Transformer-Coupled Push-Pull: the classic power stage
Before complementary transistors were cheap, this is how every radio and record player made power.
A complementary push-pull stage needs a matched NPN and PNP pair, which for decades was expensive or simply unavailable at power levels. The alternative uses two identical transistors of the same polarity plus a centre-tapped transformer — and it turns out this arrangement has a genuinely elegant property: it cancels even harmonics automatically.
How it works
Input transformer (centre-tapped) splits the drive into
two signals 180° out of phase.
Positive half-cycle → transistor Q1 conducts
Negative half-cycle → transistor Q2 conducts
Each pushes current through HALF the output transformer
primary, in OPPOSITE directions. The secondary sums the two
half-cycles into a complete waveform.
"Push-pull" is literal: one device pushes current into the
primary while the other pulls it out.
Why even harmonics cancel
Each transistor's collector current can be written as a
Fourier series with a DC term and harmonics:
I₁ = a₀ + a₁sin θ + a₂sin 2θ + a₃sin 3θ + …
I₂ = a₀ − a₁sin θ + a₂sin 2θ − a₃sin 3θ + …
(Q2 is driven 180° out of phase)
The output transformer responds to the DIFFERENCE of the
two primary half-currents:
I₁ − I₂ = 2a₁sin θ + 0 + 2a₃sin 3θ + 0 + …
→ All EVEN harmonics (and the DC term) cancel exactly
→ All ODD harmonics remain, doubled in amplitude
Two consequences follow, and both matter. First, cancelling the DC component means the output transformer core carries no net DC magnetisation, so it doesn't saturate — allowing a much smaller, cheaper core than a single-ended stage would need. Second, the even harmonics that cause the harshest-sounding distortion disappear automatically, which is why push-pull stages sound cleaner than single-ended ones of the same design quality.
Worked numerical 1 — impedance matching
A push-pull stage must deliver power to an 8 Ω speaker, and each half of the primary needs to see a 500 Ω load. Find the turns ratio for the whole primary.
Impedance reflects as the SQUARE of the turns ratio:
Z_p/Z_s = (N_p/N_s)²
For a centre-tapped primary, each half sees R_L' = 500 Ω,
so the FULL primary (both halves, i.e. 2× the turns) sees:
Z_p(total) = 4 × 500 = 2000 Ω
(4× because impedance scales with turns SQUARED, and the
full primary has twice the turns of one half)
Turns ratio, full primary to secondary:
(N_p/N_s)² = 2000/8 = 250
N_p/N_s = √250 = 15.81
So a 15.81 : 1 total ratio, i.e. each half-primary to
secondary is 7.9 : 1.
Check: each half (7.9:1) reflects 8 × 7.9² = 499 Ω ✔
Worked numerical 2 — output power and efficiency
Each transistor in a Class B push-pull stage runs from V_CC = 30 V with a reflected load of 200 Ω per half. Find output power, DC input and efficiency.
Peak collector current per device:
I_p = V_CC/R_L' = 30/200 = 0.15 A
Output power (Class B push-pull):
P_out = V_CC²/(2R_L') = 900/400 = 2.25 W
Average DC current from the supply:
I_dc = 2 × I_p/π = 2 × 0.15/3.1416 = 0.0955 A
(the 2× because both halves draw in turn)
P_dc = V_CC × I_dc = 30 × 0.0955 = 2.865 W
Efficiency:
η = 2.25/2.865 = 0.785 = 78.5% ✔ (Class B ideal)
Heat per transistor:
(2.865 − 2.25)/2 = 0.31 W each — trivially small.
Worked numerical 3 — harmonic distortion
A single-ended stage produces 8% second harmonic and 3% third harmonic. What distortion remains if it is reconfigured as push-pull?
Single-ended total harmonic distortion:
THD = √(0.08² + 0.03²) = √(0.0064 + 0.0009)
= √0.0073 = 0.0854 = 8.54%
Push-pull: even harmonics cancel, so the 2nd (8%) vanishes.
The 3rd survives:
THD = √(0.03²) = 3%
Improvement: 8.54% → 3%, a reduction of 65%.
And since the second harmonic was the dominant term,
push-pull removed the largest single contributor without
any change to the transistors themselves — purely from the
circuit topology.
💡 Exam angle: the even-harmonic cancellation proof is the high-value answer here — write the two Fourier series, subtract, and show the even terms disappear. Then state both benefits (lower distortion AND no DC core saturation). For numericals, remember impedance reflects as the square of the turns ratio, and that a centre-tapped primary's full winding sees 4× what each half sees.
Syllabus points
Push-pull operation
Create a free account to tick topics off, take notes as you read, watch the video lessons and get a day-by-day study plan built around your exam date.