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# Junior Laboratory II ECE 618

UNH

GPA 3.8

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This 8 page Class Notes was uploaded by Miss Felicita Stiedemann on Thursday October 29, 2015. The Class Notes belongs to ECE 618 at University of New Hampshire taught by Daniel Brogan in Fall. Since its upload, it has received 13 views. For similar materials see /class/231686/ece-618-university-of-new-hampshire in Computer Engineering at University of New Hampshire.

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Date Created: 10/29/15

UNHECE 518 Jumm Labmatmy I17 Sprmg 2009 7 Lecture 5 February 17 Background for Lab 5 Transformer Analysis Basic transformermodel Pf39m Secondary wmdmg winding NP mms N5 turns Primary current 1 Secondary I5 current N Primary vaitage V Secondary voltage Figure 1Abasic transfonner tt enwiki ediaor wikiTmnsformer The magnitude of the magnetic eld intensity F that is generated in the core by the primary current 1P is de ned as NPIP Ampstum 2 meter where z is the length ofthe magnetic path around the core wanYEiii 811 Dame 5 Brugan UNH ECE 618 Junior Laboratory H Spring 20097Lecture 5 February 1 7 where the permeability has units of Henriesmeter o The permeability is constant for a particular material The permeability of a vacuum ie free space is 0 4723910397 Hm o The relative permeability of a material ur is defined as u ur un1tless o A material with a high relative permeability carries a higher density of magnetic ux than a material with a low relative permeability does For example steel ur 2000 to 6000 conducts magnetic ux much better than air ur m 1 does Substituting the magnitude of the magnitude of the magnetic ux density is I W b 39339 or Tesla or Gauss m o The total magnetic ux owing in the core Figure 1 can be found by integrating the magnetic ux density over a crosssectional area A The total ux is Z N PIP1 5 Webers o The magnetomotive force mmf is the driving force that generates the magnetic ux It is defined as mmfNI Amptums 0 Using an electrical analogy the mmf is the voltage that drives the current ux around the circuit In this analogy the reluctance is the magnetic equivalent of resistance It is defined as mZMZLZL LLNPIPAZ uA Henries In other words it is easier for ux to ow around the core if it is relatively thick compared to it s path length and has a large ur Daniel S Brogan 2 UNH ECE 618 Junior Laboratory H Spring 20097Lecture 5 February 1 7 Ideal transformer equations The primary voltage can be directly related to the magnetic ux using Faraday s Law of Induction IVPNP dt Assuming that the same amount of ux ows through the primary and secondary windings the voltage induced in the secondary windings from the ux that was induced by the primary windings is IVSI NS alt The turns ratio a is de ned from the above equations as aamp NS s ls From the ux equation yNPIPA LLNSISA Z Z a1s VS 11 VPIPVSIS Finally the impedance is de ned as JPIP VPlsa2N ZS VS IP Daniel S Brogan 3 UNH EOE 51 8 Jim or Laboratory II Sprmg 2009 7 Lecth 5 szmary 1 7 Ideal transformer summary NS There is a 39 39 I 39 current ows into This can be helpful but be aware that there are variations in the way different people deline the dot notation Nonideal transformer model and equations Figure 2 I 39N S 5 4 E E V s s P Figure 2 V RP and R are the primary and secondary winding resistances respectively These are from the resistance ofthe wire Typical values 19 to 29 XP and X are the primary and secondary winding reactances respectively These 39 39 39 r pi a semblooQ RC represents the core losses from eddy cunents and hysteresis Typical values RP or R XM is the reactance due to ux ow in the core Typical values a few kHz 25169 to 50169 Note that the secondary impedances have been moved to the primary side of the model and d by 02 to show the effect on the circuit that they would have in their true position led in scale Note also that the load on the secondary side will appear to the source to be sca magnitude by a2 as well Danle s Brogan UNH ECE 618 Junior Laboratory H Spring 20097Lecture 5 February 1 7 Open Circuit Test The open circuit test requires that the secondary terminals of the transformer be disconnected from any load Recalling that RS and X S are actually on the other side of the ideal transformer from their position shown in Figure 2 no current ows through these components Thus the total impedance is N i 1N1 1 jXM R Z0C RPjXP Using a voltage meter a current meter and a power meter P V cos 9 where 9 is the angle between the voltage and the current Thus the power has the form of a dotproduct Having measured the above variables I l l Yoc m J V46 RC XM Short Circuit Test Now the secondary terminals of the inductor are shorted together The short circuit impedance is 1 ZSCRP1XP i 1 R jXM a2RS ja2XS C Ifa is small ZSC m RP 1X1 a2RS jaZXS RP aZRS jXP aZXS Z Vzt9 SC I RPa2RSjXPa2XS Daniel S Brogan 5 UNH ECE 618 Junior Laboratory H Spring 20097Lecture 5 February 1 7 Note if a is large the result is the same as for the open circuit test In this case the primary side can be shorted and the measurements made on the secondary side In this case the reversed turns ratio of la will be small An impedance meter that outputs Z 46 vs f Hz can be used to accomplish the above results directly The three voltmeter method can also be used to determine the impedance of a twoport network This technique is described at the end of this document The total impedance including the load as seen from the source is 2 ZLOADED Zsc a ZLOAD Example A highvoltage source is used to drive a load as shown in Figure 3a ZLINE 18 124 480V Z LOAD 400 j300 Figure 3a Transformer Example Original Circuit 60 Hz The loop current is 1 m m 9084 318quot mA 418j324 52943710 The load voltage is VLOAD IZLOAD m 09084 378 4001300 m 09084 378 5004 369 V LOAD m 4544 09 V The power lost in the line is PLINEiLOSS 391392 ReZ m 09082 18 148 w LINE Figure 3b shows the same circuit with ideal stepup and stepdown transformers on either end of the transmission line Daniel S Brogan 6 UNH ECE 618 Junior Laboratory H Spring 2009 iLecture 5 February 1 7 IP Is ZLINE 18 124 480V ZLOAD VL 60 Hz 400 1300 a1 110 a2 10 Step Up Step Down Figure 3b Transformer Example With Ideal StepUp And StepDown Transformers Step Up and Step Down refer to the voltage magnitudes VS VP a o The currents are I 2 480400 2 480400 P all 18j24a 400j300 00118j244001300 48040quot 48040quot P m 95944 3688quot mA 4001830024 5002943688quot 1 allp m 95944 3688o mA LINE IS azlm alaZIP 1P m 95944 3688quot mA The load voltage is I Z 095944 3688 5004 369 m 47974 001 V VLOAD s LOAD N The power lost in the line is P LINEiLOSS 391 LINE 2 Re2m m 0095942 18 m 0166 w The power lost in the transmission line is now much less than it was without the transformers Likewise the voltage delivered to the load is now much closer to the source voltage than it was without the transformers Daniel S Brogan UNHECE 618 JuniorLaboratory 11 Spring 20097Lecture 5 February 17 Three Voltmeter Method of Impedance Calculation EE 618 Measurement of the input impedance of a twopart The input impedance of the two port BC D E can be measured using the circuit shown below 1 Measure Vab the voltage across a known resistor R 2 Measure Vbc the voltage across the input of the twoport 3 Measure Vac the applied voltage from the AC source Since the branchesrabbc and ca form a closed loop the vector values of voltage Vab Vbc an V must form a closed triangle The voltage across the resistor is in phase with the input current to the two port The angle ID is the phase angle between this current and the input voltage 0 the twoport In the case of your transformer measurements t this angle is positive v a be Th equot Zin Rin 139 J Xin IR where IR VabR 2 2 2 V 7V 818007005quot I 2V1 Vb Daniel S Brogan

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