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.