Re: Dumb "current transformer" questions
- From: Cydrome Leader <presence@xxxxxxxxxxxxxx>
- Date: Wed, 14 Dec 2005 08:37:19 +0000 (UTC)
Ignoramus29580 <ignoramus29580@xxxxxxxxxxxxxxxxxxxx> wrote:
> On Tue, 13 Dec 2005 14:12:39 -0800, Jim Stewart <jstewart@xxxxxxxxxxx> wrote:
>> Ignoramus29580 wrote:
>>> On Tue, 13 Dec 2005 12:38:48 -0800, Jim Stewart <jstewart@xxxxxxxxxxx> 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/commerce/command/ExecMacro/Omron/macros/itemdisp.d2w/report?prmenbr=316&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
>
.
- References:
- Re: Dumb "current transformer" questions
- From: Jim Stewart
- Re: Dumb "current transformer" questions
- From: Jim Stewart
- Re: Dumb "current transformer" questions
- Prev by Date: Re: Colchester Student Mk 1 - some questions
- Next by Date: Re: Ping Bruce bergman -- 3 phase receptacles
- Previous by thread: Re: Dumb "current transformer" questions
- Next by thread: Re: Dumb "current transformer" questions
- Index(es):
Relevant Pages
|
