How can I measure the current of a 320watt solar panel?

Hello,
I am trying to plot the voltage and amperage of a 320W solar panel. I am using a voltage divider circuit to measure the voltage, and i am trying to use a ACS712 module rated for 20 amps.
My problem is finding getting a load that can dissipate the 40 volts @ 8 amps that the panel outputs.
My professor told me that i would have to use some type of heating element to do it. What i don't understand is how a multi-meter can read the amperage without having to use a heating element to dissipate the heat.
Any suggestions would be highly appreciated thank you.

What i don't understand is how a multi-meter can read the amperage without having to use a heating element to dissipate the heat.

I think you are having problem with basic electricity. Unless something is drawing the current, like your heater, then no current flows and their is nothing to measure.

My professor told me that i would have to use some type of heating element to do it.

This is a little tricky because you need the "right" resistance. 40V/8A is 5 Ohms, so you need a 5 Ohm 320W (or more) resistor. You'd probably need to get a handful of power resistors and wire them in series/parallel.

What i don't understand is how a multi-meter can read the amperage without having to use a heating element to dissipate the heat.

The meter has a low-value resistor. You get a low (but measureable) voltage drop across the resistor which is measured to calculate current. Since the voltage drop is low, the meter doesn't dissipate much power.

The power is getting dissipated somewhere else... In a light bulb, heater, or motor, or whatever is being powered.

Yeah i forgot to ask if a 5 ohm resistor would work. In amazon, JIANXIN-09-13-338-02, i found a 5 ohm resistor but it's rated at 100 watt. Would I just hook them up in parallel? What im also unsure of is how a variation in amperage will affect the circuit. The solar panel runs at 320watt under ideal conditions. In the real world it does about 240-260 watts due to efficiency loss from heating above ideal operating conditions. Voltage is constant what changes is the amperage.

I want to replicate the voltage drop meter with the arduino but that sounds like a project for another time.

You can make a 5 Ohm, 400 watt resistor by hooking 4 of those together. Two in series makes 10 Ohms, two 10 Ohm resistors in parallel make 5 Ohms.

jremington:
You can make a 5 Ohm, 400 watt resistor by hooking 4 of those together. Two in series makes 10 Ohms, two 10 Ohm resistors in parallel make 5 Ohms.

Oh man i completely forgot about the equations for total resistance. But will the value be affected if the total resistor is 400 watts.

You can't use a fixed resistor is you have to plot voltage and current.
You basically need a 320watt adjustable zener diode.
Or an adjustable current sink.
Or a variable resistor.

Plenty of big resistors around you.
Heaters, clothing irons, hotplates, etc.
Some of them in parallel will get you the right value.
Problem is that you need them to be adjustable to plot a graph.
Leo..

Dunking a power resistor in a container of cooking oil increases it’s power handling enormously.

Allan.

Hi,
Welcome to the Forum.

What are the specs for your PV?
Can you post a link to specs/data please?

What is its Open Circuit Voltage?
What is its Short Circuit Current?
These will govern the setup of your test rig.

Voltage is constant what changes is the amperage.

I think you will find that a PV is a current source and constant output voltage it ain't.

Thanks.. Tom.. :slight_smile:

Hmm, so it sounds like im back to square one.
heres a link to the data solar panels website you can get the datasheet from there.
http://www.canadiansolar.com/fileadmin/user_upload/downloads/datasheets/v5.531/canadian_solar-datasheet-superpower-CS6K-MS-v5.531en.pdf

When measuring the voltage of the panel using an multi meter it constantly reads 40 volts. When measuring the amperage with the multi meter it would read 6-7 amps and would increase to 8-9 as the panel decreased in surface temperature.

From that i figured that the panel power output varied through the amperage. Maybe it is because i have never had a load on the solar panel so maybe that cause the voltage to change.

Other ammeter circuits that I have looked at use a potentiometer. But im not sure if they sell ones that can take the amperage that the solar panel is putting out.

Most multimeters have a separate input to measure 10 or 20 amperes, at least for 30 seconds or so.

so it sounds like im back to square one.

Only because you are paying no attention to what is being said here.

You can not treat current like voltage, they are two different faces of the same thing. You can not just measure the current capability from something, it has to be real current you measure. Unless that current is flowing through some sort of load then there is no current at all to measure.

Grumpy_Mike:
You can not just measure the current capability from something, it has to be real current you measure. Unless that current is flowing through some sort of load then there is no current at all to measure.

What do you mean by real current? I thought that the point of the power resistors was to create a load so that i can measure the current.

Also if voltage and current are 2 faces of the same thing do you mean that i have to read the amperage and voltage together?

Kronustor:
What i don’t understand is how a multi-meter can read the amperage without having to use a heating element to dissipate the heat.

What you measure it call Short-circuit current Isc. it have zero voltage at this point.

Kronustor:
When measuring the voltage of the panel using an multi meter it constantly reads 40 volts. When measuring the amperage with the multi meter it would read 6-7 amps and would increase to 8-9 as the panel decreased in surface temperature.

The voltage you read was Voc Open circuit Voltage and Short-circuit current Isc.

From the datasheet it show CS6U-320P the optimum operating Voltage and current is

Opt. Operating Voltage (Vmp) 33.6 V

Opt. Operating Current (Imp) 6.91 A

at ambient temperature 20°C, irradiance of 800 W/m2

allanhurst:
Dunking a power resistor in a container of cooking oil increases it's power handling enormously.

Allan.

Just to be clear, you have to leave it there. :slight_smile:

Kronustor:
Also if voltage and current are 2 faces of the same thing do you mean that i have to read the amperage and voltage together?

Not simultaneously but under the same conditions. So if you have a load connected then you measure the voltage with the load connected and you measure the current with the load connected.

Grumpy_Mike:
Not simultaneously but under the same conditions. So if you have a load connected then you measure the voltage with the load connected and you measure the current with the load connected.

Then you adjust the load to give you your different voltage/current characteristics to map like this graph.

Tom... :slight_smile:

And, hope there are no moving clouds when you are adjusting the load.
Otherwise you still have to go with the adjustable 320watt zener diode from post#6.
With adjustable zener I mean the opposite of an adjustable voltage power supply (sink, not source).
I think that's something you have to build yourself though.
Leo..

Wawa:
And, hope there are no moving clouds when you are adjusting the load.
Otherwise you still have to go with the adjustable 320watt zener diode from post#6.
With adjustable zener I mean the opposite of an adjustable voltage power supply (sink, not source).
I think that's something you have to build yourself though.
Leo..

The thing with that is that solar panels lose efficiency as they heat up. If i recall correctly my panel loses about .4% per degree over 110 degrees Fahrenheit. I am testing a cooling system I made. The system holds water to the back of the solar Panel and brings the temperature back down to 110, give or take. That is why I am trying to log the power output, so that I can compare power output before and after cooling the panel.

I thought that using light bulbs would work, since light bulbs lighten or dim depending on how much electricity you pass through them, and I figured that maybe a heating element would do the same. I ended up ditching the light bulb idea because the panel runs at too high of a voltage to run any light bulbs i can find.

I attached a 2000 watt heating element and it seems to be working when attached to a power meter Ill try to run the code to see what happens.

I ended up ditching the light bulb idea because the panel runs at too high of a voltage to run any light bulbs i can find.

Not if you put two or more in series, with a positive temperature coefficient they share very nicely.