Hall effect to measure DC: howto?

Hi there. I've got a 1000 billion$ question ::slight_smile:
I'd like to measure voltage and current from a generator (Vmax=24V). surfing on the web i found that it would be possible using a Hall Effect sensor. How can i wire it to arduino and the generator? Could anyone explain to me its functioning??

thanx a lot mates :wink:

http://en.wikipedia.org/wiki/Electric_current#Electromagnetism :slight_smile:

mmm.. i don't need the physic explanation (i've already know it) but how the sensor works (pins,cables to wire, connections etc). :wink:
the other question?

Check this out:
openenergymonitor.org/emon/

specifically this link on that page:

This should explain everything and a bit more :wink:

You will have to copy/past the links, this is my 1st post and it seems like one cannot post live links on the 1st post ???

dubbleUJay

There are several kinds of Hall current sensors or current tranducers. You either connect the wire from the generator in series with the sensors input/output pins (e.g., primary current pins or terminal pins) or put the wire through the hole in the sensor.

Many of the sensors have ICs that do the conversion from current to voltage within given range, for example between 0-5V. They have typically three pins for interfacing the uC: Vcc, GND and Vout. You connect Arduino's Vcc and GND to sensor's Vcc and GND. Sensor's Vout goes to Arduino's analog input. You may need to add resistors or/and capacitors, e.g., a capacitor between sensor's Vcc and GND. Check the datasheet for application diagrams.

Choose a sensor of which amp range matches your generator. For example, if the max. current from you generator is 6 amps DC, you may choose a unidirectional sensor that can sense currents from 0 to 10A. (or bidirectional from -10 to 10 if it is an AC generator). If you choose a sensor with too wide range (say from 0 to 100A), you loose resolution and accuracy.

For unidirectional sensors you get 0V from Vout at 0A, from bidirectional sersors you get 0-2.5V when I<0 and 2.5-5 when I>0.
The sensitivity of the sensor is given as mV/A, for example 100mV/A. (0.1V change in voltage from Vout means 1A change in current.)

Arduino has 10bit ADC which gives you values from 0 to 1023. If the sensor's sensitivity is 100mV/A and analogIn reading is 123, in the case of an unidirectional sensor the current is (5.0/(10240.1))123 = 6A. For bidirectional: (5.0/(10240.1))(123-512) = -18.99A.

Choose a sensor of which amp range matches your generator. For example, if the max. current from you generator is 6 amps DC, you may choose a unidirectional sensor that can sense currents from 0 to 10A. (or bidirectional from -10 to 10 if it is an AC generator). If you choose a sensor with too wide range (say from 0 to 100A), you loose resolution and accuracy.

It seems that "unidirection" does not always mean that the sensor is for DC only. But check the datasheet for the sensing range and what is the Vout value for I=0.

There are several kinds of Hall current sensors or current tranducers. You either connect the wire from the generator in series with the sensors input/output pins (e.g., primary current pins or terminal pins) or put the wire through the hole in the sensor.

where could i buy them?
yesterday i asked something about the generator, and the one that asked me the system told me it's a AC generator: a wind turbine

where could i buy them?

http://uk.farnell.com/allegro-microsystems/acs712elctr-05b-t/sensor-current-5a-soic8-712/dp/1329623

what about this?
http://www.seeedstudio.com/depot/electronic-brick-electricity-meteranalog-p-471.html

Yes but it is not a Hall effect device like you wanted, and it is a current transformer so it will only measure AC. Again not what you asked for.

yeah...i saw that it's not a hall effect sensor, but it is another way to measure. the link you've posted

http://uk.farnell.com/allegro-microsystems/acs712elctr-05b-t/sensor-current-5a-soic8-712/dp/1329623

is another current sensor. i only find current sensors

and if I use this scheme

using different resistors to downsize the gain? and if i use also the current sensor a've posted before (or this that uses hall-effect analysis http://www.seeedstudio.com/depot/noninvasive-ac-current-sensor-30a-max-p-519.html) to calculate power generated?

but it is another way to measure

No you want to measure DC. This will not measure DC.

is another current sensor. i only find current sensors

Hall effect transducers only measure current.

Are you saying now that you don't want to measure DC current but AC power?

in one of my posts I said

yesterday i asked something about the generator, and the one that asked me the system told me it's a AC generator: a wind turbine

I got yesterday three of those: http://search.digikey.com/scripts/DkSearch/dksus.dll?Cat=1966573&k=398-1000-ND
They include ICs that levels the Vout to 0-5V and provide isoloation. They have holes you can put the wire through, but they are not split-core.

and the one that asked me the system told me it's a AC generator: a wind turbine

Sorry I did not understand that use of English. It made little sense, in fact it means that you were told it was a wind turbine, not you said it was a wind turbine.
So if you want to measure AC power why is the thread headed how to measure DC?
You can waste a lot of peoples time by not stating what you want to do.