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M AGNETIC P ROPERTIES OF A F ERRITE P OT- CORE I NDUCTOR C2–1 Experiment C2. Magnetic Properties of a Ferrite Pot-core Inductor Objectives In this experiment you will investigate the properties of an inductor and transformer en- closed in a ferrite pot-core. In addition to being strongly ferromagnetic, ferrite is a good electrical insulator. This makes it useful for constructing inductors and transformers for use at high frequencies, since it helps to avoid problems with eddy currents. You will: • measure the inductance L, resistance RL and quality factor Q of two inductor coils with an air core • repeat these measurements with one coil short-circuited to determine the leakage in- ductance • repeat the previous measurements with a ferrite core and obtain a value for the relative permeability µr for ferrite • examine the behaviour of the coils as a transformer. If you have not done experiment A5, you should ﬁrst read the short Background section of that experiment. Also, if you are unfamiliar with AC circuits you should read Appendix C in the pink section at the end of these notes. Prework Questions. 1. Consider an LCR circuit, where the inductor has an air-core. (a) How do the inductance of the inductor, and the resonant frequency of the circuit, change if the number of turns of wire in the coil is doubled? (b) How do they change if the core is replaced with ferrite? 2. Sketch a diagram for a concentric air-core transformer in operation, with the outer coil as the primary. Assume that the secondary coil is not connected, and that the current in the primary coil is increasing and near maximum. Include the ﬁeld lines in your sketch. (Don’t worry about the gap in between the primary and secondary coils.) 3. Sketch a similar diagram, assuming the secondary coil is short circuited. Also include current for the inner coil, if any. Explain any differences between this sketch and your previous one. C2–2 G ROUP C E XPERIMENTS Background For a long air-cored inductor with N turns, length , radius a and cross-sectional area A = πa2 , the inductance is given by: L = µ0 N 2 A/ . (1) By ‘long,’ we mean that a. If the inductor is not long relative to its other dimensions, the inductance is more accurately represented by (see Figure C2-1 ): µ0 N 2 A L= . (2) + 0.90a Fig. C2-1 Parameters of a short air-cored inductor The inductance is increased substantially if a ferromagnetic material is placed inside or around the coil. If the magnetic ﬂux is conﬁned entirely within the magnetic material (e.g., as in the toroidal geometry of Figure C2-2(a)), then the inductance is given by (see Figure C2-2) L = µr µ0 N 2 A/ , (3) where µr is the relative permeability of the core and is now the length of the magnetic circuit (dotted line in Figure C2-2(a)), not the length of the winding. magnetic circuit N S A = effective cross- Hg sectional area Hc (a) (b) Fig. C2-2 Examples of (a) a closed magnetic circuit, and (b) a magnetic circuit with an air gap If a small gap is introduced into the magnetic circuit, as in Figure C2-2(b), we need to apply the boundary conditions for B and H to determine L. If Bc , Hc refer to the core and Bg , Hg to the gap, we have Bc = B g since B · dS = 0. Since B = µr µ0 H, µr µ0 H c = µ 0 H g . M AGNETIC P ROPERTIES OF A F ERRITE P OT- CORE I NDUCTOR C2–3 Since H · dl = N i, we have N i = Hg g + Hc , where g is the gap width and is the path length in the magnetic material. Using the deﬁnition for inductance, L = N φ/i, where φ is the magnetic ﬂux, we ﬁnd that µr µ0 N 2 A L= (4) + µr g If i changes with time, the voltage across L is given by di dφ v=L =N dt dt Procedure 1. A ferrite core kit may be obtained from the service counter. It contains two halves of a ferrite pot core which can be assembled around two coils wound on a plastic former; the arrangement is shown in Figure C2-3. Please handle these carefully since ferrite is a ceramic which chips easily. Fig. C2-3 Ferrite pot core components Measure L, Q and RL for air-cored coils Select one of the two coils wound on the plastic former. Measure the inductance L of this ﬁrst coil (without the ferrite core), by using the circuit in Figure C2-4 to determine the resonance frequency 1 f0 = √ . 2π LC The non-ideal nature of the inductor may be represented with either a series or parallel resistor. Obviously, for an ideal inductor the series RL would be zero, whereas the parallel RL would be inﬁnite. In Figure C2-4, RL represents the equivalent parallel loss resistance of the inductor. The circuit should be constructed using the trainer board provided. The unused coil must be left open circuit. An accurate value for C and R1 may be obtained using the DigiBridge impedance bridge available in the laboratory - it is more convenient to measure these before wiring the circuit. Resonance should be determined by looking for zero phase shift between v in and vout . C2–4 G ROUP C E XPERIMENTS R1 = 100kΩ v in vout (to Oscilloscope) (to Oscilloscope) C ~ L RL 0.01µF Fig. C2-4 Parallel resonant circuit used to determine L and RL The sharpness of the resonance is expressed in terms of Q, the quality factor. By comparing vin and vout at resonance, you can calculate RL and hence the Q of the inductor, using RL vout RL Q= and = . ω0 L vin R1 + RL When measuring vin and vout it is best to use one oscilloscope probe to avoid cali- bration differences between two probes. If you want to use two probes, their voltage response to a common source must ﬁrst be compared over the relevant frequency range. 2. Repeat the measurements described above to determine L, Q and R L for the second coil. 3. Compare your values of L with the predictions of eqns. (1) and (2) and comment. C1 Flux linking and leakage for air-cored coils 4. With an air core, the magnetic ﬂux produced by one coil will not be strongly linked to the other coil. The proportion of magnetic ﬂux which is not linked between the coils is called the leakage ﬂux. A direct method of measuring the leakage ﬂux is to measure the inductance of one coil when the second coil is short-circuited. If the coils formed an ideal transformer the inductance would be zero. This is because the current in the secondary coil would produce a ﬂux which completely cancels the ﬂux due to the current in the primary coil (why?). Any inductance measured in the primary coil when the secondary is shorted is known as leakage inductance. Measure the inductance of the smaller (fewer turns) coil again, but now with the second coil short-circuited. Comparing this result with your earlier measurements of L (with the secondary open circuit) you should be able to estimate the fraction of the ﬂux of the primary coil which is linked by the secondary coil. 5. Make a sketch of the magnetic ﬁeld around the two coils when the secondary is open circuit. C2 M AGNETIC P ROPERTIES OF A F ERRITE P OT- CORE I NDUCTOR C2–5 Measure L, Q and RL for ferrite-cored coils 6. Insert the coils in the ferrite core. You will again use the circuit in Figure C2-4. Since Q, RL and L depend strongly on vin , you should record the input level used. Use a value of 2V p-p (peak to peak) for vin . Repeat the measurements described to in steps 1 and 2 to ﬁnd L, Q and RL for each coil. Check that L ∝ N 2 by comparing the inductance of the two coils. Determine µr for ferrite 7. Calculate the value of µr from eqn. (3) assuming that the effective area A = 198 mm 2 and the effective magnetic path length = 70.0 mm. The problem with this calculation is that the values of A and are difﬁcult to estimate and A varies in different places within the ferrite. As a result, the magnetic ﬂux varies dramatically within the ferrite. The values you have used are effective values supplied by the ferrite manufacturer. Flux linking and leakage for ferrite-cored coils 8. Repeat part 4, with the coils now inserted into the ferrite core, to determine the leak- age ﬂux with the ferrite present. Comment on the difference between the results with and without the ferrite. 9. Make a sketch of the magnetic ﬁeld around the coils when they are in the ferrite, with the secondary open circuit. C3 Importance of air gaps 10. Insert small pieces of plastic (0.10 mm thickness) on opposite sides of the ferrite cores so the two halves are separated by a gap of one thickness of plastic. Measure the new resonance frequency of the smaller turn coil and compare the measured inductance with the value calculated using eqn. (4) above. Hint: Is there one gap or two? The two halves of the core are well polished to ensure close contact in normal use. Give an explanation, in your log book, of why even a small gap makes a big difference to L. 11. Determine the leakage ﬂux with the gap between the halves of the ferrite core present by repeating the procedure used in item 4. Comment on the difference between this case and the case with no gap. Transformer 12. Set up a transformer by connecting the sinewave generator to the coil with 240 turns and measure the voltage across this coil and the voltage induced in the other coil. These measurements should be made at a frequency chosen to lie somewhere in the range 10 − 20 kHz and with an input signal ∼ 10 V p-p. Check the validity of the relation v2 /v1 = N2 /N1 for the two cases : C2–6 G ROUP C E XPERIMENTS (a) a ferrite core with no air gap, and (b) no ferrite core. How do these results relate to your previous measurements of leakage inductance for these two cases? Conclusion Collect the results of this experiment together into a table and compare the different cases you have examined. What conclusions can you draw - especially regarding the importance of the core to the performance of the coils as a transformer? C4

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Experiment C2. Magnetic Properties of a Ferrite Pot-core Inductor

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