## The 4:1 half-wave coax balun

**Balun - dimension shown for ca. 145 MHz (velocity factor of cable ca. 0.7)**
**The half-wave (4 : 1) balun**

The balun has two outputs; the first (A) connected directly to the inner of the main length of coax while other (B) is connected to the inner after a half-wave length of feeder. All the screens are connected together at one point. If we assume the half-wave section is near resonant then the impedance will be mostly resistive. We can consider the power into the balun as composed of V volts and I amps (where P = V x I). Because of the delay brought about by the half-wave section of cable the second output connection will be 180 degrees out of phase with the first. Thus if the first output connection has + V volts then the other will have - V volts. The voltage difference on the output will thus be V(b) = V - - V = 2V. The balun therefore acts as a voltage doubler. As the power through the system must be constant (ignoring losses) the output current must drop to a half, I(b) = I/2. If resonant we can estimate using Ohms the apparent change in resistance through the balun. The impedance change will be R(b) = V(b) / I(b) = (2V) / (I/2) = 4R and so we can account for the 4:1 impedance transformation for this type of balun.

**THE CREATIVE SCIENCE CENTRE**

Dr Jonathan Hare, Physics Dept. The University of Sussex

Brighton, East Sussex. BN1 9QJ.

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