Practical Transformer Model
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ECET 3500 Electric Machines Practical Transformer Model 1 Ideal Transformer Model Although the following model can be used to define the operation of an “ideal” transformer, the operational characteristics of a “real world” or “practical” transformer has it cannot be predicted using this model because it does not account for the internal losses or other non-ideal qualities of a practical transformer. For this reason, we will now consider how to model a practical transformer. ~ Ip a Np Ns ~ Ep ~ a Es Np Ns ~ Is turns ratio ~ Ep ~ Ip 1 ~ Is a ~ Es a Ideal Transformer Practical Transformer Model The circuit model for a “practical” transformer can be developed by expanding the ideal transformer model to include various additional circuit elements that account for the “losses” that typically occur when operating an actual transformer. ~ I1 ~ V1 Rp ~ Ip jX p R fe jX m Np Ns ~ Ep ~ Is ~ Es Rs jX s ~ I2 ~ V2 a Ideal Transformer 2 Practical Transformer Model Within the model: ◦ Rp and Rs account for the primary and secondary winding resistances, ◦ Xp and Xs account for the primary and secondary winding leakage reactances, ◦ Rfe accounts for the Hysteresis/Eddy-Current losses within the core, ◦ Xm accounts for the magnetization current that is drawn into the primary winding. ~ I1 Rp ~ V1 ~ Ip jX p R fe jX m Np Ns ~ Ep ~ Is Rs jX s ~ I2 ~ V2 ~ Es a Ideal Transformer Practical Transformer Model Note – The circuit shown below is used to model the steady-state operation of transformers that have relatively low operating frequencies. (I.e. – 60Hz or “power” transformers) A different model may be required for transformers that operate at higher frequencies in order to properly account for other “losses” that are negligible at 60Hz. ~ I1 ~ V1 Rp ~ Ip jX p R fe jX m Np Ns ~ Ep ~ Is ~ Es Rs jX s ~ I2 ~ V2 a Ideal Transformer 3 Practical Transformer Model In order to reduce the model’s complexity, the parallel elements Rfe and Xm can be moved to the “left-side” of the series elements Rp and Xp. Note – This change should only introduce a small decrease in the overall accuracy of the model because Rfe and Xm are typically much larger in magnitude than Rp and Xp. ~ I1 ~ V1 R fe Rp jX p ~ Ip Np ~ Ep jX m ~ Is Ns Rs jX s ~ I2 ~ V2 ~ Es a Ideal Transformer Practical Transformer Model Additionally, the series elements Rs and Xs can be referred (moved) to the primary-side of the ideal transformer model provided that their values are each multiplied by a2. Note – as defined in the figure: R’s = a2·Rs X’s = a2·Xs ~ I1 Rp jX p Rs jX s a 2 Rs a2 X s ~ Ip Np Ns ~ Is ~ I2 2 ~ V1 R fe jX m ~ Ep ~ Es xa ~ V2 a Ideal Transformer 4 Practical Transformer Model The series resistive elements Rp and R’s can then be combined into a single resistive element Req, and the series reactive elements Xp and X’s can be combined into a single reactive element Xeq, such that: Req = Rp + R’s Xeq = Xp + X’s where: Req and Xeq account for the overall resistance and leakage reactance of the transformer’s windings. ~ I1 Req R p Rs ~ V1 R fe jX eq ~ Ip Np j ( X p X s ) Ns ~ Ep jX m ~ Is ~ I2 ~ V2 ~ Es a Ideal Transformer Practical Transformer Model This circuit model, which is sometimes referred to as the “Cantilever Equivalent Circuit” or the “Steinmetz Equivalent Circuit”, is often used to predict the operation of a practical transformer. ~ I1 ~ V1 R fe jX m Req jX eq R p Rs j ( X p X s ) ~ Ip Np Ns ~ Ep ~ Is ~ Es ~ I2 ~ V2 a Ideal Transformer 5 Example Problem Given a 500VA, 60 Hz, 240V–120V transformer with the following circuit-model parameters: RLV = 0.25, XLV = 0.40, RHV = 0.80, XHV = 1.20 Rfe(LV) = 480, XM(LV) = 160 (both referred to the LV-winding) If the transformer is used to step-up the voltage from a 120 volt source while supplying a load, ZLoad =(120+j50) … Determine the actual load voltage and the overall source current. ~ I1 ~ Vsource ~ V1 R fe Req jX eq R p Rs j ( X p X s ) ~ Ip Np Ns ~ Ep jX m ~ Is ~ I2 ~ Es ~ V2 Z Load a Ideal Transformer Determining the Model Parameters To begin the problem, the transformer’s operational turns-ratio must be determined… Since the transformer is being used to step-up the voltage from a 120V source, the 120V winding is the primary winding and the 240V winding is the secondary winding. Therefore, the operational turns-ratio is: a ~ I1 ~ Vsource ~ V1 R fe jX m Req jX eq R p Rs j ( X p X s ) ~ Ip 120 V 1 240 V 2 Np Ns ~ Ep ~ Is ~ I2 ~ Es ~ V2 Z Load a Ideal Transformer 6 Determining the Model Parameters Additionally, since the 120V winding is the primary winding, the circuit parameters relating to the HV and LV windings may be redefined as: Rp = RLV = 0.25, Rs = RHV = 0.80, Xp = XLV = 0.40, Xs = XHV = 1.20, from which Req and Xeq can be determined: Req R p a 2 Rs 0.25 12 0.80 0.45 2 X eq X p a 2 X s 0.40 12 1.20 0.70 2 ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m jX eq j 0.7 ~ Ip ~ Ep j160 ~ Is ~ I2 ~ Es ~ V2 Z Load 120 j 50 a Ideal Transformer Determining the Model Parameters Note that, as defined in the problem statement, Rfe and Xm were both referred to the LV winding: Rfe(LV) = 480, XM(LV) = 160 Since the LV winding is the primary winding, both of those impedances can be directly imported into the model. Thus, the following figure shows the circuit model for the practical transformer with all of the parameters included: ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m j160 jX eq j 0.7 ~ Ip ~ Ep ~ Is ~ I2 ~ Es ~ V2 Z Load 120 j 50 a Ideal Transformer 7 Referring the Load to the Primary-Side In order to simplify the overall circuit, the “ideal transformer” portion of the model can be removed by referring the load to the primary-side of the ideal transformer, such that: ' Z Load a 2 Z Load 12 120 j 50 30 j12.5 2 ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m ~ Ip jX eq j 0.7 ~ Ep j160 ~ Is ~ I2 ~ Es ~ V2 Z Load 120 j 50 a Ideal Transformer ~ I1 ~ Vsource Req ~ V1 R fe ~ Ip jX eq ~ Ep jX m (30 j12.5) Z Load Solving for the Load Voltage Note that, although Es is the voltage across the actual load ZLoad in the original circuit, Ep is the voltage across Z’Load in the simplified circuit. Thus, the actual load voltage can be determined by solving for Ep in the simplified circuit and then applying the turns-ratio equation in order to solve for Es . ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m ~ Ip jX eq j 0.7 ~ Ep j160 ~ Is ~ I2 ~ Es ~ V2 Z Load 120 j 50 a Ideal Transformer ~ I1 ~ Vsource ~ V1 R fe Req jX m jX eq ~ Ip ~ Ep (30 j12.5) Z Load 8 Solving for the Load Voltage Since Z’Load . is in series with both Req and jXeq in the simplified circuit, the voltage Ep can be determined using a voltage-divider equation: ' Z Load ~ ~ E p V1 R jX Z ' eq Load eq 30 j12.5 1200 0.4 j 0.7 30 j12.5 117.7 0.85 volts ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m ~ Ip jX eq j 0.7 Z Load ~ Ep j160 30 j 12.5 Solving for the Load Voltage Since Z’Load . is in series with both Req and jXeq in the simplified circuit, the voltage Ep can be determined using a voltage-divider equation: ' Z Load ~ ~ E p V1 R jX Z ' eq Load eq 30 j12.5 1200 117.7 0.85 volts 0.4 j 0.7 30 j12.5 The load voltage can then be solved by utilizing the turns-ratio equation: ~ ~ ~ ~ E p 117.7 0.85 Vload V2 Es 235.4 0.85 volts 1 a 2 ~ I1 ~ Vsource 1200 ~ V1 R fe Req 0.4 480 jX m j160 jX eq j 0.7 ~ Ip ~ Ep ~ Is ~ I2 ~ Es ~ V2 Z Load 120 j 50 a Ideal Transformer 9 Solving for the Source Current In order to solve for the source current I1 , the input impedance of the loaded transformer can first be determined: Z in R fe X m Req jX eq Z ' load 1 1 1 1 24.4 15.0 480 j160 30.4 j13.2 after which the source current can be determined by: ~ 1200 ~ ~ V 4.19 31.6 amps I source I1 1 Z in 24.4 j15.0 ~ I1 ~ Vsource ~ V1 1200 R fe Req 0.4 480 jX m jX eq j 0.7 ~ Ip Z Load ~ Ep j160 30 j 12.5 Determining the Model Parameters The model parameters for a specific transformer are often unknown. But, the parameters can be determined by performing a simple pair of tests on the transformer: The Open-Circuit Test, and The Short-Circuit Test. ~ I1 ~ V1 R fe jX m Req jX eq R p Rs j ( X p X s ) ~ Ip Np Ns ~ Ep ~ Is ~ Es ~ I2 ~ V2 a Ideal Transformer 10 The Open Circuit Test The model parameters Rfe and Xm can be determined by performing an Open-Circuit Test on the transformer. The Open-Circuit Test is typically performed by applying rated voltage to the primary winding of the transformer while leaving the secondary winding open-circuited. Req rated voltage ~ V1 Rfe jX eq ~ Ip Np Ns ~ Ep jX m ~ Is ~ Es open circuit a Ideal Transformer The Open-Circuit Test If the secondary winding is kept open-circuited, then the secondary ~ current I s must be zero: ~ Is 0 ~ ~ And, if the secondary current I s is zero, the primary current I p flowing into the “ideal” portion of the model must also be zero: ~ 0 ~ I Ip s 0 a a Req rated voltage ~ V1 Rfe jX m jX eq ~ Ip Np Ns ~ Ep ~ Is ~ Es open circuit a Ideal Transformer 11 The Open-Circuit Test ~ If the primary current I p is zero, then no current will be flowing through the series impedances Req and Xeq. Thus, the only active portion of the circuit under Open-Circuit conditions is the parallel combination of branches containing Rfe and Xm. Req rated voltage ~ V1 Rfe ~ Ip jX eq Np Ns ~ Ep jX m ~ Is ~ Es open circuit rated voltage ~ V1 Rfe jX m a Ideal Transformer The Open-Circuit Test Given the parallel combination of a resistive load Rfe and an inductive reactance Xm, If a voltage, Eoc, is applied across the loads while the total current, Ioc, and real power, Poc, being supplied to the loads are measured, then the impedance values of the loads can be determined from the equations: R fe Eoc2 Poc Xm Eoc2 Eoc I oc 2 Poc2 Ioc Poc Eoc ~ V1 Rfe jX m 12 The Short-Circuit Test The model parameters Req and Xeq can be determined by performing an Short-Circuit Test on the transformer. The Short-Circuit Test is typically performed by applying rated current into the primary winding of the transformer while the secondary winding is short-circuited. *** Warning – Only a small primary voltage is required in order to *** ******** reach rated current under short-circuit conditions ******** rated current ~ V1 Req Rfe jX eq ~ Ip Np Ns ~ Ep jX m ~ Is ~ Es short circuit a Ideal Transformer The Short-Circuit Test Under short-circuit conditions, the impedance of the “ideal” transformer contained within the model is zero ohms, therefore the entire ideal transformer portion can be replaced by an ideal wire. rated current ~ V1 R fe Req jX eq jX m 13 The Short-Circuit Test Under short-circuit conditions, the impedance of the “ideal” transformer contained within the model is zero ohms, therefore the entire ideal transformer portion can be replaced by an ideal wire. Additionally, since the impedances Rfe and Xm are typically much larger in magnitude than the impedances Req and Xeq, quite often two to three orders of magnitude (100x-1000x) larger, the impedances Rfe and Xm can be neglected without greatly affecting the operation of the circuit. rated current Req jX eq ~ V1 The Short-Circuit Test Given the series combination of a resistive load Req and an inductive reactance Xeq, If a voltage, Esc, is applied across the loads while the total current, Isc, and real power, Psc, being supplied to the loads are measured, then the impedance values of the loads can be determined from the equations: Req Psc I sc2 I sc X eq Esc I sc 2 Psc2 I sc2 Req jX eq Psc E sc ~ V1 14
