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'[EE][PIC] Clamp-Ammeter'
2007\03\20@123718 by Yigit Turgut

picon face
Hello List,

I was assigned to a project which is basicly building a AC
clamp-ammeter.The only exception is that the induced voltage across
the terminals of toroid must be equal to the magnitude of the current
flowing through the conductor -  which is placed in the
toroid.Frequency of current is 50Hz.

range =>     10A>current>0A

i(t)=5.cos(50t)

For sure,rms of induced voltage will be much smaller than the rms of
input current.
I hear you guys are saying "so what ??"

Ok.The potential problem is I am expected to build this ammeter with
at least 4 digits  precisiton after the coma.For example if i rms is
4.37843,rms is of induced voltage e must be like this ;

4.37849>e>4.37840

Any suggestions ?

2007\03\20@130339 by Alan B. Pearce

face picon face
>For sure,rms of induced voltage will be much smaller than
>the rms of input current.

Umm, I think you have a problem with mixed units. You need to think in terms
of the current coming out of the secondary of the coil as being proportional
to the input current. They will be related by the number of turns on the
secondary (think of the primary as being one turn, and remember you cannot
have part turns on a toroid).

>Ok.The potential problem is I am expected to build this ammeter
>with at least 4 digits  precisiton after the coma.

What temperature range will you have? I have some doubts you will manage
that many digits of measurement precision without some fancy temperature
handling. You also do not say what your current range is.

2007\03\20@132320 by Vasile Surducan

face picon face
On 3/20/07, Yigit Turgut <spam_OUTy.turgutTakeThisOuTspamgmail.com> wrote:
> Hello List,
>
> I was assigned to a project which is basicly building a AC
> clamp-ammeter.The only exception is that the induced voltage across
> the terminals of toroid must be equal to the magnitude of the current
> flowing through the conductor -  which is placed in the
> toroid.Frequency of current is 50Hz.
>
> range =>     10A>current>0A
>
> i(t)=5.cos(50t)
>
> For sure,rms of induced voltage will be much smaller than the rms of
> input current.
> I hear you guys are saying "so what ??"
>
> Ok.The potential problem is I am expected to build this ammeter with
> at least 4 digits  precisiton after the coma.

You are joking. Do you imagine
what means 4 digit precision for a clamp-meter if the magnetic core is
not closing perfectly every time ?
And believe me, unfortunately it never's closing the same...

2007\03\20@132920 by Yigit Turgut

picon face
> Umm, I think you have a problem with mixed units. You need to think in terms
> of the current coming out of the secondary of the coil as being proportional
> to the input current. They will be related by the number of turns on the
> secondary (think of the primary as being one turn, and remember you cannot
> have part turns on a toroid).

I am talking about measuring the magnetic field produced by current in
a conductor.A voltage will be induced in the terminals of a winding
wounded around the toroid when the magnetic flux is changing in time.I
will process this induced voltage to the project requirements.That is
it.No need to deal with the current coming out of the secondary coil.

> What temperature range will you have? I have some doubts you will manage
> that many digits of measurement precision without some fancy temperature
> handling. You also do not say what your current range is.

Room temperature.

After the amplification and necessary combination of
components,precision must be done by ... ?

2007\03\20@133340 by Yigit Turgut

picon face
> You are joking. Do you imagine
> what means 4 digit precision for a clamp-meter if the magnetic core is
> not closing perfectly every time ?
> And believe me, unfortunately it never's closing the same...

Toroid will be one-piece only.There is no moving parts.Only way to
measure is to insert the cable into toroid.

2007\03\20@134710 by Vasile Surducan

face picon face
On 3/20/07, Yigit Turgut <.....y.turgutKILLspamspam@spam@gmail.com> wrote:
> > You are joking. Do you imagine
> > what means 4 digit precision for a clamp-meter if the magnetic core is
> > not closing perfectly every time ?
> > And believe me, unfortunately it never's closing the same...
>
> Toroid will be one-piece only.There is no moving parts.Only way to
> measure is to insert the cable into toroid.

OK, than is not clamp meter is ammeter using a current transformer.
That is perfectly manufacturable.


> -

2007\03\20@140851 by Alan B. Pearce

face picon face
>> Umm, I think you have a problem with mixed units.
...

>I am talking about measuring the magnetic field produced by current
>in a conductor.A voltage will be induced in the terminals of a winding
>wounded around the toroid when the magnetic flux is changing in time.
>I will process this induced voltage to the project requirements.That
>is it.No need to deal with the current coming out of the secondary coil.

Umm, wrong. The magnetic field induces a current in the coil, and the
voltage you get out of that will depend on what resistor you put across the
coil.

This resistor will have a temperature co-efficient that will almost
certainly not meet the accuracy spec you are being asked for.

The various current probes we have around here all suffer from some form of
temperature drift, and the best ones talk in terms of accuracy of about 2%,
which is 10^2 worse than you are asking for.

Then as Vasile points out, you will not get a consistent magnetic field in a
clamp on system. The only way to do that is to thread a wire through a
permanently closed toroid. The changing air gap will cost you 1-2%, again a
couple of orders of magnitude worse than you are wanting.

2007\03\20@150840 by alan smith

picon face
yes...but.....how are you going to make sure its always in the center of the toroid?

Yigit Turgut <y.turgutspamKILLspamgmail.com> wrote:  > You are joking. Do you imagine
> what means 4 digit precision for a clamp-meter if the magnetic core is
> not closing perfectly every time ?
> And believe me, unfortunately it never's closing the same...

Toroid will be one-piece only.There is no moving parts.Only way to
measure is to insert the cable into toroid.

2007\03\20@155834 by Yigit Turgut

picon face
> Umm, wrong. The magnetic field induces a current in the coil, and the
> voltage you get out of that will depend on what resistor you put across the
> coil.

What you are saying is that magnetic field induces a current in the
coil and the coil becomes an ideal current source ?

2007\03\20@161203 by Sean Breheny

face picon face
Alan,

I'm pretty sure that he is correct. There is no physical law which
says that the secondary current equals the primary current times the
inverse of the turns ratio. That is an approximation which is valid
for a certain range of load impedances. Fundamentally, you are
producing some changing flux in the secondary. This will produce a
given open-circuit voltage according to Faraday's Law. As you load it
down, and allow current to flow in it, that current will act to reduce
the flux change. As you approach a short-circuit (load impedance much
less than the reactance of the secondary) in the secondary, the
current will begin to match the approximation you mention.

Sean


On 3/20/07, Alan B. Pearce <.....A.B.PearceKILLspamspam.....rl.ac.uk> wrote:
{Quote hidden}

> -

2007\03\21@045952 by Alan B. Pearce

face picon face
>> Umm, wrong. The magnetic field induces a current in the coil,
>> and the voltage you get out of that will depend on what resistor
>> you put across the coil.
>
>What you are saying is that magnetic field induces a current in the
>coil and the coil becomes an ideal current source ?

The coil will never be an ideal current source because of leakage inductance
and other loss effects, but within the useful range of load resistance that
will be essentially how you need to treat it.

Don't forget that the load resistance you put on the secondary will be
reflected back to the primary causing a finite, but small, voltage drop in
the wire that forms the primary winding.

But you still haven't said what current range you are looking to measure.

2007\03\21@082905 by Dave Tweed

face
flavicon
face
Alan B. Pearce <EraseMEA.B.Pearcespam_OUTspamTakeThisOuTrl.ac.uk> wrote:
> Don't forget that the load resistance you put on the secondary
> will be reflected back to the primary causing a finite, but small,
> voltage drop in the wire that forms the primary winding.

That virtual resistance appears in *parallel* with the primary
inductance, so it can only *reduce* the primary voltage (assuming
fixed primary current). The primary voltage, multipled by the turns
ratio, is the voltage you see across the secondary.

A lot of people seem to believe that the secondary voltage of a CT
can become arbitrarily high. It can't -- it's limited to a value
determined by the the primary inductance (impedance), the primary
current and the turns ratio.

-- Dave Tweed

2007\03\22@040602 by Vasile Surducan

face picon face
There is one problem you probably didn't feell it on your own skin...
:)

The current transformer secundar MUST work in short circuit. Else
become a high voltage transformer. As high voltage you'll have as big
is the transformer ratio.

Example: 1KA to 5A transformer gave around 5KV in secondary if it's
open and 100A flows in the primary.

All the theory below is just theory.

Vasile

On 3/20/07, Sean Breheny <shb7spamspam_OUTcornell.edu> wrote:
{Quote hidden}

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