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'] Current Transformers'
2004\02\19@122303 by Ian McLean

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Hi all,

Before anyone picks me up on this, I have been doing some more research on
current transformers.  I don't need to rectify the secondary of the CT.
Just put a 'burden resistor' (high watts, low resistance) straight across
the secondary winding - then read the voltage across this resistor, op-amp,
A/D input etc.

Have I got this right ?

Anyone know where I can get a suitable 50:5 CT from ?

Or is there a "better way" ?

Rgs
Ian

> {Original Message removed}

2004\02\19@124205 by Ake Hedman

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You can also have a look at  the hall effect current sensors from allegro.
http://www.allegromicro.com/sf/0750/

Regards
/Ake

-----Ursprungligt meddelande-----
Från: pic microcontroller discussion list
[spam_OUTPICLISTTakeThisOuTspamMITVMA.MIT.EDU]För Ian McLean
Skickat: den 19 februari 2004 01:43
Till: .....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU
Ämne: [EE:] Current Transformers


I would like to measure AC current using PIC A/D but am not sure what would
be the best way to do it.

I need to measure current at line voltage and frequency (240VAC, 50Hz).  I
would like to be able to measure up to 50A current (yes, I know it is big,
12kW max., but that is what I need), and convert that into a 0-5V DC range
for the PIC A/D, so I can display the current on an LCD display.  I have no
problem with the PIC side, A/D, LCD, etc., but am not sure how to do the
current conversion.  I have done DC current reading on the PIC before up to
20A, simply by using a sense resistor and op-amp, so I am half way to
knowing what I have to do.

I figured I should probably use a CT (current transformer) in series with
the hot wire of the AC supply, that does a 50A:5A ratio conversion, but
having trouble finding one of these.  Can anyone point me in the right
direction ?  On the secondary side of the CT, I would then 1/2 wave rectify
the output and pass the resulting 0-5A DC through a sense resistor in
parallel with the CT secondary winding, take the voltage drop across the
sense resistor, amplify it with a general purpose op-amp, and feed that to
the PIC A/D.  This is the only way I know (or at least think I know)to do
this, and it does seem a rather roundabout way of reading large AC currents.

Alternatively, how hard would it be to wind my own CT for this purpose, or
can someone suggest a clever alternate solution ?

PS: This is for a personal project.  Accuracy is not paramount.  If I can
get it accurate to within an amp or two, I am happy.

Rgs
Ian.

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2004\02\19@141009 by Steve Smith

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Try http://www.hobut.co.uk
Current transformers ect

Regards Steve...


-----Original Message-----
From: pic microcontroller discussion list
[.....PICLISTKILLspamspam.....MITVMA.MIT.EDU] On Behalf Of Ian McLean
Sent: 19 February 2004 00:43
To: EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU
Subject: [EE:] Current Transformers

I would like to measure AC current using PIC A/D but am not sure what
would
be the best way to do it.

I need to measure current at line voltage and frequency (240VAC, 50Hz).
I
would like to be able to measure up to 50A current (yes, I know it is
big,
12kW max., but that is what I need), and convert that into a 0-5V DC
range
for the PIC A/D, so I can display the current on an LCD display.  I have
no
problem with the PIC side, A/D, LCD, etc., but am not sure how to do the
current conversion.  I have done DC current reading on the PIC before up
to
20A, simply by using a sense resistor and op-amp, so I am half way to
knowing what I have to do.

I figured I should probably use a CT (current transformer) in series
with
the hot wire of the AC supply, that does a 50A:5A ratio conversion, but
having trouble finding one of these.  Can anyone point me in the right
direction ?  On the secondary side of the CT, I would then 1/2 wave
rectify
the output and pass the resulting 0-5A DC through a sense resistor in
parallel with the CT secondary winding, take the voltage drop across the
sense resistor, amplify it with a general purpose op-amp, and feed that
to
the PIC A/D.  This is the only way I know (or at least think I know)to
do
this, and it does seem a rather roundabout way of reading large AC
currents.

Alternatively, how hard would it be to wind my own CT for this purpose,
or
can someone suggest a clever alternate solution ?

PS: This is for a personal project.  Accuracy is not paramount.  If I
can
get it accurate to within an amp or two, I am happy.

Rgs
Ian.

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2004\02\19@143038 by Steve Smith

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That's ok but will get a wave shape approximately the same as the
current consumed (a sine wave) you would be better using a rectifier as
part of the amplifier circuit. For instance using a 1R resistor 0-5A
will give an output of 0-5V (AC) if you use just a bridge rectifier it
will be non linier in the 0-1.4v range due to the forward voltage of the
diodes. Google for precision rectifier its an op amp with a diode in the
feedback loop and output (cant remember circuit) use this with a low
gain to generate 0-5v DC

Regards Steve,

{Original Message removed}

2004\02\25@163226 by John N. Power
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{Quote hidden}

. the PIC A/D.  This is the only way I know (or at least think I know)to do
{Quote hidden}

The typical current transformer is made from a toroid core. The primary is
a single wire passing through the center hole of the toroid. A secondary
winding is wound of N turns wrapped around the toroid in the conventional
manner. If the secondary is shorted, the current in the secondary will be
1/N of the current in the primary. A small value resistor can replace the
short; this provides for sensing the secondary current. The insulation
on the primary wire, and on the plastic transformer case, if there is
one, provides excellent voltage isolation.

This type of transformer is a high pass filter; the 3db point is located at
a frequency of

f(cutoff) = R / (2 * PI * L)    PI = 3.14159...

where L is the inductance of the secondary winding, and R is the total
resistance of the secondary winding Rs and the terminating resistor Rt.
Since this sum will never go to zero (the winding resistance will always
be there), the cutoff frequency will approach a non-zero limit as the
terminating resistor approachs zero. This frequency is characteristic
of the transformer, since Rs and L are determined by the properties of
the core and the number of secondary turns.

Proper choice of the core is necessary if the cutoff frequency is to be
brought down below line frequency. Note that decreasing N in order
to decrease Rs actually decreases L by N squared, since L goes
like the square of the number of turns. This increases f(cutoff). A
large number of turns, on the other hand, is a problem with toroids,
especially if you are winding your own.

Commercial current transformers consist of a toroid with a machine
wound secondary. N is important, because it determines the scale
factor. Smaller values of N pass larger current in the secondary. For
reasonably small primary current, the secondary current can be as
small as several milliamps with sufficiently large N. This opens the
door for using a virtual ground to sense the current, reducing Rt to
zero.

The smaller transformers are used to sense primary switch current
in switching power supplies. Typically, these are less than 1 inch
in diameter, with the toroid mounted in a plastic case. The largest
value of N I have seen is 1000, with 100, 200, and 400 more
common. For 50A use, the minimum core size will most likely be
determined by the size of the primary wire.

John Power

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