[SOLVED] Simple current sensing question

Hi guys, i face some question here for measuring the current out from the solar panel. The datasheet of the solar panel show that it is 1W, 7.5V and 135ma.

I know it would be slightly higher than the datasheet in fact, so i had a measurement with it. I get 10v with open loop, but im facing question during measuring the current. I know I = V/R , so i put in different resistor to try, below is the result.

R V I=V/R
1) 1.6ohm 0.12V 75mA
2) 2.7ohm 0.25V 93mA
3) 99ohm 8.90V 90mA


So,which 1 is the correct 1? i thought the result should be same right? BTW my current is lower than the datasheet 1, but from what other people said there will be 20% higher than the datasheet for the actual 1, may i get some advice from u guys ?

Thank u very much =)

Forget to mention that, my 1.6ohm resistor is 1/2W, my 2.7ohm resistor is 1W, and 99ohm resistor is 1/4W

Current sensing is using a small resistor in series with the load and then measure the voltage drop across it to calculate the current going to the load, in that case you also want the resistor to be as small as possible (0.1 ohm or 0.02 ohm) so that it doesn't add an additional voltage.

What you are doing is changing the actual load of the solar panel,
because of the internal resistance of the panel (low current supply capability) the lower the load you use the lower voltage you will have and from some point the current measured will drop too.
A panel having specification of 7.5v 135mA should be able to have that output with a resistor of 7.5v/0.135A=55 ohm.

Alex

Sorry for I am not so understand with your answer, I am still a newbie perhaps,haha.

I don understand about the 55ohm? Do u mean that i should use a 0.1ohm 1/2W instead of the resistor above then the answer will be correct?

No, it was just an explanation of what is a current sense resistor, it is a resistor connected in series with the load to measure the current,
it is used in multimeter (ampere-meter), power supply over current protection circuits etc.

This doesn't apply in your case, if i understand correctly you only want to check the amount of power that your source can give,
the only solution i can think of is to check if the panel can give the specified current at the voltage specified by the manufacturer.
So when it says 7.5v 135mA it should be able to give that current to a resistor that pulls that exact current in the specified voltage.
Using the ohms law I=V/R you can transform this to R= V/I to find what is the resistor value that will pull the specified current in the specified voltage.
so R=V/I= 7.5V/0.135A =55 ohm (use 56 which is standard value) ,
to double check that you can do I=V/R = 7.5v/56=0.134A so it is correct

You should be able to measure a voltage of 7.5v across the resistor (which would also mean 0.134A current through it because I=V/R) if the manufacturer specification is correct.

Also note that this specification is probably the max that the panel can give in strong sunshine and directly facing the sun.

Alex

how about the actual measure voltage is 10v which is slightly higher than 7.5v when open loop?

The panel as a source has probably the same behavior as an unregulated power supply ,
the rated voltage is specified with a load that uses the specified output current,
when you use more current the voltage drop of the internal source impedance increases and the output voltage gets lower,
as your load gets "closer" to the rated current the voltage should be the same as the specification.

I have never used a solar panel so i can't give you much help but maybe you can search for a few articles like

HowStuffWorks "How Solar Cells Work"
You can find many more using google.

Alex

The point completely missing in the consideration is that the output of a solar panel also changes with irradiation. Apparently all three measurements have been taken with different light intensity, so they can't be compared, neither to each other, nor to the datasheet specification. Generally, the output current can be expected to drop monotonically from short circuit (RL = 0) to open circuit. See: File:Jvcharacteristic.jpg - Wikipedia, the free encyclopedia

Yes I agree with alexan_e.
The solar panel rating 1W 7.5V and 135ma.
means that the solar panel given the best exposure to sunlight,
can only generate 0.135mA. The voltage at that instance will be about 7.5V. If you draw less current, it is typical for the voltage to be higher.

If you test it with a lower resistor like 1ohm.
Compute:
Current 7.5V / 1ohm = 7.5A
We know that it is not possible to generate that 7.5A from the solar panel as specified in the specification. Which is why you get those low voltage reading. You are stressing the solar panel, by squeezing out the power dry.

Current 7.5V / 1ohm = 7.5A
We know that it is not possible to generate that 7.5A from the solar panel as specified in the specification. Which is why you get those low voltage reading. You are stressing the solar panel, by squeezing out the power dry.

Click to expand...

Yes, because it isn't a voltage source. But it's neither a current source, the current will slightly increase if you reduce the load resistance.

File:Jvcharacteristic.jpg - Wikipedia, the free encyclopedia