leakage Flux and Coupling Coefficients Electronics Help

One of the assumptions made for an ideal transformer is that all of the magnetic flux generated by one winding is coupled to the other winding. In fact. some of the flux generated by the primary winding of a practjeal transformer “escapes,” in the sense that it is not confined entirely to the core, and therefore does not reach the secondary winding. Magnetic flux is concentrated in a high-permeability material such as iron. but disperses in low-permeability air. Thus, two windings on one iron core have a good chance of sharing essentially the same flux, but are less likely to share flux in an air core. As illustrated in Figure 11-35, flux that is shared by the two windings of a transformer is called mutual flux 1/1″,. and that which escapes is called leakage flux, CPl’ The (‘(}(‘fjiciellf o] C’o/lpllll){ k between the two windings of a transformer is ,,!dined to be.

In an ideal transformer, 1/11 = 0 and k = I. In a practical iron-core transformer, the leakage flux is very small, typically about I% of the mutuaillux. Consequently. the coupling coefficient for iron-core transformers is very nearly equal to the ideal value of I. Such transformers are said to be tightly COlip led. On the other hand, air-curt: 11 austormers have a large leakage flux, and the coefficient of coupling may be as small as 0.01. Transformers with small values of k are said to be loosely co up led. The coupling coefficient can be increased by wrapping the secondary turns physically very close to the primary turns (i.e., overlapping them). However, in some high-frequency applications, loosely coupled air-core transformers are desirable. As an aid in analyzing the effect of leakage. flux on the performance of a transformer. it is convenient to think of the total inductance of the primary winding. I’ll’ as being composed of two components: one, the primary magnetizing intluctance, Llm” that is subjected to the mutual flux, and. two, the primary leakage inductance, 1’11(‘ that is subjected to leakage flux only. Thus,

                          Lp = Lpm/ + Lpl

If the coefficient of coupling were equal to I, then Lpm would be zero, since 1/1( wouid be zero. In that case, we would have Lp = L/lm• In general,

The secondary winding is also responsible for a certain amount of leakage. For example, if it were very loosely wound on the core and had large air gaps between its turns, we would attribute the reduced coupling to it. We can therefore regard the total inductance of the secondary winding, L” to be similarly composed of a secondary magnetizing inductance, L,nlt and a secondary leakage inductance, L,I’ Thus, L = Lpm + L

Figure 11-36 shows the equivalent circuit of a practical transformer in which the winding resistances and leakage inductances are identified as separate components. In reality. these quantities are intimate, inseparable parts of the entire windings and arc said to be distributed throughout the windings. However, we can analyze their effects on the transformer by considering them to be separate entitle- (called lumped parameters). Note that the equivalent circuit contains an idea! transformer that has inductances LIIIII and Lsm and that has zero winding resistance To use the transformer theory studied earlier, we now regard the primary   HAS that which appears across the terminals of the ideal primary in the equivaient circuit, and the secondary voltage as that which appears aeross the ideal.

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Loading Effects

It is apparent from Figure 11-36 that the full value of einwill not appear across the primary winding. The primary voltage, ep, will be less than einbecause of the voltage drop across the impedance in series with the winding. Consequently, the full value of ein will not be multiplied by the turns ratio, and the secondary voltage will be less than would be predicted for an ideal transformer. Similarly, the full value of e, does. not appear at the load, because of the voltage drop appearing across the impedance in series with it. We see that winding resistance .and leakage inductance make the ratio ein/vi. smaller than the ratio Np/Ns. We say that RI. loads the circuit,
in much the same way that a load on the output of a real voltage source reduces the terminal voltage.

To calculate the actual load voltage in a practical transformer, we must’ first consider the voltage divider formed by the impedance in series with the primary and the impedance reflected from thesecondary. Recall that we showed how a load resistance is reflected through a transformer by the square of the turns ratio. Similarly. any complex impedance in series with the secondary is .retlected to the primary side by the square of the turns ratio: Z (reflected to ‘primary) : (Np/ Ns)2Z” where Z, is the total impedance across the secondary winding. It is this reflected impedance that we must use in our voltage-divider computation on the primary side. The next example illustrates the computation.

The transformer shown in Figure 11-37(a) has a total primary inductance of 0.3 H and a total secondary inductance of 50 ml-l, The winding resistances and R, . The transformer has a coupling coefficient of 0.99.
1.’ Draw the equivalent circuit of the transformer.
2. Find the magnitude of the load voltage

Solution

1. We first find the mutual and leakage inductance in each winding. From equations (11-70) through (11-73),

L”m = kL” = 0.99(0.3 H) =: 0.297 H
L,J’ = (1 – k)LI’ = (0.1)1)(0.31-1) = 3 mH
L””, = kL, = (0.99)(50 mil) = 49,5 mH
L” = (I – k)L, = (0.01)(50 mH) = 0.5 ml l

2. Since the frequency of elll is 1.5 kHZ, the reactance corresponding to each inductive component in the circuit is X,. = jwL = j(2TTf) I. -= j(21TX 1.5 X 10.1)L =:. Thus

Figure 11-37(d) shows the equivalent circuit with the reflected impedance drawn on the primary side. We can now use the voltage-divider rule to find the value. We will neglect the impedance of, 2799 H) in parallel with the reflected impedance. because it is so much greater than the reflected impedance. Thus .

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We see that there is a discrepancy of 1.62 Y, or about 37%: between the actual and ideal magnitudes of the load voltage. This rarhcrlargc discrepancy is due to the small value of load resistance (10 H) in the example. Note that the smaller the load resistance. the greater its loading. effect, due to voltage’ division on both sides of the transformer.

Posted on November 19, 2015 in MULTISTAGE AMPLIFIES

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