Transformer coupled class B: 78.5% and maximum loss

Circuit diagram

       T1 (input)                                   T2 (output)
                        +--- B1  Q1 (NPN)  C1 ---+
                        |         E1             |
   o----+  ||  +--------+         |              +--------+  ||  +----+
        )  ||  (                  |                       (  ||  )    |
  Vi    )  ||  (-- CT -- GND     GND          VCC -- CT --(  ||  )   [RL]
        )  ||  (                  |                       (  ||  )    |
   o----+  ||  +--------+         |              +--------+  ||  +----+
                        |         E2             |
                        +--- B2  Q2 (NPN)  C2 ---+

The centre-tapped input transformer drives Q1 and Q2 in opposite phase. The centre tap of the output primary goes to VCC, and the two collectors go to its ends. Q1 conducts on one half-cycle and Q2 on the other; the secondary delivers the full waveform to RL.

Characteristic graph (composite characteristic)

  iC1 ^
      |
   Im *          Q1 conducting
      | \
      |   \
      |     \      ac load line,
      |       \    slope -1/R'L
    0 +---------*---------+-------> vCE1
      |           \
      |             \      Q2 conducting
      |               \    (Q2's axes drawn inverted)
      |                 \
  -Im +                   *
      0        VCC       2VCC
             (Q point)

The dc load line is nearly vertical through VCE = VCC (the dc resistance of the primary is negligible). The Q point is at (VCC, 0) because class B is biased at cut-off (ICQ = 0). The ac load line of Q1 has slope −1/R'L and runs from the Q point up to the maximum current Im at VCE = 0 (full swing Vm = VCC). Q2's characteristic is identical and is drawn inverted, so the same straight line continues to (2VCC, −Im).

Proof of 78.5 % from the graph

From the graph the swing is limited by VCEmin = 0, so the peak voltage swing is Vm = VCC and the peak current is Im = VCC/R'L.

• Po = Vm Im/2 = VCC Im/2
• Each transistor supplies a half-sine pulse of peak Im. The average current per transistor is Im/π, so for the two together Idc = 2Im/π.
• Pi = VCC · Idc = 2 VCC Im/π

η = Po/Pi = (VCC Im/2) / (2 VCC Im/π) = π/4 = 0.785

ηmax = 78.5 %

Condition for maximum loss

Power lost in the two transistors: P2Q = Pi − Po = 2 VCC Vm/(π R'L) − Vm²/(2R'L).

For the maximum, dP2Q/dVm = 0:

2 VCC/(π R'L) − Vm/R'L = 0 → Vm = 2 VCC/π ≈ 0.636 VCC

At this condition, P2Q(max) = 2 VCC²/(π² R'L), so each transistor dissipates VCC²/(π² R'L). The loss is maximum at about 63.6 % of the full output swing, not at full output or at no signal. At this point η = (π/4)(2/π) = 50 %.