I think the problem at your point is the resistance of your ammeter and
the voltage drop is too great for your purposes.
An ammeter -in every case- creates an insertion impedance, even if you
measure current via magnetic side-effects. Even current clamps insert a
parasitic inductor in place of the wire you are clamping it.
The magnetic field created by such low currents as you said is really
small. I don't know if there is a special IC out there.
But hey, there's a way:
If you choose a good op amp with low offset this may work well for you.
R1 = R2 = e.g. 1k for 1V per mA or 1Meg for 1V per µA
R3 and R2 make up a voltage divider to compensate for the offset
The offset correction voltage can be derived from a pot wiper but the
pot VCC has to be stabilized e.g. by a pair of LM431s.
Note that GND, V+ and V- of the op amp are not be conected to the
circuit, where you want to measure your current, if the current is not
flowing into GND.
Note also the limited frequency bandwidth of the op amp and the inverted
Hi Mini Float,
thank you for your detailed answer.
I think this circuit does not work for my case. I want to measure the
current consumption of a micro processor. So I thought about inserting
my shunt into the supply line of the uP and compensate the voltage drop
across it using a voltage regulator so that the voltage at the input of
the uP is constant...
This method works but as you said it is difficult to implement, because
the current range is so wide.
I don't really know how shall I connect your circuit to my uP? Is the op
amp used there a transimpedance amplifier which converts a current into
@ Thomas (Guest)
>I think this circuit does not work for my case. I want to measure the>current consumption of a micro processor.
For this, nobody uses a hall current sensor. Use a simple shunt and an
differential Amplifier, done.
>my shunt into the supply line of the uP and compensate the voltage drop>across it using a voltage regulator so that the voltage at the input of>the uP is constant...
Useless. Use a low value shunt, that about it.
>This method works but as you said it is difficult to implement, because>the current range is so wide.
Use different shunts and measurement ranges.
@ Thomas (Guest)
>No but as it is possible, so why not do it?!
Ahh, very clever. Do something totally useless and complicated, because
it is theoretical possible.
>I want to cover as many possibilities as possible in my internship,
Try to think like an engineer. As simple as possible, as complex as
necessary. Or like the american say.
http://en.wikipedia.org/wiki/KISS_principle>that's why I wonna try to understand the use of hall efect current>sensors as well.
Get a usefull application for hall effekt sensor and work on it.
Why do people use them?
galavanic isolation - not needed here
measure big currents with hundreds or thousand of ampere - not needed
measure without generating much heat - not needed here
Thank you for your kind reply :-)
I'm pretty much convinced by what you say, but do I have to go to my
supervisor and tell him, hey its too comlicated and forget about it?!
And he said another intern did it with the part I mentioned above, and
he said it worked very good for low frequency signal (which is the case
He guaranteed that for the hall effect sensor a lot of other problems
get saved, for example it has a very low voltage drop...
@ Thomas (Guest)
>I'm pretty much convinced by what you say, but do I have to go to my>supervisor and tell him, hey its too comlicated and forget about it?!
If you got the balls . . .
>And he said another intern did it with the part I mentioned above, and>he said it worked very good for low frequency signal (which is the case>for me)
>He guaranteed that for the hall effect sensor a lot of other problems>get saved, for example it has a very low voltage drop...
Nobody cares about 100mV of voltage drop when using a microcontroller.
If you are concerned, go for 10mV and use a better differential
>Somebody told me he used the following hall current sensor to measure>very low currents.>http://www.asahi-kasei.co.jp/ake/en/product/hall/f...
This ist a magnetic flux sensor. Far away from a complete and ready to
use hall current sensor.
>To my understanding this is dedicated to measure magnetic flux (T) and>convert it to a voltage. Out of the voltage, how can I deduce the>current?
Calculation and calibration. First you must fix it near the current
driven wire. And it must be FIXED, no more movement! And take care about
the field orientation (you do know how the hall effect works, do you?)
Then you ca do a simple calibration by measuring the driving current and
the output voltage.
But you will not get a super wide measurement range. You hat to handle
offset voltages, temperature drift etc.
If it would be so easy to measure currents down to milli or even
microamperes, there would be a lot of devices to do that for a good
Thx Falk for the details,
Actually I know how does the hall effect work (Wikipedia), so the
measured voltage is proportional to the current multiplied by the
magnetic flux density.
Is the magnetic flux density, which is created because of the flowing
current, proportional to the current? If yes then the hall voltage will
be proportional to I square!!!!
Am I right or misunderstood the proportionality?
Falk has his head up his a$$, and has a lousy arrogant attitude - a bad
It IS possible, and isnt as difficult as you might think. Fluke
currently sells at least 3 clamp-on meters/probes using Hall sensors
that are capable of measuring down to 5mA!
Try some of the more sensitive linear ratiometric sensors offered by
Allegro (eg, the A1321 at 5mV/G) , or a magnetoresistive sensor such as
the AAH002-02E at 11mV/G by NVE Corp. You'll probably have to use a flux
concentrator such as a ferrite ring for low currents.