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### Topic: Measuring joules/watts? (Read 3266 times)previous topic - next topic

#### cloister

##### Mar 01, 2011, 07:15 am
I would like to do a little project with a solar panel, something to measure the actual watts coming out of the thing, and an Arduino to integrate the panel's output over time.

However, I haven't any idea of how I would build a circuit to measure the actual amount of power the panel is producing.  Is there an off-the-shelf component that will measure that for me?  Or measure some  related value(s) through which I can calculate the watts?

Any pointers to circuit diagrams, wikipedia pages, or anything to help me get going would be greatly appreciated.

#### weirdo557

#1
##### Mar 01, 2011, 11:09 am
a voltage divider to scale the voltage so it can be measured, and a current sensor to measure the current. you could use a shunt resistor and measure the voltage drop, or you can use a hall effect sensor to measure it.

#### cloister

#2
##### Mar 02, 2011, 12:20 am
Cool.  So, measure the volts and amps coming out of the panel, because W = V*A.  Makes sense.

Being an electronics NOOB, can anybody recommend some Arduino-friendly parts / circuit diagrams for the ammeter portion?

#### weirdo557

#3
##### Mar 02, 2011, 01:22 am
do you know the power your panels are rated at?

#### Artem_F

#4
##### Mar 02, 2011, 01:39 am
you've got the right answer in #2.

Here's the circuit for current measurement:
RRRRR = 1Ohm resistor
Measure the voltage at this point, the current will be the voltage / 1ohm

#### weirdo557

#5
##### Mar 02, 2011, 02:50 am
1 ohm is kinda high for any significant current though. a smaller shunt and an op-amp would work nicely, or a hall effect.

#### cloister

#6
##### Mar 02, 2011, 05:21 am
Sweet! I should have known SparkFun would have something. That looks just about perfect.  Thanks so much!

#### MarkT

#7
##### Mar 02, 2011, 01:20 pm
Although its rather expensive - look for other current sensors if you dont require the high-side monitoring feature.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### janeik

#8
##### Mar 03, 2011, 11:10 am
P= (U*U)/R
P=(I*I)*R
A static known resistance and measure of U or I, willl give a simple equation to calculate P. No need to measure both.

regards

#### MarkT

#9
##### Mar 03, 2011, 08:28 pm
But the requirement is to measure the power delivered into the actual load, not a resistor!
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### weirdo557

#10
##### Mar 07, 2011, 01:04 am
by measuring the voltage drop across a resistor you calculate the current flowing through the circuit, which is equal to the current going into the battery.

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