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Help interfacing Current Transformers to ADC

J

John Popelish

Jan 1, 1970
0
beananimal said:
Here is what I gather my circuit is suposed to look like:
http://tinyurl.com/f6et7

Here is what I understand that needs to be done (In the event I have
made no errors).

1) I am still unsure of the best choice for the CT (the 500:1 Triad, or
the 1000:1 Avemco)

Best is not clear to me at this point, except for price. One of the
problems created by the Triad is that the load current has to be
connected through circuit board traces. This may produce some
practical problems. The Amveco just has a hole that any wire can pass
through. This allows a few connection options that do not include
passing multi ampere current through traces. The connections may cost
more than either transformer. The lower output current from the
Amveco may also be better suited to be absorbed by an opamp output.
2) the choice of CT will dictate the size of R1
3) Once R1 is calculated, then R2 needs to be selected to provide the
A/D with a suitable voltage range.

You still have not quite gotten your head around the concept of a
current source. The opamp responds to the input current, by producing
an output voltage that causes the input current to pass through R2,
while holding the input at zero volts (because the input pin does not
draw any significant current, so there is no other place for it to
go). R1 has no effect on this process, since the current into it must
also be the current out of it. It just adds extra voltage drop the
transformer must supply, lowering its accuracy a little. The optimum
value for R1 from the current transformer accuracy standpoint is zero.
Jim added it to improve the stability of the opamp feedback process,
to suppress any tendency to oscillate. That is its only purpose. So
R2 is the only part that determines the output volts per amp sensed,
and that calculation includes the transformer ratio.
4) The value of R2/R1 gives the Amplification ratio of the inverting
amplifier that is formed by the OP-AMP. In this case it would be 7.87.
So 1mv on the output of the Bridge will provide 7.87mV to the D/A
through R3. 10mV would provide 78.7mV. and 100mV would provide 787mv?

Not when R1 is fed by a current source.
5) The inverting amp is being fed from the negative side of the bridge,
so the output of the amp will be a positive voltage?

Right. This works as long as the transformer current can be sucked
into the opamp output through R2, while balancing the inverting input
at zero volts, to match the voltage applied to the non inverting input.
6) I think I have the diodes oriented correctly to protect the A/D
input pin from the transients. The more I look at them the more I
doubt myself. I thought I understood what was going on here.

I think you have them right. They prevent the output from swinging
more than a diode drop more positive than Vcc or below ground. But I
would add a second resistor after this clamp, since this is the same
voltage that turns on diodes in the PIC inputs, sharing this clamping
operation in an unknown way. The second resistor puts almost all the
clamping current through your diodes. This is pretty important, since
forward biasing, even a little bit, the internal clamping diodes,
fouls up the accuracy of all the analog input channels. For this
reason, a single current transformer that is producing an excessive
output will ruin the accuracy of all the other channels, unless you
take care of this detail. I normally use Schottky diodes (they
conduct with half a diode drop), to keep the input clamping current
very close to zero. This also applies to the overload clamp at the
opamp input. If one of these comes on, it will affect the operation
of the other 3 opamps in that package. So I would go with a Schottky,
there, also.
7) From the datasheets I see that the recomended BAV99 package is
faster and has a lower capacitance than the through hole 1n4148s I
chose. I hope this is not an issue for the intended purpose, as I am
not good with SOTs and certainly don't want to do 24 of them at 3 pins
each!

If I am totaly off base, please let me know and I will give this up and
buy more electronics books before I waste anybody elses time.

This is not at all a waste of time. This is producing something worth
archiving, that others can find with a search.
Lastly: There will be 24 of these in my house running 24/7. After the
numerous open secondary warnings, is there anything else that I need to
do to ensure safety for myself and our home? Honestly, after all of
the grave warnings, I actually feel like (in comparisson) that the act
of poking around the grid voltage and gain stages of my tube amps is
safe!

If the rectifier and burden (R1 and input clamp diode, in this case)
are soldered together, then there is little risk of an open secondary.
 
H

Homer J Simpson

Jan 1, 1970
0
I looked several years ago, but was afraid to purchase something that I
was not sure of. I started out looking 5MHz stuff and ended up looking
at 100MHz stuff. I gave up in frustration and lack of the most basic
ability to choose a good used unit. I have shelfs full of impulse eBay
buys that I did not research well.

UK or US? I wound up with 4 of the same model because they were so freakin'
cheap.

Item 180026570369 looks like a good price and would be most useful.

220025063526 was a good price although the S&H was a tad high and would bear
thinking about.

320025623173 really isn't worth anything unless you collect these.

170028648004 looks a bit grubby but if it would clean up would be fine. I
bet it cost mucho $$$ when new.

170027374925 would be worth looking at for sure.

I'd keep away from the modular scopes and rack mounted scopes unless you
really know what you are buying!
 
J

John Woodgate

Jan 1, 1970
0
In message <[email protected]>, dated Thu, 14
I'm analog... what's a "wave digital filter" ?:)

When you are washing vegetables, you put your hand in the water with
spread fingers and move it about to trap the veg as the water drains
away.
 
B

beananimal

Jan 1, 1970
0
Homer said:
UK or US? I wound up with 4 of the same model because they were so freakin'
cheap.

Item 180026570369 looks like a good price and would be most useful.

220025063526 was a good price although the S&H was a tad high and would bear
thinking about.

320025623173 really isn't worth anything unless you collect these.

170028648004 looks a bit grubby but if it would clean up would be fine. I
bet it cost mucho $$$ when new.

170027374925 would be worth looking at for sure.

I'd keep away from the modular scopes and rack mounted scopes unless you
really know what you are buying!

Thanks, That gives me someplace to start. I am in Pittsburgh, PA :)


Are any of the "sound card" or similar PC based scopes worth the
trouble?
 
B

beananimal

Jan 1, 1970
0
John said:
Best is not clear to me at this point, except for price. One of the
problems created by the Triad is that the load current has to be
connected through circuit board traces. This may produce some
practical problems. The Amveco just has a hole that any wire can pass
through. This allows a few connection options that do not include
passing multi ampere current through traces. The connections may cost
more than either transformer. The lower output current from the
Amveco may also be better suited to be absorbed by an opamp output.


You still have not quite gotten your head around the concept of a
current source. The opamp responds to the input current, by producing
an output voltage that causes the input current to pass through R2,
while holding the input at zero volts (because the input pin does not
draw any significant current, so there is no other place for it to
go). R1 has no effect on this process, since the current into it must
also be the current out of it. It just adds extra voltage drop the
transformer must supply, lowering its accuracy a little. The optimum
value for R1 from the current transformer accuracy standpoint is zero.
Jim added it to improve the stability of the opamp feedback process,
to suppress any tendency to oscillate. That is its only purpose. So
R2 is the only part that determines the output volts per amp sensed,
and that calculation includes the transformer ratio.


Not when R1 is fed by a current source.


Right. This works as long as the transformer current can be sucked
into the opamp output through R2, while balancing the inverting input
at zero volts, to match the voltage applied to the non inverting input.


I think you have them right. They prevent the output from swinging
more than a diode drop more positive than Vcc or below ground. But I
would add a second resistor after this clamp, since this is the same
voltage that turns on diodes in the PIC inputs, sharing this clamping
operation in an unknown way. The second resistor puts almost all the
clamping current through your diodes. This is pretty important, since
forward biasing, even a little bit, the internal clamping diodes,
fouls up the accuracy of all the analog input channels. For this
reason, a single current transformer that is producing an excessive
output will ruin the accuracy of all the other channels, unless you
take care of this detail. I normally use Schottky diodes (they
conduct with half a diode drop), to keep the input clamping current
very close to zero. This also applies to the overload clamp at the
opamp input. If one of these comes on, it will affect the operation
of the other 3 opamps in that package. So I would go with a Schottky,
there, also.

I have replaced the Diodes with the Schottkys. I chose common 1n5 type.
I have decided to use the Amveco 1000:1 CT for the reasons you have
stated (PCB connection) The docs specify a 100 ohm burden to get .1
Volts per Amp (and something about a Ratio Correction Factor of 1.02)

Yes, I am still having a hard time with the current source as opposed
to the voltage source, esp when it comes to the op-amp. R2 is still
bothering me and I am not sure how to determine it's value. I suppose
I need to revisit some of the better OP-AMP and active filter websites
and start from scratch with regards to what I understand.

Here is the updated schematic:
http://tinyurl.com/f3joa

Bill
 
B

beananimal

Jan 1, 1970
0
Homer said:
170027374925 would be worth looking at for sure.

"Kikusui COS6100M 100MHz 5 Channel Oscilloscope"

Wow, there are a lot of these $100 100MHz scopes (I looked around a bit
and some folks are saying they are every bit as good as the Tek 465? I
think I do understand that there are really only 2 full featured
channels.

This may be just what I want! Anybody have any opinions on them?
 
J

Jim Thompson

Jan 1, 1970
0
On 14 Sep 2006 23:00:16 -0700, "beananimal" <[email protected]>
wrote:

[snip]
Here is what I gather my circuit is suposed to look like:
http://tinyurl.com/f6et7

Here is what I understand that needs to be done (In the event I have
made no errors).

1) I am still unsure of the best choice for the CT (the 500:1 Triad, or
the 1000:1 Avemco)

I'd do the 1000:1, since the OpAmp has to source the current delivered
by the CT (see below).
2) the choice of CT will dictate the size of R1

Sort of, I'd use about 1/2 of the specified burden resistor.
3) Once R1 is calculated, then R2 needs to be selected to provide the
A/D with a suitable voltage range.

No. R2 is selected such that the current from the CT, filtered, times
R2 is your desired full-scale voltage.
4) The value of R2/R1 gives the Amplification ratio of the inverting
amplifier that is formed by the OP-AMP. In this case it would be 7.87.
So 1mv on the output of the Bridge will provide 7.87mV to the D/A
through R3. 10mV would provide 78.7mV. and 100mV would provide 787mv?

Just think currents
5) The inverting amp is being fed from the negative side of the bridge,
so the output of the amp will be a positive voltage?
Yes

6) I think I have the diodes oriented correctly to protect the A/D
input pin from the transients. The more I look at them the more I
doubt myself. I thought I understood what was going on here.

Your drawing is correct.
7) From the datasheets I see that the recomended BAV99 package is
faster and has a lower capacitance than the through hole 1n4148s I
chose. I hope this is not an issue for the intended purpose, as I am
not good with SOTs and certainly don't want to do 24 of them at 3 pins
each!

Just use the 1N4148... speed isn't an issue here.
If I am totaly off base, please let me know and I will give this up and
buy more electronics books before I waste anybody elses time.

Lastly: There will be 24 of these in my house running 24/7. After the
numerous open secondary warnings, is there anything else that I need to
do to ensure safety for myself and our home? Honestly, after all of
the grave warnings, I actually feel like (in comparisson) that the act
of poking around the grid voltage and gain stages of my tube amps is
safe!

Build it in a metal box ?:)

One last caution, and I hope it doesn't confuse you. It's possible
that the LM324 won't be able to source output current sufficiently to
support the CT current, if your full scale is over 10 Amps. In that
case add an NPN emitter follower to the output of the LM324 and drive
the feedback network with the emitter.

...Jim Thompson
 
J

John Larkin

Jan 1, 1970
0
On 14 Sep 2006 23:00:16 -0700, "beananimal" <[email protected]>
wrote:

[snip]
Here is what I gather my circuit is suposed to look like:
http://tinyurl.com/f6et7

Here is what I understand that needs to be done (In the event I have
made no errors).

1) I am still unsure of the best choice for the CT (the 500:1 Triad, or
the 1000:1 Avemco)

I'd do the 1000:1, since the OpAmp has to source the current delivered
by the CT (see below).

Umm, err, I think someone has already mentioned this as a potential
problem. Not to mention the fact that ct accuracy heads for the pits
as burden voltage increases.

John
 
J

Jim Thompson

Jan 1, 1970
0
On 14 Sep 2006 23:00:16 -0700, "beananimal" <[email protected]>
wrote:

[snip]
Here is what I gather my circuit is suposed to look like:
http://tinyurl.com/f6et7

Here is what I understand that needs to be done (In the event I have
made no errors).

1) I am still unsure of the best choice for the CT (the 500:1 Triad, or
the 1000:1 Avemco)

I'd do the 1000:1, since the OpAmp has to source the current delivered
by the CT (see below).

Umm, err, I think someone has already mentioned this as a potential
problem. Not to mention the fact that ct accuracy heads for the pits
as burden voltage increases.

John

I've already said, several times now, use a resistor somewhat SMALLER
than the specified burden resistor ;-)

I've actually done this without the rectifier, and R1=0, with a fast
OpAmp, for current limit on-the-fly (cycle-by-cycle) applications.,
but I was reluctant to have the OP do R1=0, for fear that a cheap
OpAmp will sing, and the OP is inexperienced.

...Jim Thompson
 
J

John Popelish

Jan 1, 1970
0
beananimal said:
I have replaced the Diodes with the Schottkys. I chose common 1n5 type.
I have decided to use the Amveco 1000:1 CT for the reasons you have
stated (PCB connection) The docs specify a 100 ohm burden to get .1
Volts per Amp (and something about a Ratio Correction Factor of 1.02)

That just means that it looses 2% of its turns accuracy when driving a
100 ohm burden.
Yes, I am still having a hard time with the current source as opposed
to the voltage source, esp when it comes to the op-amp. R2 is still
bothering me and I am not sure how to determine it's value.

All the current transformer current has to pass through R2 and into
the opamp output. The input pins are very high resistance, so they
respond to applied voltage(amplifying the difference voltage between +
and - inputs) without drawing any current. Since the + input is at
zero volts, and the input current from the CT is negative (because it
comes from the - end of the bridge) the output has to go positive to
suck that current through R2. But the voltage to drive that current
through R2 comes from the opamp output, rather than from the CT, so
its accuracy is kept high. If you want the output to be 5 V when the
average of the absolute line current is 1 A, calculate what R2 value
will drop 5 volts when 1/1000 A passes through it. That would be 5 k.

The capacitor across R2 is what converts the reading from
instantaneous absolute current to average of the absolute current, if
R2*C is longer than a cycle.
 
H

Homer J Simpson

Jan 1, 1970
0
Wow, there are a lot of these $100 100MHz scopes (I looked around a bit
and some folks are saying they are every bit as good as the Tek 465? I
think I do understand that there are really only 2 full featured
channels.

Yes. Often the sellers don't know what they are either. More than two real
channels is not common.
 
J

John Perry

Jan 1, 1970
0
Jim said:
I've already said, several times now, use a resistor somewhat SMALLER
than the specified burden resistor ;-)

I've actually done this without the rectifier, and R1=0, with a fast
OpAmp, for current limit on-the-fly (cycle-by-cycle) applications.,
but I was reluctant to have the OP do R1=0, for fear that a cheap
OpAmp will sing, and the OP is inexperienced.

....Or, better, use R1=10 ohms between the bridge output and ground, and
Rin=100 ohms between the bridge output and the op amp input. You get
both a good, low resistance for the CT and low input current for the op
amp. You would do well to use a precision or auto-zero op amp.

Also, for input protection I like four hefty LEDs, series pairs, each
pair back-to-back for CT output protection. You get insignificant shunt
current, a pretty sharp limiting voltage, and a bonus is that if you can
see them, you failure indication. Just have to be sure the selected
LEDs can handle the temporary inrush currents.

I've seen an easy-to-use schematic module that would put out pdf's, but
I can't find it right now. I'd like to illustrate some of my comments
on abse. Any pointers?

John Perry
 
J

Jim Thompson

Jan 1, 1970
0
...Or, better, use R1=10 ohms between the bridge output and ground, and
Rin=100 ohms between the bridge output and the op amp input. You get
both a good, low resistance for the CT and low input current for the op
amp. You would do well to use a precision or auto-zero op amp.

That works. Allows scaling of the CT current that is applied to the
OpAmp, but, as you point out, aggravates VOS effects.
Also, for input protection I like four hefty LEDs, series pairs, each
pair back-to-back for CT output protection. You get insignificant shunt
current, a pretty sharp limiting voltage, and a bonus is that if you can
see them, you failure indication. Just have to be sure the selected
LEDs can handle the temporary inrush currents.

I've seen an easy-to-use schematic module that would put out pdf's, but
I can't find it right now. I'd like to illustrate some of my comments
on abse. Any pointers?

John Perry

I'm basically using the original (non-crispy :) version of MicroSim
Schematics as a front-end for PSpice.

Surfing around you can probably find it on the web, and it doesn't
require licensing. Unlike OrCAD Capture, it's VERY easy to use, VERY
intuitive, and it EASILY supports hierarchical drawings.

As for outputting PDF's, all you need is a printer driver for PDF...
PDFWriter (Adobe, $) or CutePDF Writer (free).

...Jim Thompson
 
B

beananimal

Jan 1, 1970
0
Jim said:
That works. Allows scaling of the CT current that is applied to the
OpAmp, but, as you point out, aggravates VOS effects.


I'm basically using the original (non-crispy :) version of MicroSim
Schematics as a front-end for PSpice.

Surfing around you can probably find it on the web, and it doesn't
require licensing. Unlike OrCAD Capture, it's VERY easy to use, VERY
intuitive, and it EASILY supports hierarchical drawings.

As for outputting PDF's, all you need is a printer driver for PDF...
PDFWriter (Adobe, $) or CutePDF Writer (free).

...Jim Thompson
--

Every time I think I this worked out, you guys improve the design. I
honestly have to say I feel somewhat helpless. Though I am learning,
this is one of the hardest projects I have had to wrap my head around
(code makes sense to me, current sources and analog stuff is a
struggle). This has gone a long way to expose my gaps in knowledge
and basic understanding of some the basic fundamentals of electronics
and the building blocks of circuits and how they interact.

John, I would love to see your ideas (as it is becoming more and more
obvious that my grasp of all of this is just glimmers of understanding
that I am trying to sort out from post to post).

With regards to R2, I will be using a PIC with a 5V supply rail. From
what I can decipher from the datasheet, the MAX input voltage to the
A/D is the rail voltage +0.3V

I would also gather that I want to scale the voltage to cover as much
of that range (0-5V) as possible to garner the most resolution I can.

If I were to decide that 20A was the max input to the CT, it would
follow that the 1000:1 ration would give me 20ma at the output and 2
volts on the bridge? It would follow that to scale the 2V to 5V I
would need a 250 Ohm resistor for R2?

In similar, If I decided to use the 10A CT and wanted full scale (5V)
at 10A, then R2 would be 500 Ohm?

With regards to the emitter follower. I understand the basic concept
but again lose some of the details of the exact method to connect the
transistor. It's not something that is intuitive to me and looking at
the emitter follower circuits I have seen in the past and in my "book"
the emitter is usually tied to ground via a resistor. I would assume
that would hold true in this case. The output (between the emitter and
ground) would feed the network and the A/D. The base is connected to
the output of the opamp and the collector to Vcc?

I am not sure if the (4) LEDs are to replace D1 (but would assume so)

Again, here is my interpretation of the circuit with the values I have
calculated (er guessed at).
http://tinyurl.com/h5jq4
 
B

beananimal

Jan 1, 1970
0
John said:
...Or, better, use R1=10 ohms between the bridge output and ground, and
Rin=100 ohms between the bridge output and the op amp input. You get
both a good, low resistance for the CT and low input current for the op
amp. You would do well to use a precision or auto-zero op amp.

Also, for input protection I like four hefty LEDs, series pairs, each
pair back-to-back for CT output protection. You get insignificant shunt
current, a pretty sharp limiting voltage, and a bonus is that if you can
see them, you failure indication. Just have to be sure the selected
LEDs can handle the temporary inrush currents.

I've seen an easy-to-use schematic module that would put out pdf's, but
I can't find it right now. I'd like to illustrate some of my comments
on abse. Any pointers?

John Perry

Would the LM318 be considered suitable (precision) or maybe one of the
Maxim MXL1014 types?

You can also use a PDF printer like "win2pdf" as a print driver. Most
of the free PDF printers do put a visible watermark or add a page to
the PDF.

Bill
 
J

Jim Thompson

Jan 1, 1970
0
On 15 Sep 2006 12:16:14 -0700, "beananimal" <[email protected]>
wrote:

[snip]
Every time I think I this worked out, you guys improve the design. I
honestly have to say I feel somewhat helpless. Though I am learning,
this is one of the hardest projects I have had to wrap my head around
(code makes sense to me, current sources and analog stuff is a
struggle). This has gone a long way to expose my gaps in knowledge
and basic understanding of some the basic fundamentals of electronics
and the building blocks of circuits and how they interact.

Well... I've only been a tinkerer for 50 years... it takes time to get
good at this stuff. ;-)
John, I would love to see your ideas (as it is becoming more and more
obvious that my grasp of all of this is just glimmers of understanding
that I am trying to sort out from post to post).

With regards to R2, I will be using a PIC with a 5V supply rail. From
what I can decipher from the datasheet, the MAX input voltage to the
A/D is the rail voltage +0.3V

What supply do you have available for the OpAmp?
I would also gather that I want to scale the voltage to cover as much
of that range (0-5V) as possible to garner the most resolution I can.

If I were to decide that 20A was the max input to the CT, it would
follow that the 1000:1 ration would give me 20ma at the output and 2
volts on the bridge? It would follow that to scale the 2V to 5V I
would need a 250 Ohm resistor for R2?

In similar, If I decided to use the 10A CT and wanted full scale (5V)
at 10A, then R2 would be 500 Ohm?

With regards to the emitter follower. I understand the basic concept
but again lose some of the details of the exact method to connect the
transistor. It's not something that is intuitive to me and looking at
the emitter follower circuits I have seen in the past and in my "book"
the emitter is usually tied to ground via a resistor. I would assume
that would hold true in this case. The output (between the emitter and
ground) would feed the network and the A/D. The base is connected to
the output of the opamp and the collector to Vcc?

I am not sure if the (4) LEDs are to replace D1 (but would assume so)

Again, here is my interpretation of the circuit with the values I have
calculated (er guessed at).
http://tinyurl.com/h5jq4

Not correct. Pin 1 of rectifier goes to ground. 10 ohms goes from
pin 3 to ground.

If VCC for OpAmp is 5V, you will NOT be able to output 5V to the ADC.
A minimum of 8V for the OpAmp is needed.

...Jim Thompson
 
B

beananimal

Jan 1, 1970
0
Jim said:
What supply do you have available for the OpAmp?

I can provide the OpAmps whatever they need. I think the max on the
PIC was 5.5V supply. I suppose I will need to power the PIC with a
different supply rail than the filter. I had counted on supplying the
filter section with the supply rail from the the 15V rail that powers
the other electronics in the project.
If VCC for OpAmp is 5V, you will NOT be able to output 5V to the ADC.
A minimum of 8V for the OpAmp is needed.

I knew this part (at least!) but somehow neglected to remember it as my
brain tried consume all of the new information.

fixed:
http://tinyurl.com/znjvc

Bill
 
J

Jim Thompson

Jan 1, 1970
0
I can provide the OpAmps whatever they need. I think the max on the
PIC was 5.5V supply. I suppose I will need to power the PIC with a
different supply rail than the filter. I had counted on supplying the
filter section with the supply rail from the the 15V rail that powers
the other electronics in the project.


I knew this part (at least!) but somehow neglected to remember it as my
brain tried consume all of the new information.

fixed:
http://tinyurl.com/znjvc

Bill

But now you have a scale factor issue ;-)

Suppose you have 20A RMS or 20A*rt(2) peak, AVERAGE current into
summing node is...

20*rt(2)/1000*10/110*2/pi = 1.637mA * 250 (feedback R) = 409mV :-(

So feedback resistor needs to be ~3K for a 5V full-scale

Of course, since you used the 10 ohm, the OpAmp is no longer supplying
wads of current, so you don't need the follower ;-)

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
But now you have a scale factor issue ;-)

Suppose you have 20A RMS or 20A*rt(2) peak, AVERAGE current into
summing node is...

20*rt(2)/1000*10/110*2/pi = 1.637mA * 250 (feedback R) = 409mV :-(

So feedback resistor needs to be ~3K for a 5V full-scale

Of course, since you used the 10 ohm, the OpAmp is no longer supplying
wads of current, so you don't need the follower ;-)

...Jim Thompson

I'd probably go back to R2=787, and change R6=53.6

...Jim Thompson
 
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