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Home Impedance Matching The Matching Transformer  
See also: The Microphone Matching Transformer, Matching, Input Transformers, Transformer Coupling, Output Transformers  
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The Matching TransformerAuthor: N.H. Crowhurst A matching transformer for a loudspeaker is just like one used to step up the output from a microphone, except that it is bigger, and designed to handle much more power. Suppose that a transformer is to match a 16ohm loudspeaker to an amplifier requiring, not 16 ohms, but 6400 ohms, to which it is to deliver its power.
Suppose 9 watts are to be transferred. The formula for power is W = El, where W is power in watts, E is voltage in volts, and I is current in amperes. (Remember that the relation between voltage and current in a resistance is E = IR, where R is the resistance in ohms.) Combining the two formulas, W = El = E X E/R = E2/R. Then by multiplying both sides by R» WR = E2. In the example, WR = 9 X 6400 = 57,600 = E2. Hence, E = 57,600  240 volts. Also I = E/R = 240/6400 = .0375 ampere. At the voicecoil resistance, WR = 9 X 16 = !*4 = E2. Hence, E = Y144 = 12 volts. I = = 12/16 = 0.75 ampere. But what would happen if a matching transformer were not used? The amplifier is only designed to deliver ,0375 ampere into a 6400ohm resistance or load as it is called. If it is connected to 16 ohms, it will probably not deliver much more than .0375 ampere  maybe .05 ampere  and the waveform will be very distorted. When ,05 ampere is delivered to 16 ohms, the voltage is only E = IR = 16 X *05 = 0.8 volts, and the power delivered is only W = El = 0.8 X 05 = .04 watt, in place of the expected 9 watts!
Notice what the transformer does: it reduces the voltage by the ratio of turns in the windings (called the turns ratio) and it also increases the current by the same ratio. It thus effectively multiplies the resistance (or impedance) connected to the secondary by the square of the turns ratio. In this case, the ratio was 20:1. The impedance connected to the secondary is multiplied by 20 X 20 = 400 (16 X 400 = 6400 ohms). The input voltage is 240 volts and the output is 12 volts; the input current is .0375 ampere and the output 0.75 ampere. The input and output power are the same (except for any losses due to the inefficiency of the transformer, which we have conveniently ignored; in practice, an output transformer would be more than 90% efficient, so this is no very great error).
When the amplifier supplies 240 volts to the "high" winding of the transformer, the core will be magnetized, and, due to its high inductance, very little current will be drawn from the amplifier, unless the voice coil is connected to the low winding. The high winding must have 20 times as many turns as the low winding. This way, 240 volts induction in the primary will cause 12 volts in the secondary. When the voice coil is connected across a 12volt source, it will draw 0.75 ampere. If no current flowed in the primary of the transformer, this secondary current would destroy the induction by saturating the core, and the 12 volts (as well as the 240 volts) would disappear. To sustain the 12 volts, the amplifier must supply current to the primary to neutralize the effect of the 0.75 ampere in the secondary. As the primary winding has 20 times as many turns, it will only require onetwentieth the current, or .0375 ampere, to have the same effect and neutralize the effect of the secondary current. Thus the transformer causes the primary winding to take .0375 ampere from the amplifier at 240 volts, when the secondary is connected to a voice coil of 16 ohms that takes 0.75 ampere at 12 volts. To the amplifier, it is the same as connecting a voice coil with a resistance of 6400 ohms, which it "wants," This is matching. What would happen if the voicecoil resistance were 20 ohms instead of 16 ohms? If the transformer secondary voltage were still 12 volts, the voice coil would only take 0.6 ampere in place of 0.75 ampere. The turns ratio would still produce 240 volts across the primary winding, but the primary current required to balance the new secondary current of 0.6 ampere will be 0.6/20, or 0.03 ampere, in place of 0.75/20, or 0.0375 ampere. This is the same as if a load of R = E/I = 240/0.03 = ohms were connected to the amplifier directly. 8000 ohms is just 400 times the 20 ohms connected to the secondary winding of the transformer. The 20:1 turns ratio of the transformer thus always multiplies the resistance, or impedance, connected to its secondary winding by a factor of 400, or 20 squared (20*20).


Home Impedance Matching The Matching Transformer 