AC and meter movements
in [Basics]
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From: Phil Allison on 12 Apr 2010 04:11 "Alfred E. Schoen" >>> AC current is best measured with a CT, and there are some small >>> inexpensive ones that will handle currents of 5 to 30 amps with a ratio >>> of about 500:1 such as this: >>> http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=237-1103-ND >>> >>> The output can be rectified and provides a current of 10 mA for 5 A with >>> enough voltage to produce a linear output through a silicon diode >>> bridge. >> >> ** Absolute BOLLOCKS!! >> >> That CT is speced to use a 60 ohm load resistor and it outputs 1.65V at >> 15 amps - however a silicon diode bridge will not even *begin >> conducting* until the AC voltage reaches 0.7V rms. > > The resistor needs to be placed after the rectifier. ** Shame you did not say that before. The behaviour of the CT with such a load is not defined. Big shame that any DC component is not read as it would be with a simple MI meter. The CT would be screwed completely by a significant DC component. > Germanium or Schottkey diodes would perform much better, of course. ** Of course. Beam me up - Schottky !!!!!!!!!!! > And there is also the problem of variations due to temperature. ** Blah, blah, blah ... ..... Phil
From: Paul E. Schoen on 12 Apr 2010 13:47 "Mike Cook" <mcham(a)NOTyahoo.com> wrote in message news:0001HW.C7E813D6005CAE3CB01AD9AF(a)news.eternal-september.org... >> For the OP's purposes of voltage measurement, a silicon or germanium >> diode >> bridge will work well, and can be used with most commonly available >> meters. >> AC current is best measured with a CT, and there are some small >> inexpensive >> ones that will handle currents of 5 to 30 amps with a ratio of about >> 500:1 >> such as this: >> http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=237-1103-ND >> >> The output can be rectified and provides a current of 10 mA for 5 A with >> enough voltage to produce a linear output through a silicon diode bridge. >> >> Paul > > Excellent. Thank you Paul for the CT explanation. That looks like my > solution > for ampere measurement. > > Cheers. Please read my reply to Phil for more information. This CT circuit with silicon diodes is not very linear and it may not even show any current below about 1 ampere. It should be much better if you use germanium diodes or Schottkys. You may also consider using something like an old small audio output transformer with a 4 ohm secondary used for current, and the primary may have enough turns to drive a bridge rectifier. Best of luck, Paul
From: Paul E. Schoen on 12 Apr 2010 15:09 "Phil Allison" <phil_a(a)tpg.com.au> wrote in message news:hpukke$4u6$1(a)news-01.bur.connect.com.au... > "Alfred E. Schoen" > >>>> AC current is best measured with a CT, and there are some small >>>> inexpensive ones that will handle currents of 5 to 30 amps with a ratio >>>> of about 500:1 such as this: >>>> http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=237-1103-ND >>>> >>>> The output can be rectified and provides a current of 10 mA for 5 A >>>> with enough voltage to produce a linear output through a silicon diode >>>> bridge. >>> >>> ** Absolute BOLLOCKS!! >>> >>> That CT is speced to use a 60 ohm load resistor and it outputs 1.65V at >>> 15 amps - however a silicon diode bridge will not even *begin >>> conducting* until the AC voltage reaches 0.7V rms. >> >> The resistor needs to be placed after the rectifier. > > > ** Shame you did not say that before. > > The behaviour of the CT with such a load is not defined. That is the standard connection for rectifier type AC current meters, and I did not think it needed to be stated. But this is a basics newsgroup and I should have been more specific. When I worked as a meter technician I assembled and calibrated many meters which contained a small CT and copper oxide rectifier directly to the meter movement, or with a small resistor to provide adjustment and damping. A CT with a bridge rectifier on its output very quickly generates enough voltage to overcome the forward drop of two diodes. For a 500:1 ratio CT, the primary voltage only needs to be about 3 mV. It also depends on the VA rating of the CT (in this case about 2 VA), and the coupling of the primary winding to the core and the core to the secondary, and the magnetic characteristics of the core. These are complex and not well defined, as you stated. > Big shame that any DC component is not read as it would be with a simple > MI meter. > > The CT would be screwed completely by a significant DC component. Yes, but the OP wanted to read AC current from a variable transformer and most likely without any DC component. A search for an iron vane AC amps panel meter turned up this one, from Simpson, distributed by Grainger, for about $90: http://www.grainger.com/Grainger/items/1T865?Pid=search Iron vane meters tend to be fairly rare and expensive, as well as being bulky and possibly not as rugged as most DC meters with a D'Arsonval movement. Here are some cheap AC panel meters. They might be iron vane or rectifier type: http://www.mpja.com/prodinfo.asp?number=17683+ME http://www.mpja.com/prodinfo.asp?number=8735+ME http://www.mpja.com/prodinfo.asp?number=17255+ME http://www.mpja.com/prodinfo.asp?number=16558+ME The following information about Yogogawa meters states that iron-vane *voltmeters* may exhibit inaccuracy of as much as 75% at the extremes of rated frequency from 20 to 400 Hz, so the frequency effect is quite significant. But for iron vane *ammeters*, the frequency effect is very small: http://www.yokogawa.com/us/mi/pdf/Panel_Meters_250260PM_01E_A_Apr2007.pdf The specs for Triplett iron vane meters indicate an effect of 1% over the range of 25-133 Hz. http://www.electro-meters.com/Assets/PDF_files/Triplett/APM/Characteristicsr.pdf >> Germanium or Schottkey diodes would perform much better, of course. > > ** Of course. > > Beam me up - Schottky !!!!!!!!!!! Spelling correction acknowledged. Proceed... Paul
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