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'AC Current Measurement'
2000\05\19@220830 by R. William Kichman

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face
> Maybe this has been covered before, but please bear with me...
>
> I need to measure AC current at 60 Hertz, 0 to 35 amps to approx. 0.1A or so
> precision.  How might I most easily do such measurement using a
> microcontroller?  I understand that a voltage measurement across a precision
> resistor, or a few wraps of coated wire around a conductor will allow a
> current to be calculated, but with AC, isn't the measurement time varying?
> Won't the A/D read a constantly varying event and get something between 0A
> and peak current?
> What am I missing here?
> Any links to URLs with similar projects would be welcome.
> Thanks for your time.
>
> Bill K.

2000\05\20@103231 by Thomas C. Sefranek

face picon face
"R. William Kichman" wrote:

> > Maybe this has been covered before, but please bear with me...
> >
> > I need to measure AC current at 60 Hertz, 0 to 35 amps to approx. 0.1A or so
> > precision.  How might I most easily do such measurement using a
> > microcontroller?  I understand that a voltage measurement across a precision
> > resistor, or a few wraps of coated wire around a conductor will allow a
> > current to be calculated, but with AC, isn't the measurement time varying?
> > Won't the A/D read a constantly varying event and get something between 0A
> > and peak current?
> > What am I missing here?
> > Any links to URLs with similar projects would be welcome.
> > Thanks for your time.
> >
> > Bill K.

You need a current transformer.  The output should be rectified/filtered into a DC
voltage
which the PIC can read with it's A-D.  A simple 8 bit A-D will get you 255 levels,

so you can expect to get .13725 amp resolution.  (I'd scale it to .2 amp.)
At the low end some non linearity can occur due to the rectifiers.
(But do you REALLY need to know the absolute value of the current  at the low
end?)


--
 *
 |  __O    Thomas C. Sefranek   spam_OUTtcsTakeThisOuTspamcmcorp.com
 |_-\<,_   Amateur Radio Operator: WA1RHP
 (*)/ (*)  Bicycle mobile on 145.41, 448.625 MHz

hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

2000\05\20@120854 by rottosen

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face
MagneTek makes a really cute 60Hz current sense transformer (part number
CSE187-L). It about the size of and looks like a transistor radio output
transformer. It can handle up to 30 amps with an output of 110mv/amp
into a 60 ohm load resistor.
The best part is that it is that it is available from Digi-Key (PN
237-1103) for only $2.89 (US) in single quantity, $1.60 each in
hundreds!

-- Rich



"R. William Kichman" wrote:
{Quote hidden}

2000\05\20@162739 by Harold Hallikainen

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On Sat, 20 May 2000 10:31:49 -0300 "Thomas C. Sefranek" <.....tcsKILLspamspam@spam@CMCORP.COM>
writes:
> "R. William Kichman" wrote:
>
>
> You need a current transformer.  The output should be
> rectified/filtered into a DC
> voltage
> which the PIC can read with it's A-D.  A simple 8 bit A-D will get
> you 255 levels,
>
> so you can expect to get .13725 amp resolution.  (I'd scale it to .2
> amp.)
> At the low end some non linearity can occur due to the rectifiers.
> (But do you REALLY need to know the absolute value of the current
> at the low
> end?)
>
>

       I like the current sense transformers from Toroid Corporation of
Maryland (http://www.toroid.com/currents.htm). Also, I believe the
nonlinearity due to diodes can be minimized by putting the terminating
resistor after the diode bridge. The transformer secondary tends to be a
current source and will "do what it takes" to drive the appropriate
current thru the terminating resistor, getting over the diode drops.
       Another idea is to just have the transformer drive the terminating
resistor directly, then use a current limit resistor from the top of the
terminating resistor directly into a PIC A/D input. The PIC clamp diodes
will clip the negative half of the waveform. The positive half will get
thru undistorted. You'd then double the calculated RMS (assuming the
waveform is symmetrical).

Harold


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