Rlead represent the resistances of the wires connecting the battery to the load. As part of the model, all leads in this diagram are considered to be ideal wires (zero resistance) to which the actual resistance of the wire has been added as a lump resistance. As these resistances can influence your voltage measurements, you'll want to connect the meter as close to the load resistor as possible. Do not connect the meter at the battery itself.

If your load is (or can be made) a simple resistance, use a load resistor whose value is known to the accuracy demanded by your experiment. If you know the load resistance accurately, you can use the voltage measured across the resistor to computer the energy dissipated by it. Consider Ohm's Law (units in volts, amps, ohms, seconds):

E (voltage) = I (current) * R (resistance)

R is known, and E is measured, therefore:

I = E/R, so

I = 1.000 volt / 1000.00 ohms

I = 1.00 * 10^-3 A, or 1.00 mA.

Since you want to know energy dissipated by your load, you can derive it:

P (power) = E²/R, so

P = (1.000 volt)²/1000.00 ohms

P = 1.00 * 10^-3 W, or 1.00 mW (milliwatts).

As power is the rate of energy transferred, the total energy dissipated by the load is the power integrated over time. Assuming the voltage is constant over the measurement interval, integrating the power is reduced to multiplying the power times the duration of the interval:

E = P * T (time, in seconds)

E = 1.00 * 10^-3 W * 100.0 seconds

E = 100.0 * 10^-3 J, or 100.0 mJ (where 1 watt-second = 1 joule)

Perform your experiments at the same ambient temperature, if possible, so that your temperature-dependent load resistance remains constant (and at a temperature close to which the resistor's published value was determined, if known). As a chemist, you already know that temperature affects battery chemistry and reaction rates.

If you can use a precision resistor as your load, you might consider selecting a value between 200 and 2000 ohms so that you don't load the battery too much. Measurements of a lightly-loaded battery tend to more accurately reflect the battery's actual capacity. Published battery capacity figures aren't necessarily accurate because actual battery capacity is heavily influenced by the discharge rate - but many battery mfrs don't tell you this.

Using a higher value of resistance also reduces the tendency of the resistor to self-heat, which may change its actual resistance over the measurement interval. Attaching the load resistor to a generous heat sink will stabilize its temperature so that it better tracks ambient temperature.

Hope this is of some help.

-e

Your circuit is the most practical,simplest,best.

User must mind the reading on the DMM and adjust a series resistor(marked in your circuit'load resistor' of 1000 Ohm("Pot,volume Control") 1Watt+ rated to keep the load current steady. Your battery now better be 3 Volts.

DMM will flicker every 1/10 second or so--0never mind--take your mental average reading.DMM ranging will naturally be set to DC Volts

"There are ways to measure a battery's instantaneous internal resistance, so let me know if you're interested in making such measurements."

Yes, I would love to know about that. I have wondered how to measure the resistance of a battery. When I take the simple-minded appraoch and simply clip my multimeter' s probes to the battery terminals, I get two different resistance readings depending on which of the two possible directions I attach the probes. And one of the measured resistance values is negative! Obviously something is wrong with this approach. Electron flow originating from the battery EMF confuses my meter?

My intuition tells me that the correct method must involve something like using a potentiometer to apply a potential to the battery, slightly greater in magnitude but opposite in sign, causing a small current to flow in the direction opposite to the battery's normal flow. Then we could calculate the battery resistance from the voltage drop. Am I even close?

At those current levels there really isn't any need to go to a lot of trouble with external shunts. Buy or rent a good digital multimeter. You can get quite good results with a three and one half digit meter (one part in 2000 resolution) or go for a six and one half digit version if you need that degree of resolution and can pay the extra money. If you need you can get a lab to calibrate it traceable to standards.

There is actually nothing called a DC Current as it is all electrons when you come down to lowest limit and high precision etc. Just count their numbers for a time interval, correct for some escaping the counter device and add up that statistical error on numbers and number of times you repeat the experiment for averaging. I have seen some meters going down to 0.1fA limit. I already have 1fC charge amplifier for 1us pulse so it should measure.

Can any one suggest a way to count the electrons moving through a conductor in numbers? I have method of counting only free electrons.

What current range did the poster say? 1 to 5 milliamps?

Shyam intones: "There is actually nothing called a DC Current as it is all electrons when you come down to lowest limit and high precision etc."

Uh huh. Thanks for the clarification. I was confused.

And there's no such thing as a beach. Right? Just individual grains of sand.

Ain't no such thing as a herd, either. Just individual animals.

No rivers. Just water molecules and a bunch of other crap.

Nor oceans. Ditto.

Nor air.

Not even this computer I'm typing on. Just individual thingies.

As a matter of fact, there's no such thing as a collective noun (except perhaps, "A horde of prostitutes.")

Shyam wisely suggests: "Just count their numbers for a time interval,..."

You count them.

"...correct for some escaping the counter device and add up that statistical error on numbers and number of times you repeat the experiment for averaging. I have seen some meters going down to 0.1fA limit. I already have 1fC charge amplifier for 1us pulse so it should measure.

Can any one suggest a way to count the electrons moving through a conductor in numbers? I have method of counting only free electrons."

This is soooo relevant, like. I'm sure the poster now knows exactly what to do.

Great post.

For another thread.

-e

After reading you collection of ASCII characters I'm ROTFLMAO.

Shyam asks: "Can any one suggest a way to count the electrons moving through a conductor in numbers? I have method of counting only free electrons."

----------

If you know how to count free electrons, and if you know how to measure electronic charge, and if you understand conservation of charge, then you know all you need to know to answer your own question.

-e

Hi John

Congratulations on becoming a Guru!

I do not trust that capacitor bucket too much as it gives an uncertainity of about 1000 electrons in 1us measurement time interval. I do not have a way to make the electrons free from conductor at low currents. Do you have a way out? Can you make them to bump out n the air like electrons from a beta source? Electrons in copper that move at room temperature have multiple wavefunction occupancy as energy levels are very close packed in metal for the so called free electrons. I am not sure if we can pull them out using static field to count them correctly. Actually pulling out is more a function to the electrons hitting from outside and the electroc field. See what happens in Channel Electrons Multiplier.

I never trust those constant current ideas. It is hard to believe that electrons will keep equal distance and will arrange themselves like a crystal lattice or like Military march past while they move inside jam packed metal atoms. They may not know each other and perhaps may be bumping in each other like unmanaaged out crowd. In supper conductors these electrons can make pair and walk uninterruped, and it sounds like a magic show.

Perhaps under tunneling we may be able to count electrons through conductor. Any other idea?

John?

Only the horde of prostitutes calls me that.

-e

Are you making any money at all?

Fantasies do not bring in cash.

Special abilities bring a lot of money and brain storming in new thinking works much better. While it all right to just do what is needed for you so every one does small things.

If I have to measure the battery capacity then I will kick peak current using 100A MOSFET having lowest possible resistance. This will require only 100ns for me to measure the impulse current. Having done it multiple times I can get reasonable estimate of the battery capacity to handle charge.

Most of the batteries run into problem if drained fully. Hence, set a limit on the minimum charge or voltage that it may hold on.

If you do not have the above technology then simple methods work fine. I use DEI MOSFETS that have 2ns turn ON time and can run huge current.

I care for each electrons when I talk about small battery. Most of the specialists using low power circuit watch out for nA or even pA current.

I do not need money for food as Govenment will pay for my entire life now. I have done enough work for them for it. I agree that many of you should do something that can bring bread on the table. In early time knowledge was not sold and now it has become a way to survive which is not wrong at all.

Fuel gauge ICs were developed by Maxim and you can find these at www.maxim-ic.com

Remember that your friends are electrons and all that DC or AC stuffs are cooked up stories in simplicity in which many people build their houses now. I am not free from it.

Money devil has taken over the mind's of the people is a known fact. Watch out.

MM wrote: "Fantasies do not bring in cash."

Sure they do. Maybe not in engineering or the sciences, but think of all the science fiction, fantasy, pseudo-science/religious cults Out There. Entertainment and Dysinformation = Big Business.

Some of these people even believe their own stuff, may the Farce be with them.

-e

For your application I would use a purely resistive load that you knew the characteristics of. With a purely resistive load you will not need to know the current but can rather calculate the power as

Power = Voltage² ÷ Resistance

Keep in mind though that there will be a certain amount of non-linearity to the current through a resistor due to things like thermal expansion etc. These errors can however be factored in to the fial calculation provided you know the characteristics of the resistor you use as a load.

Finally the best way to measure the voltage would be to use a Digital Multi Meter with either a USB or RS232 output so that the readings could be taken automatically by a PC. By automating the process you can use a much shorter time base and take more readings than by doing it manually. The more accurate the time factor the more accurate your final calculation of energy will be.

Yep! With a handle on the method, the next step is to automate the process. Automating it makes life worth living.

If I were the chemist, I'd let the battery drain all the way down while the instrument continuously samples the voltage at regular, short, known intervals, hands the values off to the computer to compute power and energy at each sample, then sums it to running total. Let the experiment run until some time period has passed or until the battery voltage drops below some minimum value.

Better yet, use two instruments: one set to measure voltage across the battery (yes, The Battery, not the load resistor), and one set to measure current through the load (and lead resistances, of course, since they're all in series). The computer then multiplies the two instrument readings at the end of each measurement interval to get the power, multiplies the power by the duration of the measurement interval to get the energy, and sums energy slice to the running total - the total energy supplied by the battery. This approach is relatively independent of temperature variations insofar as the resistor is concerned, and although a resistor is still used for the load, it need not be a precision resistor. The chemist may still wish to perform the experiment at a relatively constant ambient temperature insofar as the battery chemistry is concerned.

-e

E. Energy = E. Power * Time.

E. Power =

= Voltage * Voltage / Resistor =

= Intensity of DC Current * Intensity of DC Current * Resistor =

= Voltage * Intesity of DC Current.

I have that $4 multimeter which measures the current by filling the capacitive bucket number of times and emptying out each time and storing the numbers for a period of about 100ms. You can call it a dual slope ADC and this should be available on street shop anywhere. You can also use galvanometer type current meter for 0-10mA range for about $1.

For more accurate measurement try SANWA PC5000 multimeter. If you want to be 99% sure of the reading then take it to NIST for their help.

The original poster wrote: "I would like to know the most accurate approach..."

Shyam informs: "I have that $4 multimeter... and this should be available on street shop anywhere." and "You can also use galvanometer type current meter for 0-10mA range for about $1."

I understand that NIST is looking for a Senior Metrologist. Have you submitted your application?

-e

I must apologize for missing your upgrading to Guru, well done. This is no trivial achievement and is born out by the fact that you are Guru number 7 out of some 3,900 CR4 members.

Thanks, Masu!

Quite honestly, I just assumed this "standing" was simply related to the number of posts. A motor-mouth like me sooner or later reaches a threshold that flips a bit somewhere and voila, I'm now a "guru."

I gather it doesn't work this way?

-e

Leo,

The 2 or 4 terminal inline resistor of say .001 Ohm value is useful for this measurement and a DVM or DMM (Digital Volt or MultiMeter is placed on the terminals and the I=V/R). Also a chart recorder could be used for trends and changes recording over time. I am sure you can instrument for computer monitoring/logging as well.

I historically as a power supply design engineer have used successfully to measure all levels of current. Sometimes a clipover ammeter is used.

Companies making said resistors are Ohmite, Vishay, IRC for example. Should be able to locate via DigiKey, Allied, Newark, or TTI/Mouser for example. Not likely via Radio Shack.

Ron

I want to thank bhankiii, Guest, europium, rcapper, masu, Errikos, Shyam, and RonLohrbach for their very helpful replies. After considering the feedback from this group and others, I have concluded that I have 3 or 4 feasible ways to accurately measure the electrical current produced by my experimental battery:

-----

Multimeter/Ammeter -- just place an ammeter in series with the load. Digital meters are a few percent more accurate than the analog type, but the analog meter has the advantage of being less affected by electrical noise. If I use this approach, then I need to correct for the internal resistance of the multimeter's ammeter circuit.

http://en.wikipedia.org/wiki/Multimeter

-----

Current Shunt Resistor -- high precision resistor made especially for measuring current. For a load of about 5 mA, I would use a 10 ohm resistor (whose voltage drop would be insignificant compared to that of the load). Then since current equals resistance divided by potential (amps equals ohms divided by volts), I need only measure the voltage drop across the resistor to calculate the amperage flowing through the circuit. I hear that the "Watt's Up" and "Doc Wattson R102" digital meters contain built-in precision current shunts.

http://www.rc-electronics-usa.com/current-shunt.html
http://www.powerwerx.com/category.asp?CtgID=1014

-----

LEM Current Transducer -- a solid-state device which very accurately measures DC, AC, pulsed, and mixed electrical currents by relying on the Hall Effect. I don't know much about this technology, but it sounds very cool! And Digi-Key seems to have the appropriate low-amp version at a price of only $36 US.

http://www.lem.com/docs/products/LA%2025-NP%20SP14%20E.pdf
http://www.digikey.com/scripts/DkSearch/dksus.dll?Filter

----------

The three methods above sound entirely feasible. The second and third methods (maybe all three) should be sufficiently accurate. I plan to try them side-by-side and see how well they agree. It might turn out that even a simple ammeter will suffice since my application (an experimental photovoltaic battery) produces a steady current of pure DC. I don't need to worry about transients and RMS (root-mean-square) issues. And I don't work for NIST (National Institute of Standards and Technology), so I don't need super precision at this point anyway. If I do rely mostly on ammeters, I will get mine calibrated, or at least compare it to other calibrated meters.

I want to mention some other suggested methods (including some from another forum). These are less practical for my application, but interesting enough to consider

----------

Calorimetry -- feed the battery current into a resistor to convert all of the electrical energy into heat, use the heat to raise the temperature of a known mass of water inside a thermal insulator, and calculate the average power from the temperature-vs-time data. This should work if the thermal losses are kept very low, and if the heat capacity of the calorimeter is well calibrated. However, despite the low-tech appearance, this method (used in chemistry to measure heats of reaction) is very difficult to implement for low-current outputs like mine. At 5 mA and 1 V, my power output per cell is only 5 mW. Heating 10 g of water by 1 deg C would require over 2 hours! And achieving good thermal insulation on such a small volume of water would be difficult (the surface area-to-volume ratio of a container general increases linearly as size is decreased, so small containers tend to have larger relative rates of heat loss). Perhaps this method could work for many cells in parallel to get an average power output for the ensemble. But in any case, I would end up with average power measurements when I really want instantaneous readings (or at least short-term averages) of the current.

-----

Electro-Mechanical Approach -- use the output current to drive an electric motor that lifts a known mass. Calculate the average power output from the height-vs-time data. Again, this could in principle succeed, but would in practice entail energy losses due to electrical resistance and eddy currents in the motor, and friction in the gears/cables/chain. Maybe this could work for very large current outputs. For my application the measurement apparatus would need to be tiny, thereby magnifying the energy losses (perhaps unless I use many cells in parallel). And again, this sounds like a way of measuring power rather than current.

-----

Electro-Chemical Approach -- use the output current to electrolyze water into hydrogen (H2) and oxygen gas (O2). Collect the gases separately in graduated cylinders, and use Faraday's Law and the gas volume-vs-time data to calculate average power outputs. Naturally, as a chemist, I had to mention this idea. Unfortunately hydrolysis of water occurs at typical maximum energy efficiencies of roughly 80%. And the actual energy efficiency of a given electrolyzer is tedious to calibrate. Sure would look cool though! One of my ultimate aims is to incorporate hydrogen production into my system, so I might as well start practicing.

http://en.wikipedia.org/wiki/Electrolysis

-----

Thanks again to all who helped teach me something new

I have never seen better summing out of the discussion at CR4. I am sure there may be more ideas and if you are fully satisfied and want to start your research then I will say, best of luck in your very happy new year 2007.

I will give full 5 points to this posting now.

Thank you. However, I do want to make one correction. I wrote:

"Then since current equals resistance divided by potential (amps equals ohms divided by volts)"

But of course I=V/R, so my statement was incorrect (written too fast and with insufficeint proof-reading). I should have written:

"Then since current equals potential divided by resistance (amps equals volts divided by ohms)".

Also, I forgot to include the "capacitor bucket" method of measuring current. But I have my doubts about it. I have experimented with double layer supercapacitors, and have found that, even though most of the charge can be released very quickly, a non-negligable portion of residual charge bleeds out gradually over several minutes (due to the need for reorganization of the electrolyte ions in the double layer). And of course the charging voltage increases as the supercap is charged. These two issues maybe interfere with good quantitative measurement of current?

You're welcome.

Besides expressing an interest in accurately measuring your battery's current, you also mentioned: "I am trying to quantify the energy output of an experimental battery." If this is your ultimate objective, there are more approaches (as you've seen) to measuring energy output besides an accurate knowledge of the current produced by the device.

I guess I'm just curious about your emphasis on current measurements only. Also curious about the nature of your battery's mystery load.

I know, curiosity killed the cat, but I still have four lives left.

-e

I would be glad to hear more about quantifying the energy output of a battery. My original emphasis on current came from my concern that the current was changing over the lifetime of the battery (a variable I should therefore track). However since I am looking at photovoltaic cells, the current is fairly constant as long as the power source for my light source lamp is stable (easily arranged with UPS battery backup). I had worried that the cell current might degrade over time due to decomposition of chemical components (heterogeneous liquid/solid system), but this problem has not surfaced yet.

Regarding my "mystery load", nothing very interesting: a bank of white LEDs. Of course the amount of current used depends on the applied voltage. But by using an excess of them, they seem to naturally settle to 2.60 V, just above the minimum diode foward voltage. But after learning from this group, I realize that I could just as easily -- and more accurately -- be using shunt resistors instead.

LEDs (and diodes in general) are highly non-linear loads. The current through a resistor also varies as a function of applied voltage, but the relationship is highly linear (yes, there are second, third, fourth-order effects, etc., but the first-order behavior swamps the others by many orders of magnitude). To properly characterize your cell's behavior, eliminate as many spurious variables as possible. Avoid wire-wound resistors if possible, to avoid picking up hum from your A.C. mains. Metal-film resistors are preferable.

If your photovoltaic battery is based on a new, untested, uncharacterized technology, one thing you might wish to do in addition to measuring your device's output would be to independently monitor your lamp's intensity with a well-characterized silicon photovoltaic cell, and record it's output in sync with your new battery's other measurements. A silicon reference cell may have a different intensity and wavelength response than your new battery, but both will reflect relative changes in your light source and will allow you to compensate for variations in the test battery's measurements as a function of input. Without the reference cell, imperceptible changes in the source will go unnoticed (even though you're using a UPS to maintain a constant intensity, constancy is not guaranteed) but may show up in your test cell's measurements. Without a reference cell, you're introducing a spurious variable in your test results. Measure your reference cell's output in sync with your other measurements to eliminate this variable as much as possible. It might not be a bad idea to first establish a relative-response baseline between the reference cell and the DUT (device under test) by incrementally varying the light source intensity over some range and making measurements of both outputs at each step.

By using a fixed resistance (or multiple discrete resistances which you switch in and out of the circuit to cover a wider range of currents) as your load, a shunt resistor (as such) is not needed. The load resistor itself serves this purpose. The way I'd approach this problem is to simultaneously measure both the current through the load and the voltage across the battery using separate, identical DMMs, in addition to measuring the output of the reference cell, as mentioned. As long as your load resistance is in the range of several hundred to several thousand ohms, the low impedance of the load will mitigate the effects of noise pickup, hum pickup, and so forth by the DMMs' leads. If you're really concerned about this, you can shield your DMM's leads by wrapping them in aluminum or copper foil and grounding the foil. You really shouldn't need to do this, however, because noise picked up by the DMM's leads is likely to be of the "common mode" variety. That is to say, both leads will see the same external "signal" (at the same polarity) and so there is no net difference between what each lead sees in terms of spurious noise. Your DMM measures only differences in signal between each lead (when set as a voltmeter). To better ensure that your leads see the same noise, tape them together so that they're adjacent to each other over as much of their length as practical, as this further ensures that both leads are exposed to the same external electrical environment. This goes for your current-measuring and reference-cell instruments as well. Do your tests away from anything using a transformer (such as some soldering irons, high-intensity halogen desk lamps, wall warts, etc.) Shield the leads if you're still concerned about noise, but I'd bet you won't see any difference. Certainly not with a galvanometer (even in an electrically noisy environment).

If you use galvanometers, you're stuck with making manual measurements. If possible, automate your measurements as much as you can (your budget is your ultimate constraint, of course!). Galvanometers have other problems as well (including mechanical and calibration ones). I would not recommend this approach, but that is my opinion. DMMs are superior in nearly every respect and that's what I'd use. Most definitely.

I have some other ideas if you'd like to characterize your battery's output versus wavelength. Let me know if you want my two cents' worth.

-e

I will need some time to better understand of all of the specifics you mention. Someone previously complimented me on my understanding of electronics, but I am really just starting to get a grasp thanks to you guys (I guess my training in physical chemistry helps too). I totally agree with the suggestion that I calibrate my light source with a well-characterized silicon PV (silicon PVs being very stable). I will also consider what you suggested about shielding RF noise. Thanks for taking the time to share such detailed information.

Do these discussion threads get archived permanently so I can refer to them later?

Which specifics?

One other question: as your photovoltaic battery's output is a function of both wavelength and intensity, how are you characterizing its energy output as a function of energy input?

-e

To keep the wavelength and intensity of the light source constant, I use electric lamps powered a UPS battery backup (lamps kept at nearly constant operating temperature by regulated air flow). At this point, I am not attempting to quantify the light energy impinging on the PV cells. Ultimately I need to do this in order to know the cell's energy conversion effciency. But first I simply want to know how stable the cell's output current is over time (i.e., how stable the chemistry is). My emphasis on accurately measuring current in real time derives from the need to investigate stability, rather than the need to know the absolute values.

there is not one wavelength from a lamp, so what is your wavelength?

Correct, I am not using a laser source, so of course my light source consists of a mixture of wavelengths. I have tried different brands of "sunlamps" meant to mimick our sun's spectrum. Looks yellowish-white to my eyes. The precise wavelength composition and relative intensities of the component wavelengths do not concern me at this point. For now I just need something approximating a constant-output source so that I can investigate the stability of the chemical cycle.

A note of caution: In an earlier post I mentioned varying the intensity of your light source to establish a relative-response baseline between your battery and a PV reference cell. If you are using a "sunlamp," be aware that some of these sunlamps (at least all of them that I'm familiar with) are basically mercury-arc lamps with a UV-transparent glass bulb. Mercury-arc lamps can be finicky about variations in input voltage. If you decrease the voltage much below its design range you'll extinguish the arc (you then have to let the bulb cool down awhile before it will start again). Even with a constant applied voltage, the intensity of a mercury-arc lamp can vary suddenly and unpredictably.

If you wish to test your battery's response to real sunlight, you can probably use it to good effect so long as you are also using a reference PV cell to correlate your battery's output with variations in sunlight intensity. If you try this approach, you may wish to do your tests when the sun is at its highest point in the sky to minimize spectral shift due to atmospheric absorption (which varies most rapidly at dawn and dusk). Spectral shifts during the course of your tests can (and probably will) affect the relative response between your battery and the PV reference cell.

-e

The last sentence does not make a sense in my mind (head) of mein.

No effence.

Are you a ....?

Pleaseeeeeeeee!!!!!!!!!!!

If you can refr.. for me to love you more.

Your best unfliently friend. (hopefully one day).

(can't give what you doesn't, can't, won't, ... take, if you know what I mean)

You may find this link helpful in further characterizing your battery beyond tests of current stability vs constant input. Various PV technologies are shown together with test methodologies and choice of test parameters (such measuring I-V response while measuring total irradiance using a pyrometer, etc.).

-e

You mean whats the photovoltaic - battery - load systems performance?

wavelength₁ & frenqency₂ depends from the photovoltaic stuff

with 5000 km = wavelength₂

& with 60 Hz = frenqency₂

E = m * c² = p * t = V * I * t = m * wavelength₁² * frenqency₁² =>

=> I / wavelength₂² = m * frenqency₂² / V / t =

I'm comfused,

don't know. Is it going to answer?

Huh?

For someone who said they know little about electrical things you have summed up the situation extremely well. I would go for solution No.1. The simplest and cheapest. I would probably add a voltmeter. You can always up-date to something sophisticated if you have problems.

I assume you will take readings of amps and minutes by hand. The accuracy of a digital meter is enhanced by the speed in which you take a correct reading. With an analogue meter, your reading is no better than your 'estimate' where the pointer is. Of course you can arrange for a computer data acquisition facility - provided your meters have a suitable output port.

Apply LabView where possible. You will be fine.

The most unhealthy for the man battery is the battery with the most power or energy tranfer.

You should not care for the energy output thus this is the remeaning load carriers which makes the thing, the thing which is. At least not more from the stability, replacing the energy given, unpoluting the enviroment, and have anice day.

Voltage in a specific point is a massive quantiti of load cariers with the characteristic of the similar nature of load carriers (at least from the outside) that have overwellming the thing or point.

Such a point gives energy to the surrunding place or room from the specific point to chaos. Meaning that a hypothetical hyper small load carrier (as smaller can be in order to leave the field of your entraste unannoinged) will make it enstigues or dissappear.

But only the same load carriers gets away from each other.

The nature of battery should be as meen as it can be without ever get mean with her self. Such a battery should be big enough for every load so her inside resistor will be the intadecal with the loads resistor and accomblish the maximum power tranfer meaning with other words 'half power efficience'.

For the energy output you should connect a amperometer with the same resistor of the battery as long the battery's resistor is stable ( don't know if it's possible) and at the same time have a count of seconts that comes and goes. When the battery is dead stop counting the time and calculate the energy in Joule units.

energy = power time = current's current resistor time

Tough thing to be a chemist and not to know electro electrochemics and electrounfamiliarized in a unelectroworld. What to do in this lying world.

Please forgive me for my sisaderstendigs, and spelling.

Hope I was right with so little enformations of yours.

Happy birthday, and a happy happy new year.

sorry and thanks for nothing.

Nice idea. Try this one

I*R*R Heating bimetal ---> diflection pointer ----> meter reading of power