"vkjerkoff">> >>> 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 primary with a current, the resulting voltage drop is >>>undefined >>and very non-linear. >> > > Sorry, I dont think thats true at all.** Then you are know nothing, fucking idiot. FFS go test one !!!!!!!!!!!!! Then PISS off to hell.
Current transformer
Started by ●November 30, 2011
Reply by ●January 25, 20122012-01-25
Reply by ●January 25, 20122012-01-25
On a sunny day (Tue, 24 Jan 2012 21:38:58 -0600) it happened "vkj" <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote in <MvWdnSgCifzP5ILSnZ2dnUVZ_jKdnZ2d@giganews.com>:>What??Current transformers HAVE a core. :-)
Reply by ●January 25, 20122012-01-25
vkj wrote:>> >>"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. > > vkjYes, but transformers operate at flux levels where the flux generated by the primary current is very nearly cancelled out by the secondary current. That's what they mean when they say ampere-turns cancel out. The back emf you mention does work to keep a potential transformer's open circuit magnetizing current low. But that is because potential transformers have many turns on their primary, each providing a few volts of back emf. A current transformer has one primary turn. Its primary current is mostly dependant on the downstream load and as a result, the core flux can easily reach saturation because the back emf generated by d(flux)/dt through one turn is too small to buck the source voltage. Its self inductance is very low. On the secondary, with many turns, the product of d(flux)/dt and N turns can be very high. In some cases, higher than the insulation's rated max voltage. -- Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------------ "Yee-Ha!" is not an adequate foreign policy.
Reply by ●January 28, 20122012-01-28
On 2012-01-24, vkj <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote:> 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:> 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?works fine with linear loads which draw sinusoidal current, doesn't work for phase control, bridge rectifiers feeding capacitors, or other active loads. -- ⚂⚃ 100% natural
Reply by ●January 28, 20122012-01-28
On Mon, 23 Jan 2012 22:29:51 -0500, DJ Delorie <dj@delorie.com> wrote:> >"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)That is not all, maybe not even the most important. The flux in the core when operated properly in the CT mode is nearly zero, and if the = secondary is open that is NOT true, and the core could easily saturate. BTW the relevant IEEE standards make a distinction between metering and protection/relaying CT and their performance properties in overload. ?-)
Reply by ●January 28, 20122012-01-28
On Tue, 24 Jan 2012 21:47:41 -0600, "vkj" <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote:>>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.=20 >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 =20Wrong and wrong. The primary impedance applies _ONLY_ when the secondary is terminated with a proper load (resistor). Otherwise it becomes an ordinary voltage transformer, and the primary voltage is no longer in the mV range, but in the volts range, with corresponding kV across the secondary. Clear enough yet? ?-)
Reply by ●January 28, 20122012-01-28
On Tue, 24 Jan 2012 21:47:41 -0600, "vkj" <tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote:>>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.comAn unloaded air-core CT is called a Rogowski coil. The output voltage is proportional to the derivative of the current. If all you do is look at the terminal voltage, there is no effect on the wire at all, nothing equivalent to a primary impedance. http://en.wikipedia.org/wiki/Rogowski_coil -- John Larkin, President Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
Reply by ●January 28, 20122012-01-28
On Sat, 28 Jan 2012 18:01:11 -0800, josephkk <joseph_barrett@sbcglobal.net> wrote:>On Tue, 24 Jan 2012 21:47:41 -0600, "vkj" ><tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote: > >>>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 > >Wrong and wrong. The primary impedance applies _ONLY_ when the secondary >is terminated with a proper load (resistor). Otherwise it becomes an >ordinary voltage transformer, and the primary voltage is no longer in the >mV range, but in the volts range, with corresponding kV across the >secondary. >If I may borrow a quotation, "Wrong and wrong." http://en.wikipedia.org/wiki/Rogowski_coil -- John Larkin, President Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
Reply by ●January 28, 20122012-01-28
On 1/28/2012 7:52 PM, josephkk wrote:> On Mon, 23 Jan 2012 22:29:51 -0500, DJ Delorie<dj@delorie.com> wrote:> That is not all, maybe not even the most important. The flux in the core > when operated properly in the CT mode is nearly zero,Not true. There is flux in the core no matter what. It is what gives rise to a secondary voltage across a resistor.
Reply by ●January 28, 20122012-01-28
On Sat, 28 Jan 2012 18:12:46 -0800, John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:>On Sat, 28 Jan 2012 18:01:11 -0800, josephkk ><joseph_barrett@sbcglobal.net> wrote: > >>On Tue, 24 Jan 2012 21:47:41 -0600, "vkj" >><tranquine@n_o_s_p_a_m.n_o_s_p_a_m.gmail.com> wrote: >> >>>>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.=20>>>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 =20 >> >>Wrong and wrong. The primary impedance applies _ONLY_ when the =secondary>>is terminated with a proper load (resistor). Otherwise it becomes an >>ordinary voltage transformer, and the primary voltage is no longer in =the>>mV range, but in the volts range, with corresponding kV across the >>secondary. >> > >If I may borrow a quotation, "Wrong and wrong." > >http://en.wikipedia.org/wiki/Rogowski_coilNot a "normal" current transformer in case you hadn't noticed. ?-)
