current sensor based on magnetic induction 2

[arivel]

Hi everyone .
I want to make an alternating current sensor for audio frequencies (20Hz-20KHz) for current starting from 50mA. many very small toroidal plastic nuclei (therefore air permeability) placed next to each other to form a straight line because the current conductor passes through the central hole of each of them. it is not definitive but a prototype so each toroid is wrapped in a single turn of enamelled copper wire turns.
the output of the last turn of the first toroid is connected to the first turn of the second toroid and so on until the last one, therefore the turns of all the toroids are connected to each other. this is to exploit a greater quantity of magnetic field (which is at most in contact with the wire where the electric current passes) and to have a single total winding.
naturally there must be no empty space between the conductor and the toroidal rings. the signal coming out of the sensor will be amplified.
the measurements of the toroidal rings are as follows: internal diameter 6mm external diameter 10mm round section, copper wire section 0.02mm. measured inductance of a single wound toroid 0.0000065H.
I have two questions for you.
to have the maximum intensity of the magnetic field is it better to use a solid section conductor or a tube to pass all the current near the turns of the toroids?.
Can you calculate the voltage I get at the output of each toroid for a 50 mA current? .
Thank you

 

[arivel]

sorry but there are some errors.
each individual toroidal ring measures an inductance of 0.0000013H
the copper enamelled wire has a section of 0.2 mm

 

[hacksaw]

You need a drawing of you device and your current theoretical analaysis for any meaningful response,

good luck, you may have to build your gadget first and test it

be sure to post your results!

 

[IRstuff]

Why not use a Hall sensor?

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[arivel]

no

 

[3DDave]

Isn't the toroid current a function of dI/dt as picked up by the toroids? A constant current makes a constant magnetic field which won't induce a current in the torroid and cannot produce a voltage anywhere.

Are you turning down Hall effect because the temperature is very high or very low?

 

[hacksaw]

Turning down the suggestion, no doubt based on AI notions!

 

[arivel]

@3DDave
AC

 

[3DDave]

"Can you calculate the voltage I get at the output of each toroid for a 50 mA current?" is a DC measurement.

Ask incompatible questions, get incompatible answers.

Why not Hall?

 

[3DDave]

So, a current transformer that uses single turn units to build a multi-turn device?

A series resistor seems like a better choice.

1 Ohm at 50 milliAmps = 50 mV.

 

[arivel]

50mA peak-to-ground voltage.
each toroid is wrapped in many turns of wire but only one layer.
this is because for the moment it is a test that I want to do to see what we get. if the result is satisfactory I can even do more than one lap. remember that I have to wind many toroidal rings all by hand.

 

[hacksaw]

Show us some of your test results

 

[electricpete]

I'm not sure where 3DDave is coming from. "(20Hz-20KHz)" implies ac to me.

> to have the maximum intensity of the magnetic field is it better to use a solid section conductor or a tube to pass all the current near the turns of the toroids?.

I don't think it matters. If wave effects are neglected (quasistatic approximation) then the flux for a given current is the same outside of the conductor, regardless of whether the current is concentrated at the center or the outside or uniformly spread throughout the conductor.

> Can you calculate the voltage I get at the output of each toroid for a 50 mA current? .

V = d/dt(flux)= N * A * dB(t)/dt ... where N is number of turns and A is area of each turn.
Let B = Bpeak*cos(2*pi*f*t), then
Vpeak = N*A* 2*pi*f* Bpeak = N*A* 2*pi*f* mu0*Hpeak

By Ampere's law assuming the toroid is centered on the conductor
Hpeak(r) = Ipeak / (2*pi*r)

That much is pretty basic stuff. As a simplifying assumption you might set r as the midpoint of the toroid and assume B is constant at that value within the toroid. Note the frequency f plays a role above, so the scaling factor between current and voltage will be proportional to frequency.

 

[3DDave]

"Can you calculate the voltage I get at the output of each toroid for a 50 mA current?" seemed DC to me.

Peak 50mA or RMS 50 mA, that is AC, but were not part of the original question.

 

[3DDave]

Won't dI/dt be frequency dependent? Does this then require a correction for the instantaneous waveform?

 

[IRstuff]

I want to make an alternating current sensor for audio frequencies (20Hz-20KHz) for current starting from 50mA.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[3DDave]

I saw that part and the second statement as well.

Doesnt dI/dt vary the induced current with variation in frequency for a given RMS Amperage?

 

[arivel]

this is what I got using 15 connected toroids.
I used a 20 - 20 Khz sweep signal and as you can see the user is a speaker. no amplification of the signal coming from the current sensor.
Don't ask me how much current I passed because I didn't measure.

 

[hacksaw]

Typical speaker load is 4 ohms

 

[arivel]

its impedance changes with frequency