# current transformer application

Discussion in 'Electronic Design' started by Craig, Oct 21, 2006.

1. ### CraigGuest

Hi
New to the group so here goes.

I am building a system for monitoring various parameters around the
house, one of which is the electricity consumption. I have a computer
data card which has 0 to 2.5 V DC analogue inputs.
I have employed a current transformer (100A/5A type). This gives out 0
to 60mV AC on the secondary, relative to 0 - 100A on the primary . I
need to convert this 0 - 60mV AC to the 0 - 2.5V DC for the data
card. Can anyone suggest a circuit or point me to a website please?
Many thanks
Craig Taylor
Somerset UK

2. ### Rich GriseGuest

If you know the value of the burden resistor that's in place right now,
then just replace it with one whose value is (2.5/0.06) times the old
one. It's a _current_ transformer - the output voltage is determined
by the current, the turns ratio, and the burden resistor. E = IR, and
all that. ;-)

Good Luck!
Rich

3. ### scadaGuest

If you know the secondary produces 60mv @ 100A primary, then the burden
resistor must be already connected to the secondary. R=E/I the burden R =
..06/5, or 12 milli-ohms. You need to keep the recommended burden resistor, a
change in value will change the calibration of the CT. The extreme (no
burden resistor) is dangerous and self destructing because the CT becomes a
PT @ thousands of volts. You need to connect the CT secondary to a signal
conditioner. You could buy one www.ohiosemitronics.com , or make your own.
Basically it needs to amplify the signal 2.5/.06 = 41.66 gain and rectify
it. You could rectify the signal first, however you will need to use some
kind of an active rectifier.

4. ### MarkGuest

And be aware that in an AC system, the CURRENT is not necessarily
propotional to the POWER unless you are dealing with resistive loads.
In a home many loads have a reactive component and may draw currnet
that is out of phase with the voltage and hence does not carry power.
Look up "power factor".

The current measurment will be close but the power factor will be a
significant source of error de[pending upon how accrate the data must
be.

A true wattmeter is rather complex.... hey a pun....

Mark

5. ### JamieGuest

I think you will find that it's current you must work with on the
other side not voltage.
In anycase, an OP-AMP circuit will cure your problem.
http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/opampvar2.html

6. ### Jim BackusGuest

Suggest you use a dual op amp one half as an amplifer with a gain of ~
42 to increase the voltage from 60 mV to 2.5 volts and one to rectify
it. The 60 mV is probably an rms value so the peak value will be
SQRT(2) times that i.e. 84 mV or thereabouts. To increase this peak
value to 2.5 V will require less gain (2.5/0.84). Although an old
type, a 1458 would be a suitable dual op amp, but a modern single
supply one that handles signal swings down to the 0 V rail would be
better.

Look for precision rectifer circuits for op amp based rectification.

I'm slightly surprised that at 100 Amps in the primary you only get 60
mV. Other posters (e.g. scada) have given the relationship between the
current ratio and the voltage. Scada's point about current and power
is correct and you could process instantaneous voltage and current to
calculate power, but that would be more complicated for limited
benefit.

7. ### Rich GriseGuest

Actually, the first ones were quite simple - instead of a permanent
magnet meter, you use one with a field coil, and measure volts with
one coil, and current with the other.

http://en.wikipedia.org/wiki/Wattmeter

Cheers!
Rich

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