Ignoramus29580 wrote:
Let's say that I have a cable and I want to measure the AC current
going through it. Up to, say, 100 amps.

I could use a current transformer, right?

If I have a say 200:1 current transformer, then on a 100 amp AC
current it would want to produce a 0.5 amp current. Then if I stick,
say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts
AC across the resistor.

Is that right?

Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.

Furthermore, since you're probably looking to
use it on your tig welder inverter, the transformer
probably should be rated to be accurate for
pulse work rather than just 60hz sine waves.

Here's more than you want to know about CT's:

http://www.kappaelectricals.com/technical.html#current

Oh, and don't ever leave the secondary open with
a load in the primary....

Jim Stewart wrote:

Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.

Wouldn't a hall sensor do a better job?
I mean with all the square wave and changing frequency a current
transformer seems to be not the best choice.
And if he is in DC ...


Nick, just curious
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Ignoramus29580 wrote:

I would use this meter (cost $19.99 on ebay):

How about this one? You just have to push the decimal point to the
right. And it is exactly what you need.

http://us.fluke.com/usen/products/LH...og_name=FlukeU
nitedStates&Category=CLMP(FlukeProducts)


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Ignoramus29580 wrote:
On Tue, 13 Dec 2005 12:38:48 -0800, Jim Stewart wrote:

Ignoramus29580 wrote:

Let's say that I have a cable and I want to measure the AC current
going through it. Up to, say, 100 amps.

I could use a current transformer, right?

If I have a say 200:1 current transformer, then on a 100 amp AC
current it would want to produce a 0.5 amp current. Then if I stick,
say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts
AC across the resistor.

Is that right?

Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Thanks.


Furthermore, since you're probably looking to
use it on your tig welder inverter, the transformer
probably should be rated to be accurate for
pulse work rather than just 60hz sine waves.


No, I will use it for my phase converter. For measuring current on the
welder, I have a DC current meter on the welder's panel.


Here's more than you want to know about CT's:

http://www.kappaelectricals.com/technical.html#current

Oh, and don't ever leave the secondary open with
a load in the primary....


Thanks, that was all very helpful.

I have some CTs and I want to use them with some resistors and diodes
to measure several currents in my phase converter, like input current
and leg 3 current.

I would use this meter (cost $19.99 on ebay):

http://oeiwcs.omron.com/webapp/comme...6&prrfnbr=4135

My schematic would be

Diode with low voltage drop
CT pole 1 ------------||----------------------------|--
__ to Omron
Resistor Capacitor ~~ process meter
CT pole 2 -------------------------------------------|--

I need the diode because this meter is for DC readings.

Since the voltage reading is screwed up by the voltage drop of the
diode, I would need to find a diode with low voltage drop.

The problem is way worse than that. Your circuit
will measure the *peak* voltage of *1/2* of the
sine wave (less the diode drop).

The reason is that the diode will only pass one
side of the sine wave. But when it's passing
that side, the capacitor will charge up to the
peak value (rms * sqt(2)) For the nitpickers,
yes, I'm assuming a well-formed sine wave.

The input impedance of the meter would also have
an effect, but since it's an unknown, I won't
go there.

So, 10 volts in would be 7.07 volts out.

The good news is that many times these process
meters have scale and offset values that can
be programmed. You could change the scale to
correct for the 1/2 wave and cap and change the
offset to correct for the diode. Not sure it's
the best way, but if it tracked a clip-on ammeter,
it may be good enough.

Ignoramus29580 wrote:
On Tue, 13 Dec 2005 14:12:39 -0800, Jim Stewart wrote:
Ignoramus29580 wrote:
On Tue, 13 Dec 2005 12:38:48 -0800, Jim Stewart wrote:

Ignoramus29580 wrote:

Let's say that I have a cable and I want to measure the AC current
going through it. Up to, say, 100 amps.

I could use a current transformer, right?

If I have a say 200:1 current transformer, then on a 100 amp AC
current it would want to produce a 0.5 amp current. Then if I stick,
say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts
AC across the resistor.

Is that right?

Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Thanks.


Furthermore, since you're probably looking to
use it on your tig welder inverter, the transformer
probably should be rated to be accurate for
pulse work rather than just 60hz sine waves.


No, I will use it for my phase converter. For measuring current on the
welder, I have a DC current meter on the welder's panel.


Here's more than you want to know about CT's:

http://www.kappaelectricals.com/technical.html#current

Oh, and don't ever leave the secondary open with
a load in the primary....


Thanks, that was all very helpful.

I have some CTs and I want to use them with some resistors and diodes
to measure several currents in my phase converter, like input current
and leg 3 current.

I would use this meter (cost $19.99 on ebay):

http://oeiwcs.omron.com/webapp/comme...6&prrfnbr=4135

My schematic would be

Diode with low voltage drop
CT pole 1 ------------||----------------------------|--
__ to Omron
Resistor Capacitor ~~ process meter
CT pole 2 -------------------------------------------|--

I need the diode because this meter is for DC readings.

Since the voltage reading is screwed up by the voltage drop of the
diode, I would need to find a diode with low voltage drop.

The problem is way worse than that. Your circuit
will measure the *peak* voltage of *1/2* of the
sine wave (less the diode drop).

I see. I coud use a 4 diode bridge and subtract 2 diode drops.

Just get a standard 5A full scale AC current meter. There's a few
problems with this diode stuff. If you want to measure upto 100A
get a 100:5 current transformer. Connect it straight to the meter.

1) they don't less than about 0.6 volts though with one diode
2) using one diode like in the ASCII art diagram will magnetize the
current transformer core.
3) the full rectifier will have a drop of 1.2 volts. You will not be
able to read any voltages below this.

A 10VA current transformer can only output upto 2 volts at 5A. SO,
with a 100:5 transformer and 50A though the wire though the donut, you
won't even get a DC voltage out of a bridge rectifier.


The reason is that the diode will only pass one
side of the sine wave. But when it's passing
that side, the capacitor will charge up to the
peak value (rms * sqt(2)) For the nitpickers,
yes, I'm assuming a well-formed sine wave.

Yes, it will be a normal sine wave.

The input impedance of the meter would also have
an effect, but since it's an unknown, I won't
go there.

It is a "known", actually, there is a table of impedances in the
datasheet. 1 or 10 megohm depending on the range.

So, 10 volts in would be 7.07 volts out.

Yes, that's OK with me. This meter can be programmed to perform linear
conversion y=ax+b.

All I have to do is bring various measurements (input current, input
voltage) to similar scale, so that:

1) when I switch to voltage (say 240V), my meter displays "240",
2) when I switch to current (say 25A), the meter displays "25".

That means that I have to perform conversion of my voltages from the
AC line voltage and from current transformers by bringing them to the
same scale. It's a simple algebra problem and can be done with
appropriate voltage dividers and also diodes.

The good news is that many times these process meters have scale and
offset values that can be programmed.

Just as mine is.

You could change the scale to correct for the 1/2 wave and cap and
change the offset to correct for the diode. Not sure it's the best
way, but if it tracked a clip-on ammeter, it may be good enough.

I agree. I will mess around with it, hopefuly it is not going to be
too difficult.

i

Ignoramus19198 wrote:

The reason for this is that I have to go with what I can get, and I
could get a process meter with DC capability.

Then I would build a rectifier out of OpAmps that has no diode-drop.

So you want to have a R behind the xfmr and then a rectifier that
rectifies that voltage across the R. Behind that a C and then into your
DC V-meter.


Nick
--
Motor Modelle // Engine Models
http://www.motor-manufaktur.de
DIY-DRO - YADRO - Eigenbau-Digitalanzeige

Nick Müller wrote:
Jim Stewart wrote:


Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Wouldn't a hall sensor do a better job?
I mean with all the square wave and changing frequency a current
transformer seems to be not the best choice.
And if he is in DC ...

Well, since it is Iggy that posted the question,
he probably has a big bag of surplus CT's that
he bought for $.50 (:

Keep in mind that the burden (load) on a current transformer increases as
the secondary load resistance increases. Keep the resistance as low as
possible and still develop adequate measurable output. I have never seen
current transformers without a either a resistor or coil directly across the
secondary, but it might be done. I think that unloading the secondary, even
momentarily until diodes are forward biased, is not a very good idea.
Depending on diodes and or long leads to a meter seems unnecessarily risky.
An unloaded secondary can develop destructively high voltages across the
secondary and high voltage drops in the primary along with excessive heating
of the transformer. Ideally, current transformers should have zero volts
across both primary and secondary.
Don Young
Don Young
"Ignoramus19198" wrote in message
.. .
On Wed, 14 Dec 2005 08:39:24 -0800, Jim Stewart
wrote:
Nick Müller wrote:
Jim Stewart wrote:


Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Wouldn't a hall sensor do a better job?
I mean with all the square wave and changing frequency a current
transformer seems to be not the best choice.
And if he is in DC ...

Well, since it is Iggy that posted the question,
he probably has a big bag of surplus CT's that
he bought for $.50 (:

No, I only have very few CTs, I need to use them for my phase
converter measurements.

i

If I understand you corrrectly, you mean to parallel a load resistor with a
rectified meter. That certainly should be good as long as the resistor value
is low enough and it does not get disconnected. The resistor should take the
majority of the secondary current.
Don Young
"Ignoramus19198" wrote in message
...
On Wed, 14 Dec 2005 22:41:09 -0600, Don Young wrote:
Keep in mind that the burden (load) on a current transformer increases as
the secondary load resistance increases.

Yes.

Keep the resistance as low as possible and still develop adequate
measurable output.

I agree.

I have never seen current transformers without a
either a resistor or coil directly across the secondary, but it
might be done. I think that unloading the secondary, even
momentarily until diodes are forward biased, is not a very good
idea. Depending on diodes and or long leads to a meter seems
unnecessarily risky. An unloaded secondary can develop destructively
high voltages across the secondary and high voltage drops in the
primary along with excessive heating of the transformer. Ideally,
current transformers should have zero volts across both primary and
secondary. Don Young Don Young "Ignoramus19198"


I think that what I could easily do, is parallel a higher resistance
diode so that it is parallel to the rectifier bridge. Then, while the
diodes in the bridge are recovering, this higher resistor would
momentarily take on current. (which is low since we are talking about
a transition of sinewave AC)

i

wrote in message
.. . On Wed, 14
Dec 2005 08:39:24 -0800, Jim Stewart
wrote: Nick Müller wrote: Jim Stewart
wrote:


Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Wouldn't a hall sensor do a better job?
I mean with all the square wave and changing frequency a current
transformer seems to be not the best choice.
And if he is in DC ...

Well, since it is Iggy that posted the question,
he probably has a big bag of surplus CT's that
he bought for $.50 (:

No, I only have very few CTs, I need to use them for my phase
converter measurements.

i





--

Ignoramus19198 wrote:
On Wed, 14 Dec 2005 08:37:19 +0000 (UTC), Cydrome Leader wrote:
Ignoramus29580 wrote:
On Tue, 13 Dec 2005 14:12:39 -0800, Jim Stewart wrote:
Ignoramus29580 wrote:
On Tue, 13 Dec 2005 12:38:48 -0800, Jim Stewart wrote:

Ignoramus29580 wrote:

Let's say that I have a cable and I want to measure the AC current
going through it. Up to, say, 100 amps.

I could use a current transformer, right?

If I have a say 200:1 current transformer, then on a 100 amp AC
current it would want to produce a 0.5 amp current. Then if I stick,
say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts
AC across the resistor.

Is that right?

Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.


Thanks.


Furthermore, since you're probably looking to
use it on your tig welder inverter, the transformer
probably should be rated to be accurate for
pulse work rather than just 60hz sine waves.


No, I will use it for my phase converter. For measuring current on the
welder, I have a DC current meter on the welder's panel.


Here's more than you want to know about CT's:

http://www.kappaelectricals.com/technical.html#current

Oh, and don't ever leave the secondary open with
a load in the primary....


Thanks, that was all very helpful.

I have some CTs and I want to use them with some resistors and diodes
to measure several currents in my phase converter, like input current
and leg 3 current.

I would use this meter (cost $19.99 on ebay):

http://oeiwcs.omron.com/webapp/comme...6&prrfnbr=4135

My schematic would be

Diode with low voltage drop
CT pole 1 ------------||----------------------------|--
__ to Omron
Resistor Capacitor ~~ process meter
CT pole 2 -------------------------------------------|--

I need the diode because this meter is for DC readings.

Since the voltage reading is screwed up by the voltage drop of the
diode, I would need to find a diode with low voltage drop.

The problem is way worse than that. Your circuit
will measure the *peak* voltage of *1/2* of the
sine wave (less the diode drop).

I see. I coud use a 4 diode bridge and subtract 2 diode drops.

Just get a standard 5A full scale AC current meter. There's a few
problems with this diode stuff. If you want to measure upto 100A
get a 100:5 current transformer. Connect it straight to the meter.

1) they don't less than about 0.6 volts though with one diode
2) using one diode like in the ASCII art diagram will magnetize the
current transformer core.
3) the full rectifier will have a drop of 1.2 volts. You will not be
able to read any voltages below this.

I am confused. With a current transformer, I am guaranteed a 200:0.1
current even if I put a full diode bridge into the current
transformer. Voltage drop should only result in extra power disspated
on the bridge.

Current xfmr 1 ----_
/ \
+ -
\ /
Current xfmr 2 ----~

the current will flow in one direction between points marked + and
-. Right?

Then, I can simply install a resistor between + and - and measure
voltage across the resistor (with a filter cap to decrease ripple).

Am I mistaken?

yes. The drop across the diodes may exceed the power rating of the current
transformer. They're for running a shunted meter.

The reason for this is that I have to go with what I can get, and I
could get a process meter with DC capability.

Oh please. You claim you can get anything. 5A full scale AC meters are not
hard to get.


http://tinyurl.com/882gb (datasheet and all)

A 10VA current transformer can only output upto 2 volts at 5A. SO,
with a 100:5 transformer and 50A though the wire though the donut, you
won't even get a DC voltage out of a bridge rectifier.

I will check my current transformers again tonight, but I would think
that with 200:0.1 ratio, they should hopefully be able to overcome the
voltage drop. Perhaps I am completely wrong.

i