Never use a half-wave rectifier in a CT secondary. On alternate
half-cycles it presents an open circuit to the CT secondary. If this
happens, the CT secondary voltage will skyrocket to very high numbers on
those half-cycles. You will very likely damage the CT or the connected
circuitry from the high voltage pulses.

Randy


"Ignoramus29580" wrote in message
...
On Tue, 13 Dec 2005 23:58:59 +0000,
wrote:
On Tue, 13 Dec 2005 20:23:52 GMT, 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?



While this is correct it's not the best way to use a
current transformer as a measurement device. This because most
meters capable of reading a few volts AC FSD are messed up by the
forward drop of diodes used to rectifiy the AC.

The trick is to use it as a true current transformer without
an intermediate voltage transformation. Feed the output of the
current transformer directly into a full wave rectifier - silicon
diodes are OK because forward voltage drop is not important.

Thanks, after some thinking I realized that it is excellent advice.


Short circuit the rectifier output directly through a DC
AMMETER. The ammeter will then read directly the MEAN value (0.90
x RMS) of the secondary current. The voltage drop of the
rectifier diodes will slightly increase the voltage drop at the
current transformer primary but will not affect the current
transformation ratio. For sine wave input waveform the equivalent
RMS current should be read as 1.11 x the indicated DC value.

That's very nice. I looked at my current transformers today and
realized that they are 200:0.1, not 200:1 as I incorrectly reported.

So, 50 amp current would translate into 0.025 amp current on the
transformer.

With, say, a 50 ohm resistor it would amount to 1.25 volts across the
resistor, or 0.03 watts of power dissipated on the resistor. I can get
away with using a small cheap 1/8 watt resistor.

i