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Topic: Measure current (Read 1 time) previous topic - next topic


I want to measure the current going through a string of LEDs on a reef aquarium.  The current will be anywhere between 500ma and 1500ma.  I already have an Arduino as the controller on the aquarium that monitors temperature and pH and controls lights, fans, and CO2 solenoids.  Can someone point me in the right direction to monitor the current going through a string of LEDs?  Thanx very much.


Is the current constant or is it pulsed with PWM.
If it is constant then a simple connection of putting a current meter in seriese will do. If it is pulsed then you need to use a small seriese resistor and measure the voltage across it with a scope.
You could also use a hall effect current sensor to produce a voltage from the current flow.
Is this a one off measurement or do you want it to be a constant monitoring?


I want to constantly measure it with the Arduino and display it on an LCD display.  The current will vary during the day with programming that varies the PWM input on the LED drivers, however the current going to the LED string is not PWM.  Currently the only connection of the LED string with the Arduino is with the PWM pin going to the LM3409 driver chip.  The chip powers the string via a separate 48V power supply.  I want to see the current going through the string and manipulate the PWM signal accordingly.


A hall effect current sensor wired in series with your led string is one way to go. The resulting DC output voltage from the sensor can be directly measured by an arduino analog input pin.




however the current going to the LED string is not PWM

Figure 17 in the data sheet seems to be at odds with this statement.

Pin 8 & 9 are the sense resistor, this has the information you want. Wire a differential amplifier across that resistor, then smooth the output with a cap and measure it.
You might need to mess about with potential dividers if this voltage is sitting above 5V.


Thanx for the info guys.  Lefty, I think your suggestion would be the simplest.  I'll give that a try. 

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