Ideas for inexpensive True-RMS Metering?
in [Design]
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From: Tim Williams on 6 Jul 2010 20:06 "whit3rd" <whit3rd(a)gmail.com> wrote in message news:c6d94b54-aa3a-4eb1-8810-4417e6f366cd(a)32g2000vbi.googlegroups.com... > The saturation of the core makes the coupling of primary and > secondary go away; only the primary side gets the fault current, > the secondary current is no longer proportional (unless you choose > a really odd winding scheme that stays highly flux-coupled when > the core is removed). Yes, which is why the energy delivered is proportional to flux as well. It only delivers power until the transformer stops transformering. Say you get 1A fault current from a CT (i.e., secondary referred), which throws your circuit into overload, so the voltage on the windings jumps to 5V (clamped by a perfect 5V supply, assuming ideal clamp diodes). The delivered power is evidently 1A * 5V = 5W, going into your supply. If the winding has a saturation flux of 1mWb, this fault current will flow for 1mWb / 5V = 0.2ms. The energy is 5W * 0.2ms = 1mJ, or 1A * 1mWb. Increase the fault to 10A. The winding is clamped at 5V, so 50W is delivered, and the fault again lasts for 0.2ms, because the flux is 1mWb. The energy is 10mJ. Increase the fault to 1kA. Now you get 5kW and 1J, beyond the capacity of a 1.5KE6. Strike it with a bolt of lightning. 100kA gives 500kW peak and 100J, assuming the transformer doesn't fail first; if it has an internal resistance of 0.01 ohm, it will drop 1kV, probably breaking down the meager insulation in a CT. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: John Larkin on 6 Jul 2010 20:39 On Tue, 6 Jul 2010 19:06:37 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"whit3rd" <whit3rd(a)gmail.com> wrote in message news:c6d94b54-aa3a-4eb1-8810-4417e6f366cd(a)32g2000vbi.googlegroups.com... >> The saturation of the core makes the coupling of primary and >> secondary go away; only the primary side gets the fault current, >> the secondary current is no longer proportional (unless you choose >> a really odd winding scheme that stays highly flux-coupled when >> the core is removed). > >Yes, which is why the energy delivered is proportional to flux as well. It only delivers power until the transformer stops transformering. > >Say you get 1A fault current from a CT (i.e., secondary referred), which throws your circuit into overload, so the voltage on the windings jumps to 5V (clamped by a perfect 5V supply, assuming ideal clamp diodes). The delivered power is evidently 1A * 5V = 5W, going into your supply. If the winding has a saturation flux of 1mWb, this fault current will flow for 1mWb / 5V = 0.2ms. The energy is 5W * 0.2ms = 1mJ, or 1A * 1mWb. > >Increase the fault to 10A. The winding is clamped at 5V, so 50W is delivered, and the fault again lasts for 0.2ms, because the flux is 1mWb. The energy is 10mJ. > >Increase the fault to 1kA. Now you get 5kW and 1J, beyond the capacity of a 1.5KE6. > >Strike it with a bolt of lightning. 100kA gives 500kW peak and 100J, assuming the transformer doesn't fail first; if it has an internal resistance of 0.01 ohm, it will drop 1kV, probably breaking down the meager insulation in a CT. > >Tim The usual practise in electronic metering is to have the CT secondary drive a low-resistance wirewound or manganin strip shunt. The shunt resistance is considerably less than the winding resistance. Outrageous CT overloads don't damage the shunt... most of the power dissipation is in the winding. The signal conditioning opamps or whatever are protected by high value resistances between them and the shunt. It's also common to buy "current sensors" which are just CTs that come with internal burden resistors. 0.25 and 0.333 volts RMS out at rated current are the two common standards. John
From: JosephKK on 7 Jul 2010 00:25 On Tue, 6 Jul 2010 12:29:12 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> wrote: >On Jul 5, 3:53 pm, "Tim Williams" <tmoran...(a)charter.net> wrote: >> "John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in messagenews:5bg436lfrq2a05j86fdn3s3t0opi82fr3p(a)4ax.com... >> > A CT driving a shunt isn't bad. You can use series resistance after >> > the shunt to protect an opamp. At extreme currents, the CT will >> > saturate, limiting the power dumped into the shunt. >> >> No, limited flux, unlimited power/energy. The amps delivered during the pulse are proportional to the fault current. Only the voltseconds are limited. > >The saturation of the core makes the coupling of primary and >secondary go away; only the primary side gets the fault current, >the secondary current is no longer proportional (unless you choose >a really odd winding scheme that stays highly flux-coupled when >the core is removed). Basically it becomes a Rogowski coil, and the transformation ratio changes drastically.
From: Tim Williams on 7 Jul 2010 00:49
"JosephKK" <quiettechblue(a)yahoo.com> wrote in message news:1a083654uhmf3cuhh6j5njgv5a9t0lisan(a)4ax.com... > Basically it becomes a Rogowski coil, and the transformation ratio > changes drastically. Or even less, if you wind over only a segment. This makes leakage worse under operating conditions, but could be convienient under such fault conditions. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms |