Electronics Forums > Current Transformer to 0-10 Volts

# Current Transformer to 0-10 Volts

Tom
Guest
Posts: n/a

 02-05-2007, 08:01 PM
I want to take a current measurement of 0 to 20 Amps AC using a
current transfomer, and then convert the current siginal to input into
a Analog input of a PLC (programmable logic controller). I know they
sell current transducers that do this alreay, but I want to make about
20 of these.

The PLC will take 0-10V DC, or 4-20mA

I was thinking about using a 50:5 Current Transformer, and a 5 ohm
resistor.

20 Amps ----> 50:5 ----> 2 Amps

Voltage = 5 * 2 = 10 V

Am I on the right track or does the current transformer output a
sinusoidal value.

Homer J Simpson
Guest
Posts: n/a

 02-05-2007, 08:37 PM

"Tom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...

> Am I on the right track

Yes.

> or does the current transformer output a sinusoidal value.

Yes.

Be careful. An open secondary on a CT is a nasty thing. Take no chances -
ask an expert if you are unsure.

Guest
Posts: n/a

 02-05-2007, 10:42 PM

"Tom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...
> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V
>
> Am I on the right track or does the current transformer output a
> sinusoidal value.
>

The 5 ohm resistor may be to big. You need to match or be less than the CT
secondary "Burden" rating. Typically that value is less than 1 ohm. The
voltage developed across that resistance is then amplified by a wide band
fixed gain amp., thus the signal conditioner. As Homer pointed out, do not
run the secondary into an open circuit or at high impedance. It will then
act as a PT possibly creating thousands of volts, doing all kinds of nasty
things!

Guest
Posts: n/a

 02-05-2007, 11:02 PM

"scada" <(E-Mail Removed)> wrote in message
news:OnOxh.274\$(E-Mail Removed)...
>
> "Tom" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed) ups.com...
> > I want to take a current measurement of 0 to 20 Amps AC using a
> > current transfomer, and then convert the current siginal to input into
> > a Analog input of a PLC (programmable logic controller). I know they
> > sell current transducers that do this alreay, but I want to make about
> > 20 of these.
> >
> > The PLC will take 0-10V DC, or 4-20mA
> >
> > I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> > resistor.
> >
> > 20 Amps ----> 50:5 ----> 2 Amps
> >
> > Voltage = 5 * 2 = 10 V
> >
> > Am I on the right track or does the current transformer output a
> > sinusoidal value.
> >

>
> The 5 ohm resistor may be to big. You need to match or be less than the CT
> secondary "Burden" rating. Typically that value is less than 1 ohm. The
> voltage developed across that resistance is then amplified by a wide band
> fixed gain amp., thus the signal conditioner. As Homer pointed out, do not
> run the secondary into an open circuit or at high impedance. It will then
> act as a PT possibly creating thousands of volts, doing all kinds of nasty
> things!
>
>
>

Also keep in mind, the secondary waveform will be the actual current
waveform, AC. If you need RMS, or DC output you further need to refine that

john jardine
Guest
Posts: n/a

 02-05-2007, 11:16 PM

"Tom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...
> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V
>
> Am I on the right track or does the current transformer output a
> sinusoidal value.
>

It's all AC and variable waveshapes must be considered.
You'll need an additional power supply and some electronics to rectify,
smooth and amplify the signal to 0-10Vdc.
E.g. 100:1 CT to a 1ohm resistor, then to an AD737 true RMS converter, then
to a x50 opamp, then to the PLC.
john

--
Posted via a free Usenet account from http://www.teranews.com

Rich Grise
Guest
Posts: n/a

 02-05-2007, 11:27 PM
On Mon, 05 Feb 2007 12:01:02 -0800, Tom wrote:

> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V
>
> Am I on the right track or does the current transformer output a
> sinusoidal value.

Well, I'm not a CT guy, but I'd think that the output current would
look a lot like the input current, like in any transformer.

I'd use a big resistor - 2A at 10V is 20 watts! =:-O

Good Luck!
Rich

Tom Bruhns
Guest
Posts: n/a

 02-05-2007, 11:42 PM
On Feb 5, 12:01 pm, "Tom" <(E-Mail Removed)> wrote:
> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V
>
> Am I on the right track or does the current transformer output a
> sinusoidal value.

Note that 10V at 2A is 20 watts you'll be dissipating in the load
resistor! I'd suggest that you go to a much higher turns ratio.
100:1 (500:5) might be reasonable. Then the output is 0.2A. You could
use an even higher ratio. If the transformers are easy to get, I'd
look at even 1000:1, though you may have some trouble finding those.
If you used your original 10:1 ratio, the ten volt drop would reflect
back as a 1 volt drop in your line, which is way more than you need to
allow. A ten volt output at 1000:1 reflects only ten millivolts drop
along the monitored line--actually somewhat more because of less than
perfect coupling, but still not a lot. A ten volt output at 1000:1 or
even 100:1 is also much more likely to be a reasonable burden for the
transformer. At 1000:1, a ten volt output with 20A in the primary is
20mA secondary current and 200mW dissipation.

You'll need to rectify the output; an op-amp precision rectifier is
appropriate. Then you need to convert the output to whatever the PLC
wants. The precision recitifer can easily be made to put out 0-10
volts, even if the input is only 1 volt instead of 10, but then you
need power to run the op amps. If you _know_ that your circuit will
always have some minimum current in it and the ratio between min and
max isn't too great, you could probably arrange to run the op amps on
the current transformer output, but that's not a wonderful idea from
the standpoint of precision, since the amplifier power will appear as
additional transformer load. A better idea would be to use the 4-20mA
loops, especially if these sensors will be some distance from the
PLC. You can find example 4-20mA circuits in op amp manufacturers'
data sheets. I believe that Linear Technology is one good source for
such circuits. Try their ap notes, too.

Too bad you didn't have this need a year or so ago. Marlin P. Jones
had some nice split-core 4-20mA AC current transducers with a jumper
for, um 10A, 20A and 50A full scale as I recall, for about \$10 each.
But even if you have to pay \$100 or more each for them, you'll
probably be better off buying them. I have a feeling from the last
sentence of your posting that you'll be in trouble trying to build
them.

Cheers,
Tom

Tim Dunne
Guest
Posts: n/a

 02-05-2007, 11:50 PM
"Tom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com
> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer,

....

> Am I on the right track or does the current transformer output a
> sinusoidal value.

You need a module from LEM. They're current transformers with built in
signal conditioners (often self powered) giving a PLC analog compatible
signal - 0-10, 0-5 or 4-20mA

http://web4.lem.com/hq/en/component/.../output_type,/

HTH

Tim
--
Sent from Birmingham, UK... Check out www.nervouscyclist.org
'I find sometimes it's easy to be myself, but sometimes I find it's
better to be somebody else.' - Dave Matthews 'So Much To Say'

jasen
Guest
Posts: n/a

 02-06-2007, 08:05 AM
On 2007-02-05, Tom <(E-Mail Removed)> wrote:
> I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V

And 20 Watts!

there's not (m)any CTs with that sort of output. a 20VA 60Hz transformer weighs a
few kilograms...

typicaly curent transformers produce milliwatts.

aim for the sort of current you can feed to an op-amp
eg: a 1000:1 transformer into a virtual earth or a small resistor.

then you probably want rectify and average it
(if you want RMS that'll add complexity)

> Am I on the right track or does the current transformer output a
> sinusoidal value.

it sure does.

Bye.
Jasen

Paul E. Schoen
Guest
Posts: n/a

 02-06-2007, 03:13 PM

"Tom" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...
>I want to take a current measurement of 0 to 20 Amps AC using a
> current transfomer, and then convert the current siginal to input into
> a Analog input of a PLC (programmable logic controller). I know they
> sell current transducers that do this alreay, but I want to make about
> 20 of these.
>
> The PLC will take 0-10V DC, or 4-20mA
>
> I was thinking about using a 50:5 Current Transformer, and a 5 ohm
> resistor.
>
> 20 Amps ----> 50:5 ----> 2 Amps
>
> Voltage = 5 * 2 = 10 V
>
> Am I on the right track or does the current transformer output a
> sinusoidal value.
>

You can get CTs with 100 mA rated output. There are also high ratio PCB
mounted (or not) CTs with primaries from 10 to 200 amps, and output up to
10 VRMS. Digikey has them TE1020-ND for \$7.80, or you can get a wide range
of CTs from CR Magnetics:

http://www.crmagnetics.com/8300.pdf

They also have transducers that might do what you need.

Paul

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