Measuring Current using a a data logging Multimeter
edlikestoboogie
Posts: 71
This question is general, and doesn't involve the prop specifically. but eventually it will help me with my prop project. I need to measure and log current from a small toy solar panel. The kind you can buy from radioshack. It is rated at about 400ma max and 0.5 volts DC. How do I measure the current off this solar panel? I have a data logging multimeter with an autoranging Anmeter. I was told that hooking the solar panel directly to the multimeter is not the correct way to measure the current output of the solar panel, even though my multimeter has a mA DC setting. I was told that a series of resistors connected to a load is required to measure the current that the load is taking from the solar panel. Is this correct? If so, how do I exactly do this? and what ohm resistors should I use?
Comments
Solar panels are interesting devices. The most interesting thing to me (who has a bunch of them on my roof) is that the power they produce has a "maximum" which is at a particular current draw. In plain english that means that the amount of power you get out of the panel depends on how much current you draw. To get the optimum amount of power out of a panel you need to vary the current draw while measuring the voltage and find the point where the product of the two creates the most net power.
Now on to your questions. So a solar cell that is not connected to anything generates no power. It produces a voltage potential (known as its open circuit voltage) but until there is a path between its + and - leads that can conduct current it will not generate any power. If you are not interested in doing any useful work with this solar cell (like charging a battery or running a toy), then you can connect a 1.0 ohm resistor between the + and - leads to provide a current path. A resistor converts the current going through it into heat, the amount of heat it can safely generate is measured in Watts. So a 1/4 watt resistor can safely dissipate one fourth of a watt of power as heat. So how much power will is dissipate? Remember from above that Power was I * E ? well E is also I * R so substituting that value of E in for E in the power equation we get another common form P = I * I * R (or P = I^2 * R). So your 1.0 ohm resistor is only 1 ohm which makes the math simple, P = I^2. If the solar cell can generate .4 amps (400 mA = .4 A) then the most power a 1 ohm resistor connected to it would have to dissipate would be .4 * .4 or 0.16 watts. Since this is less than .250 watts the 1/4 watt 1.0 ohm resistor would over heat in this configuration.
Now you have a circuit that is safe in all situations, solar cell that can generate at most 400mA connected to a 1.0 ohm resistor. Now to measure how well the solar cell is working we go back to the start with ohm's law. If you measure the voltage that appears across this resistor, which is to say the + probe of your voltmeter is connected to the + side and the - probe is connected to the other side of the resistor. You can use the voltage measurement to calculate the current that is flowing through that resistor. Since E = I * R by ohms law, re-arranging and solving for I you get I = E / R is the current. Since the resistance is 1.0 you can see that the current will be I = E / 1.0 which we all know anything divided by 1.0 is simply itself so the current through the resistor in amps will be exactly the same as the voltage reading across the resistor (see why we picked 1.0 ohms? )
To measure this current with a microcontroller like the propeller or a BASIC Stamp, you need an analog to digital converter which can convert a voltage into a binary number. There are lots of those around but they basically take a reference voltage which is used as the maximum voltage (0 is of course the minimum on the converter you want to use) and divide it up into n parts where n is 2^bits (256 for an 8 bit converter, 4096 for a 12 bit converter, etc). Each time you read the voltage you will know how much current (because you are using a 1 ohm resistor) the solar cell is putting out.
HTH
--Chuck
there is one question to clear:
do you use the solar panel to supply something (device)?
if so you have to INSERT your multimeter in serial to the supplying circuit
plus of solarpanel
plus of Multimeter in mA-Measuring mode
GND of multimeter
plus of supplied device
GND of device
GND of solarpanel
you can use the multimeter in ma-measuring mode as a dummy-device
plus of solarpanel
plus of Multimeter in mA-Measuring mode
GND of multimeter
GND of solarpanel
but make sure that your the measuring-range of your multimeter is more than 400 mA !
Otherwise you have to put a resistor in serial to the multimeter to limit the current to a value lower the measuring-range of the multimeter
if this was not understandable feel free to post another question
greetings
Stefan
I am using the solar Panel to recharge two AA 1.5volt batteries. I'm assuming I should use NI-CAD batteries instead of NI-MH? I am going to directly hookup the batteries to the solar panel, so that the panel recharges empty batteries. And then I will hook up the multimeter inline with the batteries and solar panel, with no resistor. Also the multimeter is autoranging and can handle 400mA. So I think this should work, right?
In the mean time, I will read your post and chuck's more carefully to fully understand this situation.
So you have a source of .5v solar cells, start with 7 of them hooked together in series.
--Chuck
So when you want to learn something about your system you should messure the voltage with a (resistor) load of - say - 1, 3.3, 10, and 33 Ohms and sketch the curve in such a diagram
You also followed up that you were going to try and charge two 1.5v batteries with this solar cell. That won't work, it won't even charge one battery because it does not generate a voltage that is higher than 1.5v (which you would need to charge one battery). Lots of fun things to learn from solar cells though so I wouldn't throw it out, just don't expect it to charge a battery by itself.
--Chuck
you could unplug the led's (not sure of the circuit), run wiring in from all 12 batteries from the back yard into the house for a good 14.4v for lighting or whatever anytime you want it.
Sean
Post Edited (bassmaster) : 10/26/2008 4:09:56 AM GMT
@deSilva: Nice to see you're still around - we've missed you here
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Cheers,
Simon
www.norfolkhelicopterclub.com
You'll always have as many take-offs as landings, the trick is to be sure you can take-off again
BTW: I type as I'm thinking, so please don't take any offence at my writing style
Post Edited (simonl) : 10/29/2008 12:08:23 AM GMT