# Current Transformer to 0-10 Volts

Discussion in 'Electronic Design' started by Tom, Feb 5, 2007.

1. ### TomGuest

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.

Tom, Feb 5, 2007

2. ### Homer J SimpsonGuest

"Tom" <> wrote in message
news:...

> 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.

Homer J Simpson, Feb 5, 2007

"Tom" <> wrote in message
news:...
> 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!

news:OnOxh.274\$...
>
> "Tom" <> wrote in message
> news:...
> > 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

5. ### john jardineGuest

"Tom" <> wrote in message
news:...
> 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

--

john jardine, Feb 5, 2007
6. ### Rich GriseGuest

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

Rich Grise, Feb 5, 2007
7. ### Tom BruhnsGuest

On Feb 5, 12:01 pm, "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.

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

Tom Bruhns, Feb 5, 2007
8. ### Tim DunneGuest

"Tom" <> wrote in message
news:
> 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...sk,displayserie/serie,AK ££ AKR/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'

Tim Dunne, Feb 5, 2007
9. ### jasenGuest

On 2007-02-05, 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

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

jasen, Feb 6, 2007
10. ### Paul E. SchoenGuest

"Tom" <> wrote in message
news:...
>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

Paul E. Schoen, Feb 6, 2007
11. ### martin griffithGuest

On 6 Feb 2007 08:05:44 GMT, in sci.electronics.design jasen
<> wrote:

>On 2007-02-05, 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.

snip

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

LTC1967 and it is simple, and cheap

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

>
>it sure does.
>
>Bye.
> Jasen

martin

martin griffith, Feb 6, 2007
12. ### TomGuest

Thanks for the info. I still can't justify the overpriced
transducers. I was looking at analog.com at some of their power
metering IC's like the ADE7757. Are there other products like this to
simplify getting the info into some form analog signal or
communication protocol like RS232?

On Feb 5, 5:42 pm, "Tom Bruhns" <> wrote:
> On Feb 5, 12:01 pm, "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.

>
> 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- Hide quoted text -
>
> - Show quoted text -

Tom, Feb 6, 2007
13. ### DaveMGuest

"jasen" <> wrote in message
news:eq9cso\$fq5\$-a-geek.org...
> On 2007-02-05, 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

>
> 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

You can get the Triad CSE187L current Xfmr from Mouser for \$2.47 ea in qty of
10.
http://www.mouser.com/search/ProductDetail.aspx?R=CSE187Lvirtualkey55310000virtualkey553-CSE187L

It has 1:500 turns ratio, current range of 0.1A - 30A. More details at

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the

Some days you're the dog, some days the hydrant.

DaveM, Feb 6, 2007
14. ### Tom BruhnsGuest

On Feb 6, 11:43 am, "Tom" <> wrote:
> Thanks for the info. I still can't justify the overpriced
> transducers. I was looking at analog.com at some of their power
> metering IC's like the ADE7757. Are there other products like this to
> simplify getting the info into some form analog signal or
> communication protocol like RS232?
>
> On Feb 5, 5:42 pm, "Tom Bruhns" <> wrote:
>
> > On Feb 5, 12:01 pm, "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.

>
> > 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- Hide quoted text -

>
> > - Show quoted text -

If measuring the average absolute value of the waveform is acceptable
(that is, if you don't need "true RMS"), and if the input resistance
of the PLC's 0-10V input is high (say >100kohms), there's an easy way
to do this, and the accuracy will be quite good, even down to currents
under an amp, for 20A full scale.

You need to get current transformers (CTs) that can output a fairly
high voltage--at least 12 volts RMS. Expect these will have a high
turns ratio. For example, the 2000:1 CR8350-2000 from CR Magnetics
(thanks, Martin, for the link) should do the job. The catalog page
lists the effective turns ratio, Te, as 2037:1. We'll be detecting
the average of the absolute value of the waveform. Assuming it's a
sine, the average is about 0.900 times the peak (2*sqrt(2)/pi). We'll
put a resistive load on the CT, such that 20A RMS sinewave in the
monitored wire results in 10.0V average magnitude across the
resistor: 10V*2037/(20A*0.900) = 1131 ohms. Actually, we'll make it
a bit more, and add a way to trim it down to the right value to
calibrate it: say 1.21k ohms 1%, paralleled with (10k fixed in series
with a 20k trimmer). We won't connect that load directly to the CT
output, but rather through a bridge rectifier. There is some
advantage to using Schottky diodes, but it's minor, and you can just
as well use 1n4148 or similar silicon diodes. The resistive load goes
across the bridge output. Since the CT looks like a current source,
the diode voltage drop is unimportant, so long as it's not so large
the CT can't put out the voltage. Now we just need to average the
output; what's across the load is the absolute value of a (nominally)
sine wave. We could just put a capacitor across the load, but that
requires the CT output to slew quite a few volts rapidly; better to
isolate the capacitor with at least a resistor, and even better, an
inductor. For the very low DC current involved, you can use the
winding of a small audio transformer to get high inductance, tens of
henries for a nominal "20k ohm" winding. Then use something like a
47uF cap at the output to smooth out the voltage. You may want to add
a little resistance in series with the inductance to reach critical
damping; then the response of the circuit will be reasonably "crisp"--
much more so than if you just use a resistor instead of the inductor.

After construction, calibrate it: put a known 20A through the
monitored wire, read the voltage on the PLC, adjust the trimpot so it
reads 10.00V. Now reduce the current to 10A, and make sure the PLC
reads 5.00V. Check the linearity further by putting in, say, 1A, and
checking for 0.500V from the PLC. If it's not as linear as you'd
like, consider applying a calibration in the PLC. But with the
rectifier connected to the CT and the load to the rectifier output,
you've removed a significant source of nonlinearity.

Cheers,
Tom

Tom Bruhns, Feb 8, 2007