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Current through a coil in a DC circuit

UneXisted

Apr 3, 2012
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I would like to measure the current through a coil in a DC circuit, but I'm not sure how to accomplish this. I cannot put a sense resistor in between and the coil is my only option to indicate for the current. I was thinking about measuring the voltage over the coil, but I'm not sure how the voltage relates to the current for this coil.

Do you have any ideas?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Clamp meter? What sort of current are we talking about? Is it AC or DC?
 

duke37

Jan 9, 2011
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You could measure the voltage across the coil but the current will depend on the coil resistance which will change with temperature. If the coil fails, the voltage will be high but the current zero.

You could use a Hall effect device to measure the magnetic field of the coil.
 

UneXisted

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In this case we are talking about DC current. I want to design a circuit that enables me to log the DC current that flows through the coil.

Therefore I need a measurement method to measure the current.

You could measure the voltage across the coil but the current will depend on the coil resistance which will change with temperature. If the coil fails, the voltage will be high but the current zero.

You could use a Hall effect device to measure the magnetic field of the coil.
Do you know any hall effect chip that can be easily assembled near a coil. I dont have a lot of space to assemble it.
 

duke37

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I have never used a Hall effect device so I cannot recommend.
Ebay have A1301 chip advertised which is a linear device and is in a similar package to a small transistor.
You will need to specify the accuracy you need and the magnetic field that the device will detect.
 

timothy48342

Nov 28, 2011
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...
I was thinking about measuring the voltage over the coil, but I'm not sure how the voltage relates to the current for this coil.
...
I think your on the right track right there. Test the resistance of the coil outside the circuit. (I assume your designing this or that you can remove it, test it, and put it back in.) Once you know the resistance, that is how the voltage relates to the current. It's a coil, so it has special AC properties, but when the current is stable (DC) it acts just like any length of wire or any resistor.

(For the most part, that is. If you need to be precise you could actually put various different volatges across it and measure the current to see if is linear enough for a simple resistance value to be reliable. All this done outside of the circuit.)
Edit: I think I remember being told that a uniform length of wire is very close to an ideal resistor. Is that true? Does anyone know? :endEdit

Once it's back in the circuit, you just measure the voltage like you said.

--tim

Also I did look briefly at Hall Effect sensors an they are quite small, some are cheap (a couple bucks or so), but some are designed for sensing position. (Like the position of a magnet) They don't provide a numerical output. (The ones I saw) And they won't be set up specifically for your coil. However the leads to your coil which carry the same current might be something more standard that can be measured with an off-the-shelf chip. (You would have to do some research and see what you find.) You'll would also have to make some room and extend the leads to the coil so that the sensor can sense just that and is not affected by other components. (Even though the sensor itself is small.) Did not know such a thing existed.. Cool.

But, I think my further above suggestion makes more sense.
 
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UneXisted

Apr 3, 2012
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Thanks Tim! I'll provide more information to clarify my goal.

The circuit is a VRM of a motherboard and the coil is part of a buck-buck converter. I want to know the current that flows through the coil so I have an indication of the energy usage of certain components.

My initial thought was to unsolder the coil, but its throughhole and with a 7-layer PCB im not sure if that's a good idea. If the coil can be seen as a very low resistor then I could use Ohms law, but I dont know the coil's resistance. I can also use a hall effect sensor.

I'm not sure what to do... still contemplating
 

timothy48342

Nov 28, 2011
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Ok, I deffinatly would not remove it from the circuit.

How acurate of a measurement do you need? Are the turns uniform? And countable?

I am thinking you could get an estimate by measuring the diameter of the turns, count them and do the math. It is probably copper. The resistivity of copper is well documented as well as the ohms per length for various thinknesses.

The only thing your missing is the thinkness of the wire. (Either to get the crosssectional area or to look it up by the wire gauge) You might be able to compare it with a sample of known gauge wire to identify it.

This wouldn't give a super accuate answer for the resistance, I'm thinking that it would be pretty close.

Anyone? This is a situation where I actually feel I might be giving bad advice, sooo...
?
--tim

Also, a pic of the coil in place might help paint a picture.
 

duke37

Jan 9, 2011
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If it is in a buck converter, it is unlikely to be DC but a mixture of AC and DC. It is likely to be an inductor to smooth the current flow. If it is a torroid, there will be little exterior magnetic field.
 

BobK

Jan 5, 2010
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If it is an inductor in a buck converter, then the voltage / current in the coil is not DC! You cannot measure the current by measuring the voltage across an indcutor when the current is changing. The voltage across the coil is proportional to the instantaneous change in current in the coil. You will need to put a resistor in series and measure the voltage across the resistor.

Bob
 

UneXisted

Apr 3, 2012
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If it is in a buck converter, it is unlikely to be DC but a mixture of AC and DC. It is likely to be an inductor to smooth the current flow. If it is a torroid, there will be little exterior magnetic field.
If it is an inductor in a buck converter, then the voltage / current in the coil is not DC! You cannot measure the current by measuring the voltage across an indcutor when the current is changing.

Bob
Why is it not only DC, but also AC? The polarity of the current doesnt change, right?

The voltage across the coil is proportional to the instantaneous change in current in the coil. You will need to put a resistor in series and measure the voltage across the resistor.Bob
The VRM is assembled on the motherboard, so it's impossible to add a resistor in series. Well maybe between the legs of the coil, but the wire is really thick and there isnt really a lot of space to do these kind of modifications.


This is how the VRM basically looks like.
32f1.jpg


Another way would be to derive the current by the number of turns of the coil. Deriving the current by the resistance is not possible you guys say? due to the AC current. I have a spare motherboar with the same coil that i can unsolder. I can use it to measure the characteristics if needed.

The only option I have is to use a Hall Effect sensor I think, but thats my last solution because i have to place it inside a computer. THere is a great chance that interference will occur.

memory%20vrm.png

This is the VRM and there are three coils near each other. The PWM controller is the upper left chip. I think the 2 chips belong to each other. They are called FET FDD6680S and MOSFET 7030BL. I am quite sure that coil 3 belongs to another circuit, but coil 1 and 2 definitely different kind of coils. Can someone clarify the purpose of these coils two coils?
 
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(*steve*)

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I guess the question is "why"?

There may be other solutions.
 

duke37

Jan 9, 2011
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A better description of what you have is an inductor, made as a torroid.
The voltage on the input of the inductor switches between 12V and ground. The average voltage across the inductor will be much, much less than this so trying to get the current using voltage drop will not be possible.
The picture shows two torroids, one with a double winding. This is likely to be at the power input side of the converter and will have one wire in the live and the other in the ground connection. This is intended to stop AC interference getting in, or more likely getting out of the converter.
The other inductor is to smooth the output current.

Why do you wish to monitor the current? Is the power supply not working?
 
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