Transformer coupled class B push-pull, 25π % efficiency

A transformer-coupled class B push-pull amplifier uses two identical transistors, each conducting for one half-cycle. The input transformer splits the signal into two halves that are 180° out of phase, and the output transformer combines the two collector currents in the load.

Circuit diagram

                            Q1 (NPN)
                 +--------B   C-------------+
                 |            E             |
          T1     )            |             (     T2
 Vi o----(       )            |             (    )----+
         (       )-- CT ------+--- GND      (-- +VCC  |
         (       )            |             (    )   [RL]
 Vi o----(       )            |             (    )----+
                 |            E             |
                 +--------B   C-------------+
                            Q2 (NPN)

(T1: input transformer with a centre-tapped secondary that drives the two bases; the centre tap and both emitters go to ground. T2: output transformer with a centre-tapped primary; the ends go to the two collectors, the centre tap to +VCC, and RL is on the secondary.)

Characteristics (graph)

  IC ^
     |  Q1 line     Q2 line
     | \   |   |   /
     |   \ |   | /
     |     \   /       dc load line is
     |      \ /        vertical at VCEQ = VCC
  0  +-------+------------> VCE
                VCC

Because the dc resistance of the transformer primary is small, the dc load line is almost vertical at VCE = VCC. The ac load line has slope −1/R'L through the Q point at (VCC, 0), where R'L = (N1/N2)² RL is the load seen across half of the primary (N1 = turns of half the primary).

Derivation of maximum efficiency

Let the peak collector current in each transistor be Im and the peak voltage across the half-primary be Vm = Im R'L.

Output power: Po = Im² R'L/2 = Vm²/(2R'L)

Input power: each transistor draws a half-sine current from VCC; the average of the two together is 2Im/π.

Pi = VCC · 2Im/π

Efficiency:

η = Po/Pi = (Im² R'L/2) / (2 VCC Im/π) = (π/4)(Im R'L/VCC) = (π/4)(Vm/VCC)

The maximum swing occurs when Vm = VCC (the collector voltage swings from 0 to 2VCC):

ηmax = π/4 = 0.7854 = 78.54 %

Since 25π = 78.54, this is written as ηmax = 25π %, the standard result for ideal class B.