Transformers can be particularly useful when you are trying to ensure that a load receives maximum power from a source. Recall that maximum power is transferred to a load when its impedance is a match with the internal resistance of the supply. Even if a perfect match is unattainable, the closer the load matches the internal resistance, the greater the power to the load and the more efficient the system.
Unfortunately, unless it is planned as part of the design, most loads are not a close match with the internal impedance of the supply. However, transformers have a unique relationship between their primary and secondary impedances that can be put to good use in the impedance matching process.
Example 1 will demonstrate the significant difference in the
power delivered to the load with and without an impedance matching transformer.
Example 1:
a. The source impedance for the supply of Fig. 1(a) is $512 W$, which is a poor match with the $8-Ω$ input impedance of the speaker. One can expect only that the power delivered to the speaker will be significantly less than the maximum possible level. Determine the power to the speaker under the conditions of Fig. 1(a).
b. In Fig. 1(b), an audio impedance matching transformer was introduced between the speaker and the source, and it was designed to ensure maximum power to the 8-Ω speaker. Determine the input
impedance of the transformer and the power delivered to the
speaker.
c. Compare the power delivered to the speaker under the conditions of
parts (a) and (b).
Fig. 1: Example 1.
Solution:
a. The source current:
$$I_{s}=\frac{E}{R_{T}}=\frac{120 \mathrm{~V}}{512 \Omega+8 \Omega}\\
=\frac{120 \mathrm{~V}}{520 \Omega}=230.8 \mathrm{~mA}$$
The power to the speaker:
$$P=I^{2} R=(230.8 \mathrm{~mA})^{2} \cdot 8 \Omega\\
=\mathbf{4 2 6 . 1 5} \mathbf{m W} \cong \mathbf{0 . 43} \mathrm{W}$$
or less than half a watt.
b. $Z_{p}=a^{2} Z_{L}$
$$a=\frac{N_{p}}{N_{s}}=\frac{8}{1}=8$$
and
$$Z_{p}=(8)^{2} 8 \Omega=\mathbf{512} \Omega$$
which matches that of the source. Maximum power transfer conditions have been established, and the source current is now determined by
$$I_{s}=\frac{E}{R_{T}}=\frac{120 \mathrm{~V}}{512 \Omega+512 \Omega}\\
=\frac{120 \mathrm{~V}}{1024 \Omega}=117.19 \mathrm{~mA}$$
The power to the primary (which equals that to the secondary for the ideal transformer) is
$$P=I^{2} R=(117.19 \mathrm{~mA})^{2} 512 \Omega=\mathbf{7 . 0 3 2} \mathrm{W}$$
The result is not in milliwatts, as obtained above, and exceeds $7 \mathrm{~W} $, which is a significant improvement.c. Comparing levels, $7.032 \mathrm{~W} / 426.15 \mathrm{~mW}=16.5$, or more than 16 times the power delivered to the speaker using the impedance matching transformer.
Another important application of the impedance matching capabilities of a transformer is the matching of the 300-Ω twin line transmission line from a television antenna to the 75-Ω input impedance of
today’s televisions (ready-made for the 75-Ω coaxial cable), as shown in Fig. 2.
Fig. 2: Television impedance matching transformer.
A match must be made to ensure the strongest signal to the
television receiver. Using the equation $Z_p = a^2Z_L$ we find
$$300 Ω = a^2 75 Ω$$
and
$$ a = \sqrt{{300Ω \over 75Ω}} = \sqrt{4} = 2$$
with $N_p : N_s = 2 : 1$ (a step-down transformer)
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