Question 4 of 8: Transformer Magnetic Circuit and Maximum Power Transfer

Question 4: Transformer Magnetic Circuit and Maximum Power Transfer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The core carries both windings on one limb, and the flux returns through two parallel air paths whose lengths and cross-sections are tabulated. Because the iron is assumed to have infinite permeability, it contributes no reluctance and the entire magnetic circuit is these two paths.

Find. Every element of the equivalent circuit referred to the primary — the two winding resistances and the magnetising reactance — and then the amplifier output impedance that maximises power delivered to the transducer.

Approach. A dc test excites no flux, so it isolates the winding resistance alone; the ac magnetising branch comes instead from the magnetic circuit, with the two air paths in parallel as seen by the wound limb. Referring the secondary quantities through $a^{2}$ then reduces the whole device to a simple series–shunt network, and the maximum-power-transfer theorem asks for the conjugate of the impedance seen looking in at the primary.

1. Extract the winding resistances from the dc tests. With direct current there is no rate of change of flux and hence no induced voltage, so the applied voltage divides across resistance only: $$R_1 = \frac{10\ \text{mV}}{100\ \text{mA}} = 0.100\ \Omega, \qquad R_2 = \frac{0.1\ \text{mV}}{100\ \text{mA}} = 0.001\ \Omega.$$
2. Form the turns ratio and refer the secondary resistance. $a = N_1/N_2 = 200/20 = 10$, and resistances refer across the ideal transformer as the square of the ratio: $$a^{2}R_2 = 10^{2} \times 0.001 = 0.100\ \Omega.$$ The two windings therefore contribute equally once referred — the design is deliberately balanced.
3. Compute the reluctance of each air path. With the iron of infinite permeability the only reluctance is that of the two paths, $\mathcal{R} = L/(\mu_0 A)$: $$\mathcal{R}_1 = \frac{3.77\times10^{-2}}{(4\pi\times10^{-7})(0.02)} = 1.50\times10^{6}\ \text{A}\,\text{Wb}^{-1},$$ and since $L_2$ is exactly twice $L_1$ over the same area, $\mathcal{R}_2 = 3.00\times10^{6}\ \text{A}\,\text{Wb}^{-1}$.
4. Combine the paths and find the magnetising inductance. The two paths offer alternative routes for the flux leaving the wound limb, so their reluctances combine as elements in parallel: $$\mathcal{R}_{eq} = \frac{\mathcal{R}_1\mathcal{R}_2}{\mathcal{R}_1+\mathcal{R}_2} = \frac{(1.5)(3.0)}{4.5}\times10^{6} = 1.00\times10^{6}\ \text{A}\,\text{Wb}^{-1}.$$ The magnetising inductance seen from the primary follows as $$L_m = \frac{N_1^{2}}{\mathcal{R}_{eq}} = \frac{200^{2}}{1.00\times10^{6}} = \boxed{40\ \text{mH}}$$
5. Convert to a reactance at the operating frequency. $$X_m = 2\pi f L_m = 2\pi(1000)(0.040) = 251.3\ \Omega,$$ so the magnetising branch is $j251.3\ \Omega$. This completes part [a]: the equivalent circuit is $R_1 = 0.1\ \Omega$ in series with the parallel combination of $jX_m = j251.3\ \Omega$ and the referred secondary branch $a^{2}R_2 = 0.1\ \Omega$ in series with the referred load.

Part [b] — matching the amplifier. With the transducer connected, the amplifier looks into the whole network:

1. Refer the load to the primary. $$a^{2}Z_L = 100 \times 0.078 = 7.80\ \Omega,$$ so the series branch beyond the magnetising node totals $$R_1 + a^{2}R_2 + a^{2}Z_L = 0.1 + 0.1 + 7.8 = 8.00\ \Omega.$$
2. Combine with the magnetising branch. The magnetising reactance shunts the series path, so the input impedance is $$Z_{in} = \frac{jX_m\,(8.00)}{8.00 + jX_m} = \frac{j251.3 \times 8.00}{8.00 + j251.3} = 7.99 + j0.254\ \Omega.$$ Because $X_m$ is some thirty times the series resistance, the magnetising branch is very nearly an open circuit and shifts the result by only about 0.1 %.
3. Apply the maximum-power-transfer theorem. Maximum power is delivered when the source impedance is the complex conjugate of the load it drives, so the amplifier output impedance must be $$Z_{amp} = Z_{in}^{*} = \boxed{7.99 - j0.25\ \Omega \;\approx\; 8\ \Omega}$$ In practice the amplifier would simply be specified as an 8 $\Omega$ output — the small capacitive term needed to cancel the magnetising reactance is within the tolerance of any real output stage.