State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Lab 6: Transformers 1. Objectives 1) Measure the frequency response of the transformer. 2) Determine the input impedance of the transformer. 3) Use transformer to improve the source-to-load power transfer (impedance matching). 2. Introduction AC signals can be transferred from one circuit to another without actual wire connection between them. One way of doing it is to use two (or more) inductors located in proximity of each other. AC signal in one of the inductors would generate time variable magnetic field that can be used to generate AC signal in another inductor. This type of coupling between circuits is called magnetic coupling and it is realized in device called transformer. Transformers can be used to perform many useful functions, for example: 1. DC isolation between two circuits. The magnetic coupling by a transformer isolates circuit grounds, i.e. grounded signal generator can be transformed into the floating one. 2. Change the magnitude of the voltage and current. High-voltage/low-current signals can be transformed into low-voltage/high-current signals. 3. Impedance matching for the maximum power transfer from generator to the load, etc. Conversion of the input AC signal into magnetic field occurs in the input inductor of the transformer that is called primary coil. Magnetic field (sometimes it is called magnetic flux density) is generated by electrical current and, according to Amper’s law, the line integral of magnetic field around a closed loop is directly proportional to the electrical current passing through the loop, i.e. B dl I (1) The parameter µ is called magnetic permeability and characterizes the type and strength of the response of the material to external magnetic field. In vacuum the parameter µ is equal to µ0 = 4π*10-7 H/m (in SI units). Iron, nickel, cobalt and some of the rare earths (neodymium, gadolinium, etc.) metals exhibit a unique magnetic behavior which is called ferromagnetism. A small externally imposed magnetic field can cause the magnetic domains inside the ferromagnetic material to line up with each other. This results into an increase of the magnetic field inside material. The amount of increase is characterized by so called relative magnetic permeability that is defined as a ratio of the absolute magnetic permeability of the material µ and that of the vacuum µ0. In some ferromagnetics the relative magnetic permeability can be very large and, hence, can cause 1 State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ giant increase of the magnetic field inside the material. For instance, in iron µ ~ 200*µ0, in permalloy (88.5 % nickel, 21.5 % iron) ~ 8000*µ0. Ferrites (special iron oxide based ceramics materials) can be fabricated with µ in the wide range from 50 to 3000. The ferromagnetic material shaped to form a closed ring (core) with two coils of wire wound around it is a typical design of the transformer. Due to high value of the core magnetic permeability almost all the magnetic field generated by either of the coils of wire is confined inside the core. The resulting magnetic flux in the core is = BA, where A is the core cross-section area (Figure 1). This magnetic flux is nearly the same inside both coils of wire. According to Faraday’s law, a time-varying magnetic flux generates the voltage across a coil of wire. For our two coils of wire with number of turns N1 (primary) and N2 (secondary) the generated voltages will be equal to: V1 N 1 d , dt V2 N 2 d dt (3) Thus, the voltage change from primary coil of wire to secondary one is determined by the turns ratio n=N2/N1, the ratio of number of turns in the secondary to the primary. V2 N 2 n V1 N 1 (4) To find relationship for currents, one can apply Amper’s law (1) to a closed loop around the core. The transformer is usually designed to have the right part of the equation small N1 I 1 N 2 I 2 B l0 (5) where l is length of the circular integration path inside the core. In case the right part of the equation (5) is small indeed, the current ratio is inversely proportional to turn ratio: I2 1 I1 n (6) A I1 I2 N2 V1 N1 Figure 1 2 V2 State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Thus, the transformer converts the energy in the form of a time-varying large current at a low voltage into nearly the same amount of energy in the form of a small current at a high voltage (or vice versa). V1I1 V2I2 (7) Equations above define an ideal transformer with a circuit symbol shown in Figure 2. The dots indicate the sign conventions for two voltages: plus terminal should be chosen to be the dotted terminal which usually indicates the beginning of the coil. A transformer designed to output more voltage than it takes in across the input coil is called a "step-up" transformer (N2>N1), while one designed to do the opposite is called a "step-down" (N2<N1), in reference to the transformation of voltage that takes place. Figure 2 Looking into the primary the impedance is Z1 = V1/I1, looking into the secondary the impedance is Z2 = V2/I2. From these two it is easily obtained that: Z1/Z2 = (N1/N2)2 = 1/n2 (8) When RL is present across the secondary, i.e. Z2 = RL, the impedance seen looking into the primary is: Z1 = RL/n2 (9) These equations assume an ideal (no flux leakage, lossless magnetic core, perfect windings) transformer capable of infinite power delivery. A real transformer is able to deliver only a limited power. Using phasors, expression (3) can be written as V1= N1*j, then equation (5) can be rewritten as follows N1 I 1 N 2 I 2 lV1 jN1 A (10) One can see that performance of a transformer degrades with frequency decrease due to the right term in equation (10). The ideal transformer model neglects this fact since it deals only with mutual inductance between primary and secondary coils and ignores coils’ self-inductances. 3 State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Improved transformer model (Figure 3a) includes inductance LC that models the finite inductance of the primary coil. RC is resistance that models energy losses in the core. Core losses increase with frequency. In many practical cases RC is very large though and can be omitted. RW1 and RW2 stand for resistances of primary and secondary windings, respectively. They can be measured in DC mode with an Ohmmeter. VG represents a sine wave generator with internal impedance RG. The equivalent circuit for the transformed loaded with RL is shown in Figure 3 (b). The equivalent circuit is the 1st order L/R circuit with one pole. Frequency response analysis can be applied to determine LC and RC. RG RW1 TX1 2 RW2 RG RW1 2 VG LC RC Z2=RL => VG LC RC (N1/N2)2 (RW2+RL) 1 1 N1:N2 (a) (b) Figure 3. An ideal transformer without load (open circuit secondary) shows inductive impedance to signal generator, i.e. voltage in primary leads the current by 90º. Loading the transformer secondary with resistance changes the character of the input impedance of the primary to resistive one and the phase between the current and voltage in primary becomes zero. 3. Preliminary lab A. Consider a step-down transformer with 15 turns in the primary and 6 turns in the secondary windings. Calculate the input impedance of this transformer loaded to an 8 Ω resistor, assuming that series resistances of the windings are negligible. B. A signal generator with an internal impedance of 50 Ω provides an open circuit voltage of 2 V. Find the voltage drop across a 50 Ω load connected to the generator. Repeat the calculations for an 8 Ω load. Find the electrical power delivered to the load in both cases. C. Consider the generator from part B connected to a primary of the transformer from part A. The secondary is loaded to the same 8 Ω resistor. Find the voltage drop across the primary. 4 State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 4. Experiment The experiments will be performed with an iron-core transformer and a ferrite core transformer. Frequency response and turns ratio 1. Use the automatic measurement software to record the frequency response of the iron-core and the ferritecore transformers with an open secondary (see the schematic in Figure 4a). Note that the voltage across the secondary may drop at very low frequencies (several Hz) and may resonate at high frequencies due to the transformer parasitic capacitance. Identify the flat regions of the frequency responses. Set frequency of the signal generator to be within these ranges for the next experiments. 2. For the ferrite-core transformer, find the ratio of output and input voltages from the frequency response. Count the numbers of turns N1 and N2 in the primary and secondary windings. Compare the experimental value of the voltage ratio with the theoretical one and comment on the result. (a) (b) Figure 4 The 50 Ω resistor within the box represents the internal impedance of the signal generator Input impedance with open and shorted secondary (Choose either ferrite-core or iron-core transformer) 3. Insert a 1 kΩ resistor in the primary (Figure 4b). This resistor is serving as a current monitor; using the Ohm’s law one can easily convert the voltage measured across this resistor to the current through it. Measure the voltage (channel 1), current (channel 2) and the phase shift between them for the primary with open-circuit and short-circuit secondary. Calculate the complex input impedances Z = V/I; here V and I are voltage and current phasors. Present the results in table: V (V) Voltage on ch. 2 (V) Open Secondary Shorted Secondary 5 I (A) Phase Shift(º) Z (Ω) State University of New York at Stony Brook Department of Electrical and Computer Engineering ESE 211 Electronics Laboratory A 2011 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Impedance matching using transformer 4. Disconnect the signal generator from any circuit. Set Vpp of a sine wave equal to 1 V. Display the output of the generator (not loaded to anything) on the scope. Measure the Vpp of the displayed waveform. Explain the observed value. 5. Load the generator with a 50 Ω resistor. Display the voltage across the resistor on the scope and measure Vpp. Compare with part 4 and make a conclusion about the internal impedance of the generator. 6. Load the generator with an 8 Ω resistor and repeat the previous experiment. Explain the difference. Find the electrical power delivered to this resistor. 7. Load the generator with the step-down ferrite-core transformer; load the secondary with the 8 Ω resistor. Measure the Vpp across the primary. Compare with the result obtained in part 5. Make a conclusion about the input impedance of the loaded transformer. Measure Vpp across 8 Ω resistor and find the electrical power delivered to it. Compare with the value from part 6. Comment on the result of comparison. Report The report should include the lab goals, short description of the work, the experimental data presented in tables and plots, the data analysis and comparison followed by conclusions. Please follow the steps in the experimental part and clearly present all the results of measurements. 6