Hi guys, does anybody here have experience withhall effect sensors to measure low currents (uA and mA range)? It would be great if you can refer me to an IC which may be appropriate for me. Thanks!
Hi, 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:
1 | ____ |
2 | +--|_R1_|-----------+ |
3 | | |\ OP | |
4 | (Current->)----*-----------|-\ AMP | |
5 | ____ | >----*----(V_out = -R1 * current) |
6 | (Current<-)----*--|_R2_|-*-|+/ |
7 | | | |/ |
8 | | | ____ |
9 | (GND) +---|_R3_|----(Offset adjust voltage) |
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 voltage. 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 output voltage. regards, mf
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 a voltage? Thanks, Thomas
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/file/hw-300b_e.pdf 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?
@ 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. Regards Falk
No but as it is possible, so why not do it?! I want to cover as many possibilities as possible in my internship, that's why I wonna try to understand the use of hall efect current sensors as well.
@ 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. K.I.S.S. 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 here measure without generating much heat - not needed here Regards Falk
Hey Falk, 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 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... Thx
@ 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) Fairy tales. >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 amplifier. >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 prize. Regards Falk
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? Thanks!
Falk has his head up his a$$, and has a lousy arrogant attitude - a bad combination. 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.
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