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

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Experiment C2. Magnetic Properties of a Ferrite Pot-core Inductor Powered By Docstoc
					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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Description: Experiment C2. Magnetic Properties of a Ferrite Pot-core Inductor