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'[PIC] measure 30 amps of current?'
2008\03\04@164932 by

I am rather new to PIC development but I am learning fast. I would like to
build a circuit to measure the amount of AC current in AMPs flowing through
a wire. Any suggestions on how to measure the current without directly
connecting to the wire which will be carrying up to 125vac @ 30 amps.

Can the PIC ADC be used for this purpose? Eventually I would like to drive an
LCD with the value in amps.

Any suggestions on where to begin?

Thanks,
Mike
> Any suggestions on where to begin?

http://www.siliconchip.com.au/cms/A_30551/article.html

It uses an F84 and comparator, but that could easily be replaced
with a PIC that has ADC

Probably a current transformer would be your best bet. This is
basically a step-up transformer with a single turn primary ( the wire
being sensed is just fed through the core) and a multi turn (can be
100s of turns) secondary. The secondary current is then 1/n times the
primary current.

The critical requirement is that there is a "burden" or load resistor
fitted to transform the secondary current into a secondary voltage of
sensible value.

For measuring AC current you will need to include some sort of
rectifier and don't forget that the peak current is well in excess of
the rms value. One trick is to place the burden on the DC side of a
bridge rectifier so that the diode non-linearities are minimised. It
is still reccomended that a (higher value) resistor be placed on the
AC side as well though in case the bridge becomes open circuit or very
high voltages can be generated.

This method is commonly used as it not only provides a suitable
voltage range, but also a high degree of isolation can be obtained.
With appropriate design, the PIC ADC can be used without having to use

Alternative methods include measuring the voltage drop across a series
resistance (including using the cable resistance) and using magnetic
sensors. The series resistance approach is the simplist but involves
direct connection to the measured circuit with obvious safety
concerns.

RP

On 05/03/2008, Michael Cunningham <michael.cunningham.usagmail.com> wrote:
{Quote hidden}

> -
Michael Cunningham wrote:
> I am rather new to PIC development but I am learning fast. I would like to
> build a circuit to measure the amount of AC current in AMPs flowing through
> a wire. Any suggestions on how to measure the current without directly
> connecting to the wire which will be carrying up to 125vac @ 30 amps.
>
> Can the PIC ADC be used for this purpose? Eventually I would like to drive an
> LCD with the value in amps.
>
> Any suggestions on where to begin?
>
> Thanks,
> Mike
>

http://www.allegromicro.com/en/Products/Part_Numbers/0754/index.asp

Rolf
Michael Cunningham wrote:
> I am rather new to PIC development but I am learning fast. I would like to
> build a circuit to measure the amount of AC current in AMPs flowing through
> a wire. Any suggestions on how to measure the current without directly
> connecting to the wire which will be carrying up to 125vac @ 30 amps.
>
> Can the PIC ADC be used for this purpose? Eventually I would like to drive an
> LCD with the value in amps.
>
> Any suggestions on where to begin?
>
> Thanks,
> Mike
>
I would use a bridge, a cap, and a Current Sensor made by Allegro Semi.
They interface directly to a PIC ADC
and can measure up to 50A. You have to change AC to DC (the bridge), so
there is a small but correctable
error at low currents.

We use it all the time; costs about \$6 USD. Sold by Newark, among others.

--Bob Axtell

{Quote hidden}

I'd use this approach with one of these transformers:
www.toroid.com/standard_transformers/current_sensing_transformers/current_sensing.htm
. I've done this, then do RMS calculations in software.

Harold

--
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opportunities available!
>> I am rather new to PIC development but I am learning
>> fast. I would like to
>> build a circuit to measure the amount of AC current in
>> AMPs flowing through
>> a wire. Any suggestions on how to measure the current
>> without directly
>> connecting to the wire which will be carrying up to
>> 125vac @ 30 amps.
>>
>> Can the PIC ADC be used for this purpose? Eventually I
>> would like to drive an
>> LCD with the value in amps.

>
> http://www.allegromicro.com/en/Products/Part_Numbers/0754/index.asp

A Hall sensor as above is an excellent solution. It can cost
somewhat more than good enough solutions in many cases but
for 1 off jobs is liable to be far more flexible and useful
than most others.

It allows

Non contact
AC or DC
Pre calibrated by the manufacturer

Here is one of the above from Digikey

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=620-1106-ND

In stock, \$US5.50 in 1.s
Of the ones they list this is arguably one of the best as it
has higher sensitivity than most (40 mV/A) and 100A range.
Some have lower values of both. It also has a good
temperature range -40 /+150.

Good datasheet & mini-app note

http://www.allegromicro.com/en/Products/Part_Numbers/0754/0754-050.pdf

There are potentially more accurate solutions, but few
easier ones to use.

Russell McMahon

> > How about something like:
> >
> > http://www.allegromicro.com/en/Products/Part_Numbers/0754/index.asp

Thanks everyone for the advice so far..

So you just stick the large straight leads on the IC into any old
ferrite core suppressor
that fits around the ac wire I am sensing?  Is there a specific ferrite core I
should be using? Does the plastic coating on the ac line affect the results
on these sensors?

Or does this actually wire inline in the AC circuit being sensed? I thought
these type of sensors were non contact.

Then run the output of the sensor through a rectifier to ensure a
stable dc current, then to a smoothing capacitor, and finally to
the ADC converter on the PIC for measurement.

So a PIC can actually read the full output voltage of this sensor?

I am not familiar with the voltage/current handling capabilities of

Can a pic handle the full output of this IC?

Thanks,
Mike
> http://www.allegromicro.com/en/Products/Part_Numbers/0754/index.asp

> So you just stick the large straight leads on the IC into any old
> ferrite core suppressor

The ferrite is in the package

> Or does this actually wire inline in the AC circuit being sensed?

Yes, in-line

> I thought these type of sensors were non contact

Most types are. This one happens to be integrated and part of the
circuit. If you want a clamp or feed-through sensor, have a look at

> So a PIC can actually read the full output voltage of this sensor?

If Vcc of the PIC is greater than or equal to the Vcc of the sensor
IC, yes. Otherwise you'd have to scale down the sensor IC output

AFAIK, Hall sensors sit at Vcc/2. If you want just positive current
figures, that Vcc/2 to Vcc range can be expanded to 0V - Vcc
with a summing amp

As they recommend 5V for the IC Vcc, you'd probably use that
also for the PIC. Actually 5.12V would make the maths easier -
you'd have to check the error and linearity figures for the sensor IC
to see whether that's worthwhile

> I am not familiar with the voltage/current handling capabilities of

The sensor IC is quoted at 1ohm o/p at 1.2mA. The recommendation
for some PICs is 5k input impedance, others at 2.5k input impedance.
They will work at higher i/p impedances if you have longer sampling
times or buffer with a capacitor

On 3/4/08, Michael Cunningham <michael.cunningham.usagmail.com> wrote:
> I am rather new to PIC development but I am learning fast. I would like to
> build a circuit to measure the amount of AC current in AMPs flowing through
> a wire. Any suggestions on how to measure the current without directly
> connecting to the wire which will be carrying up to 125vac @ 30 amps.
>
> Can the PIC ADC be used for this purpose? Eventually I would like to drive an
> LCD with the value in amps.
>
> Any suggestions on where to begin?

See what error do you want for this measurement. The hall effect
sensors are good as "indicators" and not for precise measurements over
the full temperature range. If 5% is ok for you (1.5A error at full
scale), go with hall, but if you need a more accurate measurements go
with precision transformers.

Vasile
>> www.allegromicro.com/en/Products/Part_Numbers/0754/index.asp
>>
>> So you just stick the large straight leads on the IC into any old
>> ferrite core suppressor
>
>The ferrite is in the package

I don't think they have any ferrite in th package. It is not needed.

>> Or does this actually wire inline in the AC circuit being sensed?
>
>Yes, in-line
>
>> I thought these type of sensors were non contact

You may also like to look at a similar device made by LEM, which I learnt

See
(watch out for URL folding, try http://www.lem.com/ otherwise) for their
Minisens range. These require an external loop around the sensor to achieve
the same as the Allegro units, but it does mean that you can vary the
sensitivity, while still using the same device, by varying the number of
turns around it.

>Or does this actually wire inline in the AC circuit being sensed?

Yes

>I thought these type of sensors were non contact.

Internally there is a sensor that measures the magnetic field created by the
current being measured. Hence there is no electrical contact between the
circuit under test and the measurement circuit.

Vasile Surducan wrote:
{Quote hidden}

This really depends on the waveform, Some time ago i built a power
monitor using a current transformer
and resistor as load  i picked the resistor to give areasonable voltage
span for the PIC ADC and i connected
the other end of the winding to V/2. This mostly worked very well but i

When testing after i first built it, i tried various loads and all was
well until i tried a power drill, which at low
speeds only uses one half of the mains cycle. Because the average
voltage out of the current transformer is zero
the zero part of the waveform gets shifted below zero. This obviously
affect RMS calcualtions.

I guess one solution would be to have a very high turns ratio and use an
opamp configured to convert current to voltage.

PS I'm sure the LEM ones are nearer 1%

> >The ferrite is in the package
>
> I don't think they have any ferrite in the package. It is not needed.

There's "something"

Datasheet says -

Both tr and tRESPONSE are detrimentally affected by eddy current
losses observed in the conductive IC ground plane and, to varying
degrees, in the ferrous flux concentrator within the current sensor
package

At 06:00 AM 3/5/2008, Richard Benfield wrote:

>PS I'm sure the LEM ones are nearer 1%

The LEM current sense modules are freaking accurate.

They are also Hall-sensor based, but as part of a closed-loop
servo.  Hall sensor linearity is completely a non-issue with the LEM parts.

They have a gapped ferrous (metal) loop that the conductor passes
through.  There is a solenoid coil that the loop passes through, with
a linear hall sensor in the middle of the gap.  An op-amp causes
current to flow through the solenoid coil such that the field from
the coil balances (cancels) the field from the conductor(s) passing
through the loop.  The current needed to balance the field is made
available on the output pin.

Although the LEM modules are good down to DC, I've only ever used
them for measuring AC current.  As such, I have NO idea if the DC
zero point shifts with temperature or not - its just not been important for us.

dwayne

--
Dwayne Reid   <dwaynerplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

I'm currently doing (or attempting to do!) much the same thing using one
of these:

http://www.crmagnetics.com/pdf/3110.pdf

3000-turn split-core transformer - gives up to 10V depending on the
burden resistor - I'm using this to connect it to the PIC:

http://board.homeseer.com/showpost.php?p=627341&postcount=19

(Rest of discussion may or may not be relevant:

HTH,
Andrew

On 3/4/08, Michael Cunningham <michael.cunningham.usagmail.com> wrote:
>> I am rather new to PIC development but I am learning fast. I would like to
>> build a circuit to measure the amount of AC current in AMPs flowing through
>> a wire. Any suggestions on how to measure the current without directly
>> connecting to the wire which will be carrying up to 125vac @ 30 amps.
>>
>> Can the PIC ADC be used for this purpose? Eventually I would like to drive an
>> LCD with the value in amps.
>>
>> Any suggestions on where to begin?
>>
On Sun, Mar 9, 2008 at 5:20 AM, piclist <amerton.piclistihug.co.nz> wrote:
> I'm currently doing (or attempting to do!) much the same thing using one
> of these:
>
> http://www.crmagnetics.com/pdf/3110.pdf

Very cool.. looks like a good solution. Do you run the entire ac line
hot/neutral/ground through the unit or just the hot? I like that it doesnt
splice into the wire.  It has a 0.40" opening for the wire. If you just run
the hot through doesnt the extra space for it to move around change
the reading?  Ie varies the strengrth of the detected magnetic field?

How acurate is it?

Is this true?

"When you convert a AC to DC by a full wave rectifier
bridge you have to multiply the output voltage by 1.41
to obtain the VA"

If so I never knew this before:)

Thanks,
Mike
Michael Cunningham wrote:
> On Sun, Mar 9, 2008 at 5:20 AM, piclist <amerton.piclistihug.co.nz> wrote:
>
>> I'm currently doing (or attempting to do!) much the same thing using one
>> of these:
>>
>> http://www.crmagnetics.com/pdf/3110.pdf
>>
>
> Very cool.. looks like a good solution. Do you run the entire ac line
> hot/neutral/ground through the unit or just the hot? I like that it doesnt
> splice into the wire.  It has a 0.40" opening for the wire. If you just run
> the hot through doesnt the extra space for it to move around change
> the reading?  Ie varies the strengrth of the detected magnetic field?
>
>
You have to hook it round just one, either phase or neutral, otherwise
the pair will cancel each other out.  I bought a \$3, 1 meter three-core
extension cable and split it with a hobby knife, since the meter box
(where this thing is going to end up!) is too far away and outside...

I haven't tried moving the cable round to see if there's any effect - I
assume there would be, but have no idea how much.  Since the readings
are so close, I probably won't bother - if need higher accuracy, you
might have to investigate...
> How acurate is it?
>
I have a linear response (based on 3 data points!) which closely mirrors
the readings from the NZD20 "appliance meter" I bought locally - my
CR3110 reads slightly low, but it's off by around 20W/1000, which is
good enough for my purposes...

Admittedly, I do have a non-linear region at the bottom end (<150W -
one/two lightbulbs!), but I can live with that.  It may be something to
do with the bridge (4*4004's), but I haven't investigated too much yet -
the forum suggests that this shouldn't be the cause, though.
> Is this true?
>
> "When you convert a AC to DC by a full wave rectifier
> bridge you have to multiply the output voltage by 1.41
> to obtain the VA"
>
>
I kind of ignored that bit of the discussion since I got pretty close to
the bought meter without it.  However, I just consulted my (20-year-old)
university text (electronics was not my major!) and it seems that the
1.4*output gives the peak voltage suffered by the bridge, so I guess it
should be used to determine the diode spec?  (I think my 4004's are way
over spec'd, if that's the case!)

But I'm sure some of the Great Experienced Minds here will have

Andrew
Hi Andrew:

It's been my experience that the split type current transformers only work well
for currents near their max rating.  For small currents the error increases.
It sort of says that on the CR Magnetics web page, but not clearly.

I used a toroid type for just this reason and put it in the main breaker panel
with the burden resistor and some series resistors as a safety precaution, i.e.
the output is a low voltage into a high impedance D/A so the resistors will
normally have no effect, but if line voltage appeared on the wires it would be
current limited.

Using a seismometer type display on the computer screen was quite interesting.
I.e. to get the maximum time resolution vertical lines are drawn one pixel
wide and separated by maybe 10 pixels.  24 hours of data of zero current is a
whole bunch of straight vertical lines.

With this setup you can see when a current starts and stops.  Also since most
things draw different currents and are used at different times of day you can
pretty much see what's being used.  For example you can tell if it's a bathroom
light or the refrigerator light that caused the current late at night.

--
Have Fun,

Brooke Clarke
http://www.prc68.com/P/Prod.html  Products I make and sell
http://www.prc68.com/Alpha.shtml  All my web pages listed based on html name
http://www.PRC68.com
http://www.precisionclock.com
On 10/03/2008, piclist <amerton.piclistihug.co.nz> wrote:
> Michael Cunningham wrote:

....
> > "When you convert a AC to DC by a full wave rectifier
> > bridge you have to multiply the output voltage by 1.41
> > to obtain the VA"
> >
> >

Michael,
When measuring AC power levels the problem is that the AC voltage and
current is varying so the instantainious power delivered (the product
of voltage and current at that instant) is not constant. So it is
convienient to convert the AC readings to an equivelent constant
reading that would deliver the same power over an extended period.
(i.e greater than one cycle). This can be done by taking a large
number of readings over the course of one cycle, multiplying the
current and voltage and iintegrating the result.

Or, if we assume we are looking at sinewaves, we can perfom a
mathmatical calculation based on whatever parameter we actually know
or measure.

So for example if we have a sinewave at 325V peak applied to a
resistor, it will produce pretty much the same amount of heat as if a
230V DC voltage was applied.

With a sinewave, the ratio of the peak value  to this "average"
(otherwise known as the rms value)  is root2 = 1.414. That's where the
1.41 factor comes from and mathamatically it appears as a term in the
integration result.

So a "230V" AC supply is actually a calculated value, based on the
waveform being a sinewave with a peak value of just over 325V.

The same goes for current.

So a 30A AC current will have a peak value of ~42.4V.

If you are measuring the current, it is therefore useful to know the
type of measurement you are making - is it an instantainious reding -
or a pek reading. If it is a peak reading, how long is the peak value
held before it decays?

Since you are using a PIC, the readings are likely to be
instantainious, in which case you either need to synchronise the
reading with the incoming signal peak, or take a number of readings
and  process the results. Simply averaging the readings will give you
a value related to the correct value but there will always be some
error as the straightforward avaerage value is NOT the same as the rms
vale..

Best bet / easiest. is to take samples at 1mS intervals  (or less) and
for longer than an AC cycle and use the maximum of these as the peak
value to convert to rms by dividing by 1.414. Or even better, use the
(max-min) result  for a pk-pk reading & divide by 2.818

Always assuming a sine wave.

The complicated way is to take each value, square it, average the
results over a cycle period (17 or 20mS for 60 or 50Hz) and then
square root to get the rms value.
This will produce accurate results in the event the waveform is not a
sine wave but is a lot more calculation intensive.

Richard
Michael,
> When measuring AC power levels the problem is that the AC voltage and
> current is varying so the instantainious power delivered (the product
> of voltage and current at that instant) is not constant. So it is
> convienient to convert the AC readings to an equivelent constant
> reading that would deliver the same power over an extended period.
> (i.e greater than one cycle). This can be done by taking a large
> number of readings over the course of one cycle, multiplying the
> current and voltage and iintegrating the result....

Thanks Richard.. That was was one of the most informative emails I

Mike
Since most of these type projects seek to measure motor driven things or the
like, anyone care to address power factor? It can be quite much, even for a
kitchen stand mixer... ;)

Richard Prosser wrote:
{Quote hidden}

> On 10/03/2008, piclist <amerton.piclistihug.co.nz> wrote:
>> Michael Cunningham wrote:
>
> ....
>> > "When you convert a AC to DC by a full wave rectifier
>> > bridge you have to multiply the output voltage by 1.41
>> > to obtain the VA"

A bit awkwardly put. If I have a 10VAC (RMS) input and full-wave
rectify that with ideal diodes assembled into an ideal bridge
rectifier, and draw 1A
from the output, the input current will be 1A. 10W in, 10W out.

However, if I put a BFC (big fat capacitor) on the output, the
output voltage will rise to 14.14V  (minus a bit because of ripple),
and if I draw 1A from it, I'll be getting 14.1W out. The input power
(obviously) is also 14.1W. Typically you want to use a transformer
rated 1.6 to 1.8 x the DC output current of the power supply because
of the I^2R heating of the transformer coils. In this case, if we draw
1A (14.1W), we'd want to use a transformer rated at 1.6-1.8A
at 10V, or 16-18VA

Best regards,
Spehro Pefhany
--
"it's the network..."                          "The Journey is the reward"
s...interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Here's an interesting idea using opto isolators to directly measure true
power and apparent power.

http://www.edn.com/article/CA6531585.html?spacedesc=designideas&industryid=44217

Harold

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