Line Driver Features Synthesized Output Impedance

The common line driver uses a series resistor at the output to control the amplifier’s output impedance. The output impedance should be equal to the load impedance in order to achieve high return loss (good impedance matching) at the amplifier’s output.

The resistor dissipates half of the power that the amplifier delivers, while the load receives only half of the amplifier’s output voltage swing. This limits the maximum power that the amplifier can deliver to the load.

A line driver was developed to offset this problem (see the figure). It employs a transformer to synthesize an output impedance with negligible output voltage drop, thus providing high power-delivery efficiency. The amplifier uses voltage feedback through R_F and current feedback through the transformer to achieve a defined output impedance. To calculate the output impedance, short the input to ground, and connect a V_O source at the output. If V_I = 0, then V_N also is zero. If V_N = 0, then the voltages on the primary and secondary of the transformer also are zero. If the voltage on the primary is zero, then V_A = V_O.

The current through R_F is I(R_F) = V_O/R_F, which is equal to the current through the secondary of the transformer (since V_N = 0). The current through the primary of the transformer is I(R_F) × N and it’s also equal to I_O, therefore:

I_O = (V_O/R_F) ×N

R_OUT = V_O/I_O = R_F/N

For example, if R_F = 1K and N = 20, then R_O = 50 Ω.

Now let’s calculate the line driver gain. Because the line driver has a defined output impedance, its output voltage depends on the load value. Thus, the EMF voltage gain, which is the voltage gain with an open output, can be calculated.

A voltage generator V_IN is connected at the input. Because the output load is infinite, the output current I_O is zero, and the currents in the primary and secondary of the transformer are zero. V_N = V_IN, which also is the voltage at the secondary of the transformer. V_A = V_IN × (1 + R_F/RG ), which is the common noninverting amplifier gain. Therefore:

V_O = V_A - (V_IN /N) = V_IN × (1 + R_F/RG ) − (V_IN/N)

G_V(EMF ) = V_OUT/V_IN = 1 + (R_F/RG ) - (1/N)

For example, if RG = 100 W:

G_V(EMF) = 1 + (1000/100) − (1/20) = 11 − 0.05 = 10.95

If we load the amplifier with 50 W equal to the output impedance, the voltage gain will be halved:

V_O = 5.475 V_IN

The voltage drop on the primary of the transformer will be a mere 0.05 V_IN, less than 1% of V_O. That’s a very noticeable improvement compared to the 50% loss inherent in the typical line driver circuit.