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