PM2002DP

4/10

http://www.sames.co.za

CTTERMINATIONRESISTOR

CURRENTSENSORINPUTRESISTORSFIGURE5(USING

ACURRENTTRANSFORMER)

Provisionismadefortheuseofacurrenttransformeronthe

module. The existing shunt needs to be removed and the

current sense resistors will have to be recalculated for the

specificcurrenttransformer.ThevoltagedropacrosstheCT

terminationresistoratratedcurrentshouldbeatleast20mV.

TheCT'susedshouldhavealowphaseshiftandaratioof

1:2500.The CT is terminated witha 2.7

resistor givinga

voltagedropof864mVacrosstheterminationresistoratrated

conditions

TheresistorsR6,R7definethecurrentlevelintothecurrent

senseinputsthedevice.Theresistorvaluesare selectedforan

inputcurrentof16µAonthecurrentinputsatratedconditions.

For an 80A-ratedmetertheresistorvaluesarecalculatedas

follows:

R6=R7=(I/16µA)x R /2

=80A/16µAx2.7

/2

=2.7k

I Linecurrent

RSH=CTTerminationresistor

2500=CTratio

Ω

L

SH

L=

Ω

Ω

Figure5:Currentinputconfiguration

LIVEIN

LIVEOUT

R16

IIN

IIP

J5

L

CT1

R6

R7

PhaseCompensation(Whenusingacurrenttransformer)

ReferenceVoltageVREF

GroundGND

Phase shift caused by the current transformer may be

correctedbyinsertingacapacitorinthevoltagedividercircuit.

SeeFigure6,CapacitorC7.Tocompensateforaphaseshiftof

0.18degreesthecapacitorvalueiscalculatedasfollows:

C=1/(2 x

xMainsfrequencyxR5xtan(Phaseshiftangle))

C=1/(2 x

x50x1Mxtan(0.18degrees))

C=1.013µF

TheVREFpinoftheSA2002Disconnectedtoaresistorthat

determinestheonchipbiascurrent.

TheGNDpinoftheSA2002Distothelivephase,whichis

halfway between VDD and VSS. Note that supply bypass

capacitorsC1andC2arepositionedascloseaspossibletothe

supply pinsofthedevice,andconnected toa solid ground

plane.

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