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Current transformer

Started by Tim Williams November 30, 2011
"vkj"

> Why? I dont understand why one cant simply use the open ckt voltage on > the > secondary as a measure of the primary current.
** For the simple reason that a "transformer" always converts voltages according to the turns ratio on the windings. When you feed an iron cored transformer prmary with a current, the resulting voltage drop is undefined and very non-linear.
> If you consider the ct. transformer as a coupled inductor, then > the actual calculations are quite simple.
** An inductor could be used to measure AC current (at a single frequency) as there is a simple ratio between current and voltage drop ( long as the inductive reactance is much bigger than the resistance). A secondary winding would enlarge the voltage to be measured - but that is no longer a *current transformer* at all and loses most of the advantages. The original idea of a current transformer was to allow safe and simple measurement of high AC supply currents with a standard moving iron ammeter. No rectifiers are needed, no high voltages involved and there is virtually no effect on the circuit being measured. ... Phil
On Monday, January 23, 2012 7:17:58 PM UTC-8, vkj wrote:

> >Current transformer: > >current_in / turns_ratio = current_out. (for a low load resistor or > short). > >current_out x load_resistor = voltage_out.
> >One lesson: Do Not Leave Out The Load Resistor, as then > > Ze Foltage Kan Beacome Ferry High > > > > Why? I dont understand why one cant simply use the open ckt voltage on the > secondary as a measure of the primary current.
The function of a 'current transformer' is as a near-perfect conductor on the primary circuit (it doesn't dissipate energy, ideally). If you leave the secondary open, the primary circuit is OPEN, just like you'd turned a switch off. You aren't going to measure the current that way, I hope!
"DJ Delorie"  wrote in message news:xnd3a9aic0.fsf@delorie.com...

> "vkj" <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> writes: >> Why? I dont understand why one cant simply use the open ckt >> voltage on the secondary as a measure of the primary current. >> Heres my argument:
> Because a current transformer is normally in the 1000:1 range. > Do you really want to try to measure 120,000 volts?
Actually, the voltage would never get anywhere near that high, although = it=20 could get high enough to damage the insulation. For one thing, a 1000:1=20 ratio is a 5000:5 CT (usually), which is fairly rare. Most systems run=20 currents of 100 to 500 amps or so. If you do run current through a CT = with=20 an open secondary, it will just function as an inductor, and it will = soon=20 saturate at 60 Hz so the primary voltage drop will be only several = volts.=20 Thus the secondary will probably produce only several hundred to a = thousand=20 volts or so. However, this is often beyond the voltage rating of the = control=20 wiring and could cause breakdown of insulation, arcing, and other = damage.=20 And much higher spikes will be created when the power is switched on and = off. This could be especially bad if the CT is in a PWM motor control = line=20 where the current may be switched at 20 kHz or so.
> (also: this is why it's important to hook up the load resistor > *before* you put the current transformer on the primary wire)
Most CTs are shipped with a shorting jumper or spring clip on the = secondary.=20 And some are protected by a pair of high current anti-parallel diodes. = But=20 sometimes you may need more than 0.6V peak to drive the load, and since = CTs=20 sometimes operate at 2x to 5x nominal rating, that can be a lot of power = to=20 dissipate in the diodes. Paul=20
On Mon, 23 Jan 2012 21:17:58 -0600, "vkj"
<tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote:

>> >>I appreciate your deep knowledge of the subject, >>however maybe I should not read about it, >>as it makes something that I always perceived as extremely simple, >>sooooo complicated. >> >>Current transformer: >>current_in / turns_ratio = current_out. (for a low load resistor or >short). >>current_out x load_resistor = voltage_out. >> >>Of course there are some details, >>but those can usually be disregarded as the errors caused by those are >very small. >>NORMALLY. >> >>One lesson: Do Not Leave Out The Load Resistor, as then >> Ze Foltage Kan Beacome Ferry High >> > >Why? I dont understand why one cant simply use the open ckt voltage on the >secondary as a measure of the primary current. Heres my argument: > >B = prim. ct. * I(geometry-dependent stuff) (Biot Law); (I() => integral) >flux = B * A; A => area >Vsec = d(coupled-flux)/dt > >For sinusoidals therefore, the open circuit is proportional to the prim. >current. If you consider the ct. transformer as a coupled inductor, then >the actual calculations are quite simple. > >Right? >
Yup. http://en.wikipedia.org/wiki/Rogowski_coil John
On a sunny day (Mon, 23 Jan 2012 22:29:51 -0500) it happened DJ Delorie
<dj@delorie.com> wrote in <xnd3a9aic0.fsf@delorie.com>:

> >"vkj" <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> writes: >> Why? I dont understand why one cant simply use the open ckt voltage on the >> secondary as a measure of the primary current. Heres my argument: > >Because a current transformer is normally in the 1000:1 range. Do you >really want to try to measure 120,000 volts? > >(also: this is why it's important to hook up the load resistor *before* > you put the current transformer on the primary wire)
And even connect the sense leads below the connection of the transformer, just in case of a bad contact: ----- | |---------- CT [ ] electronics |---------- | ----
On a sunny day (Mon, 23 Jan 2012 21:17:58 -0600) it happened "vkj"
<tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote in
<NNOdnTDSQ7B7v4PSnZ2dnUVZ_jCdnZ2d@giganews.com>:

>> >>I appreciate your deep knowledge of the subject, >>however maybe I should not read about it, >>as it makes something that I always perceived as extremely simple, >>sooooo complicated. >> >>Current transformer: >>current_in / turns_ratio = current_out. (for a low load resistor or >short). >>current_out x load_resistor = voltage_out. >> >>Of course there are some details, >>but those can usually be disregarded as the errors caused by those are >very small. >>NORMALLY. >> >>One lesson: Do Not Leave Out The Load Resistor, as then >> Ze Foltage Kan Beacome Ferry High >> > >Why? I dont understand why one cant simply use the open ckt voltage on the >secondary as a measure of the primary current. Heres my argument: > >B = prim. ct. * I(geometry-dependent stuff) (Biot Law); (I() => integral) >flux = B * A; A => area >Vsec = d(coupled-flux)/dtnergy, > >For sinusoidals therefore, the open circuit is proportional to the prim. >current. If you consider the ct. transformer as a coupled inductor, then >the actual calculations are quite simple. > >Right?
Yes, simple! What impedance will you see primary? The secondary divided by n^2 (turns ratio square). But the secundary is now infinite. So you would get a large primary impedance only limited by saturation effects of the core material. There are no infinities in nature (my opinion), so that would be the limiting factor in the voltage you would see secundary.
On 2/12/2011 11:57 AM, Tim Williams wrote:
> Actually Jim, that's exactly the problem. > > A high impedance load will work perfectly.
Only as long as the primary current doesn't drive the core into saturation. Sylvia.
>>Why? I dont understand why one cant simply use the open ckt voltage on
the
>>secondary as a measure of the primary current. Heres my argument: >> >>B = prim. ct. * I(geometry-dependent stuff) (Biot Law); (I() =>
integral)
>>flux = B * A; A => area >>Vsec = d(coupled-flux)/dtnergy, >> >>For sinusoidals therefore, the open circuit is proportional to the prim. >>current. If you consider the ct. transformer as a coupled inductor,
then
>>the actual calculations are quite simple. >> >>Right? > > >Yes, simple! >What impedance will you see primary? The secondary divided by n^2 (turns
ratio square).
>But the secundary is now infinite. >So you would get a large primary impedance only limited by saturation
effects of the core material.
>There are no infinities in nature (my opinion), so that would be the
limiting factor in the voltage
>you would see secundary. >
What?? Heres a little "thought experiment": SUppose you had a small solenoid (air core) of say 10 turns. Apply a 6V 50Hz voltage across it. Im sure you will agree that it would hardly present any impedance, and the current would be limited almost entirely by the resistance. So are you saying that if I now wind a secondary winding over the coil, I would magically get "infinite" impedence on the primary? Remember there is no iron here, and it would take a lot to saturate vacuum. Answer: Think inductance. Transformer voltage equations dont directly apply in this case. The voltage across the coil is very small because of the small inducatance (ignroing resistance). If you have a secondary, the voltage across it would be gverned by the mutual inductance and the primary current. Vsec = M d(prim.ct.)/dt; M = mutual inductance. vkj --------------------------------------- Posted through http://www.Electronics-Related.com
>On 2/12/2011 11:57 AM, Tim Williams wrote: >> Actually Jim, that's exactly the problem. >> >> A high impedance load will work perfectly. > >Only as long as the primary current doesn't drive the core into
saturation.
> >Sylvia. >
Lets not confuse the focus of the issue and consider air cores only, thus eliminating saturation. Lets say your current transformer presents an impedance of 1 milli-Ohm. Lets say your current is 20A. The prim. voltage is then 20mV. As someone has pointed out the turns ratio of a typical CT could be 1000:1, so the sec. open circuit volt is still just 20V. Hardly disastrous. vkj --------------------------------------- Posted through http://www.Electronics-Related.com
> >"vkj" > >> Why? I dont understand why one cant simply use the open ckt voltage on
>> the >> secondary as a measure of the primary current. > >** For the simple reason that a "transformer" always converts voltages >according to the turns ratio on the windings. When you feed an iron cored
>transformer prmary with a current, the resulting voltage drop is undefined
>and very non-linear. >
Sorry, I dont think thats true at all. As I pointed out the fundamantal mechanisms are: Current causes flux. Flux is coupled to secondary. change of flux causes secondary voltage, Vsec = M d(iprim)/dt. In the same way, the change of flux in the primary itself causes a "back emf" and results in self inductance, V = L di/dt. The notion of a transformer follows from this. Assuming the primary inductance is "large", then you can equivalently talk of voltages. It is well known that ferro-magnetics have extremely non-linear permeabilities, and hence current )or voltage) transformers can be very non-linear. vkj
>> If you consider the ct. transformer as a coupled inductor, then >> the actual calculations are quite simple. > > >** An inductor could be used to measure AC current (at a single frequency)
>as there is a simple ratio between current and voltage drop ( long as the
>inductive reactance is much bigger than the resistance). A secondary
winding
>would enlarge the voltage to be measured - but that is no longer a
*current
>transformer* at all and loses most of the advantages. > >The original idea of a current transformer was to allow safe and simple >measurement of high AC supply currents with a standard moving iron
ammeter.
>No rectifiers are needed, no high voltages involved and there is virtually
>no effect on the circuit being measured. > > >... Phil > > >
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