Re: Dumb "current transformer" questions

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" <ignoramus29580@xxxxxxxxxxxxxxxxxxxx> wrote in message
news:9SOnf.440$CV4.427@xxxxxxxxxxxxxxxxxxxxxxxx
> On Tue, 13 Dec 2005 23:58:59 +0000, pentagrid@xxxxxxxxx
> <pentagrid@xxxxxxxxx> wrote:
>> On Tue, 13 Dec 2005 20:23:52 GMT, Ignoramus29580
>><ignoramus29580@xxxxxxxxxxxxxxxxxxxx> 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
>


.

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