Transformer coupled class A efficiency

In a transformer-coupled class A amplifier the load RL is connected to the collector through a transformer. The transformer matches the load to the transistor: the load seen by the collector is R'L = (N1/N2)² RL. The dc resistance of the primary is almost zero, so the dc drop is small and VCEQ ≈ VCC.

Circuit diagram

                    +VCC
                     |
         +-----------+
         |           |       T (N1:N2)
        [R1]         )||(------+
         |        N1 )||( N2   |
         |           )||(     [RL]
         |           |  (------+
         |           C
 Vi o-||-+----------B  Q (NPN)
         |           E
        [R2]         |
         |           +------+
         |           |      |
        GND        [RE]   [CE]
                     |      |
                    GND    GND

(R1, R2: base divider; RE with CE: stabilisation; primary N1 between VCC and the collector; load RL on the secondary N2.)

Derivation of efficiency

The ac load line has slope −1/R'L through the Q point (VCEQ = VCC, ICQ). The collector voltage swings between VCEmax and VCEmin.

VCC = (VCEmax + VCEmin)/2 (Q point in the middle of the swing)

Output power (ac):

Po = (VCEmax − VCEmin)² / (8 R'L)

Input power (dc): Pi = VCC ICQ. Since the dc resistance of the primary is zero, the ICQ for the maximum swing is ICQ = VCC/R'L.

Pi = VCC²/R'L = (VCEmax + VCEmin)²/(4 R'L)

General efficiency:

η = Po/Pi = [(VCEmax − VCEmin)² / (8R'L)] × [4R'L/(VCEmax + VCEmin)²]

η = 50 % × [(VCEmax − VCEmin)/(VCEmax + VCEmin)]²

For the largest swing, VCEmin = 0 and VCEmax = 2VCC, giving η = 50 %. So the maximum efficiency of the transformer-coupled class A amplifier is 50 %, double the series-fed value (25 %).