Impedance matching with a transformer: What turns ratio is required to match two 16 Ω loads connected in parallel (effective 8 Ω) to an output impedance of 22.2 Ω on the other side?

Introduction / Context:
Transformers are commonly used to match a source impedance to a load impedance to maximize power transfer or meet amplifier stability requirements. The impedance seen at one winding is the actual load scaled by the square of the turns ratio. Selecting the correct ratio prevents power loss, distortion, and overheating.

Given Data / Assumptions:

• Two 16 Ω loads in parallel → Z_load = 8 Ω.
• Desired matched impedance on the other side is 22.2 Ω.
• Ideal transformer approximation (no losses, perfect coupling).

Concept / Approach:
For an ideal transformer, the impedance referred from secondary to primary obeys Z_primary = (N_primary / N_secondary)^2 * Z_secondary. Let a = N_primary : N_secondary. We need the reflected impedance to equal 22.2 Ω when the real load is 8 Ω.

Step-by-Step Solution:
Compute equivalent load: Z_s = 16 ∥ 16 = (16*16)/(16+16) = 8 Ω.Set matching condition: Z_p = a^2 * Z_s = 22.2 Ω.Solve for a: a^2 = 22.2 / 8 = 2.775.a = sqrt(2.775) ≈ 1.666 → required turns ratio ≈ 1.67:1 (primary:secondary).

Verification / Alternative check:
Reflect 8 Ω through a = 1.67: Z_p ≈ (1.67)^2 * 8 ≈ 2.7889 * 8 ≈ 22.3 Ω, essentially the target 22.2 Ω within rounding. This confirms the selection.

Why Other Options Are Wrong:
1.38:1 and 0.72:1 yield reflected impedances significantly different from 22.2 Ω.

0.60:1 steps the impedance down instead of up.

2.22:1 would reflect 8 Ω to ≈ 39.5 Ω, far from the requirement.

Common Pitfalls:
Forgetting that impedance scales with the square of the turns ratio, not linearly. Also, be careful to combine parallel loads correctly before reflection.

Final Answer:
1.67:1