Air Core Transformer

As the name implies, the air-core transformer does not have a ferromagnetic core to link the primary and secondary coils. Rather, the coils are placed sufficiently close to have a mutual inductance that will establish the desired transformer action. In Fig. 1, current direction and polarities have been defined for the air-core transformer. Note the presence of a mutual inductance term M, which will be positive in this case, as determined by the dot convention.
Air-core transformer equivalent circuit.
Fig. 1:Air-core transformer equivalent circuit.
From past analysis in this chapter, we now know that
$$\bbox[10px,border:1px solid grey]{e_{p}=L_{p} \frac{d i_{p}}{d t}+M \frac{d i_{s}}{d t}} \tag{1}$$
for the primary circuit.
We found in Chapter 11 that for the pure inductor, with no mutual inductance present, the mathematical relationship
$$v_{1}=L \frac{d i_{1}}{d t}$$
resulted in the following useful form of the voltage across an inductor:
$$\mathbf{V}_{1}=\mathbf{I}_{1} X_{L} \angle 90^{\circ} \quad \text { where } X_{L}=\omega L$$
Similarly, it can be shown, for a mutual inductance, that
$$v_{1}=M \frac{d i_{2}}{d t}$$
will result in
$$\bbox[10px,border:1px solid grey]{\mathbf{V}_{1}=\mathbf{I}_{2} X_{m} \angle 90^{\circ} \quad \text { where } X_{m}=\omega M} \tag{2}$$
Equation (1) can then be written (using phasor notation) as
$$\bbox[10px,border:1px solid grey]{\mathbf{E}_{p}=\mathbf{I}_{p} X_{L_{p}} \angle 90^{\circ}+\mathbf{I}_{s} X_{m} \angle 90^{\circ}} \tag{3}$$
$$\mathbf{V}_{g}=\mathbf{I}_{p} R_{p} \angle 0^{\circ}+\mathbf{I}_{p} X_{L_{p}} \angle 90^{\circ}+\mathbf{I}_{s} X_{m} \angle 90^{\circ} $$
$$\bbox[10px,border:1px solid grey]{\mathbf{V}_{g}=\mathbf{I}_{p}\left(R_{p}+j X_{L_{p}}\right)+\mathbf{I}_{s} X_{m} \angle 90^{\circ}} \tag{4} $$
For the secondary circuit,
$$\bbox[10px,border:1px solid grey]{\mathbf{E}_{s}=\mathbf{I}_{s} X_{L_{s}} \angle 90^{\circ}+\mathbf{I}_{p} X_{m} \angle 90^{\circ}} \tag{5} $$
$$\mathbf{V}_{L}=\mathbf{I}_{s} R_{s} \angle 0^{\circ}+\mathbf{I}_{s} X_{L_{s}} \angle 90^{\circ}+\mathbf{I}_{p} X_{m} \angle 90^{\circ} $$
$$\bbox[10px,border:1px solid grey]{\mathbf{V}_{L}=\mathbf{I}\left(R_{s}+j X_{L_{s}}\right)+\mathbf{I}_{p} X_{m} \angle 90^{\circ}} \tag{6}$$
$$\mathbf{V}_{L}=-\mathbf{I}_{s} \mathbf{Z}_{L}$$
into Eq. (6) results in
$$0=\mathbf{I}_{s}\left(R_{s}+j X_{L_{s}}+\mathbf{Z}_{L}\right)+\mathbf{I}_{p} X_{m} \angle 90^{\circ}$$
Solving for $\mathbf{I}_{s}, we have
$$\mathbf{I}_{s}=\frac{-\mathbf{I}_{p} X_{m} \angle 90^{\circ}}{R_{s}+j X_{L_{s}}+\mathbf{Z}_{L}}$$
and, substituting into Eq. (4), we obtain
$$\mathbf{V}_{g}=\mathbf{I}_{p}\left(R_{p}+j X_{L_{p}}\right)+\left(\frac{-\mathbf{I}_{p} X_{m} \angle 90^{\circ}}{R_{s}+j X_{L_{s}}+\mathbf{Z}_{L}}\right) X_{m} \angle 90^{\circ}$$
Thus, the input impedance is
$$\mathbf{Z}_{i}=\frac{\mathbf{V}_{g}}{\mathbf{I}_{p}}=R_{p}+j X_{L_{p}}-\frac{\left(X_{m} \angle 90^{\circ}\right)^{2}}{R_{s}+j X_{L_{s}}+\mathbf{Z}_{L}}$$
or, defining $\mathbf{Z}_{p}=R_{p}+j X_{L_{p}}$, $\mathbf{Z}_{s}=R_{s}+j X_{L_{s}}$ and $X_{m} \angle 90^{\circ}=+j \omega M$ we have
$$\mathbf{Z}_{i}=\mathbf{Z}_{p}-\frac{(+j \omega M)^{2}}{\mathbf{Z}_{s}+\mathbf{Z}_{L}}$$
$$\bbox[10px,border:1px solid grey]{\mathbf{Z}_{i}=\mathbf{Z}_{p}-\frac{(\omega M)^{2}}{\mathbf{Z}_{s}+\mathbf{Z}_{L}}} \tag{7}$$
The term $(\omega M)^{2}(\mathbf{Z}_{s}+\mathbf{Z}_{L}) $ is called the coupled impedance, and it is independent of the sign of $M$ since it is squared in the equation. Consider also that since $(\omega M)^{2}$is a constant with $ 0^{\circ} $ phase angle, if the load $ \mathbf{Z}_{L} $ is resistive, the resulting coupled impedance term will appear capacitive due to division of $ \left(\mathbf{Z}_{s}+R_{L}\right) $ into $ (\omega M)^{2} $. This resulting capacitive reactance will oppose the series primary inductance $L_{p}$, causing a reduction in $ \mathbf{Z}_{i} $. Including the effect of the mutual term, the input impedance to the network will appear as shown in Fig. 2.
Input characteristics for the air-core
Fig. 2:Input characteristics for the air-core transformer.
Example 1: Determine the input impedance to the air-core transformer in Fig. 3.
Fig. 3:Example 1.
Solution: \begin{aligned} \mathbf{Z}_{i} &=\mathbf{Z}_{p}+\frac{(\omega M)^{2}}{\mathbf{Z}_{s}+\mathbf{Z}_{L}} \\ &=R_{p}+j X_{L_{p}}+\frac{(\omega M)^{2}}{R_{s}+j X_{L_{x}}+R_{L}} \\ &=3 \Omega+j 2.4 \mathrm{k} \Omega+\frac{((400 \mathrm{rad} / \mathrm{s})(0.9 \mathrm{H}))^{2}}{0.5 \Omega+j 400 \Omega+40 \Omega} \\ & \cong j 2.4 \mathrm{k} \Omega+\frac{129.6 \times 10^{3} \Omega}{40.5+j 400} \\ &=j 2.4 \mathrm{k} \Omega+322.4 \Omega \angle-84.22^{\circ} \\ &=j 2.4 \mathrm{k} \Omega+(0.0325 \mathrm{k} \Omega-j 0.3208 \mathrm{k} \Omega) \\ &=0.0325 \mathrm{k} \Omega+j(2.40-0.3208) \mathrm{k} \Omega \\ &=\mathbf{32.5} \Omega+j \mathbf{2079} \Omega\\ &=\mathbf{2079.25} \Omega \angle \mathbf{8 9 . 1 0}^{\circ} \end{aligned}

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