## Transmitter application

Link coupling was used widely in times past for coupling a valve RF Power Amplifier to the line. The practice has been replaced widely by use of Pi couplers or Pi-L couplers.

There has been revival of link coupling, more in discussion than practice, and the theory of operation of a link coupler is topical.

Hams often (if not usually) explain a link coupler as transforming impedance by a ratio given by the square of the turns ratio. This explanation is not valid for most constructions where there is significant flux leakage.

The circuit can be explained by quite basic linear circuit theory.

Consider a link coupling circuit for 3.6MHz for a transmitter which requires a 2kΩ resonant load impedance, and target operating Q of 10.

The capacitance to tune the tank will be around 400pF for the target operating Q, so a tank inductor of 5µH is chosen, and a link coil of 1µH. The 5µH tank inductor could be 57mm diameter, using 9 turns of 1mm wire with a winding pitch of 3mm. The 1µH is 3 turns of the same diameter, wire and pitch.

The link coil is adjusted in proximity to the tank coil to obtain 35% flux coupling, giving a mutual inductance of 0.35*(5*1)^0.5=0.8µH. A tee equivalent circuit of the coupled coils is 4.2µH, 0.8µH 0.2µH, see the figure above.

Now square of the turns ratio is (9/3)^2=9, but the impedance transformation ratio is actually 2000/50=40. This example illustrates the deficiency of the ‘square of the turns ratio explanation’, it is not valid for transformers with significant flux leakage, and not applicable to typical tuned tank circuits.

Link coupling was often used in receiver input circuits. Not all receiver input circuits are or should be designed for maximum power transfer.

Fig 2 above shows the part of the schematic of the Mighty Midget receiver from QST. The tuned input circuit consists of two coupled inductors L1 and L2, and the tuning cap C1. This circuit ought to have been designed for maximum power transfer with a 50Ω source, ie the input impedance should look like 50+j0Ω. It may not have been done, but this article explains a method of achieving that goal.

Again, the solution is found through quite basic linear circuit theory.

Given the tuning capacitor used, a 356pF variable to cover 3.5-7.3MHz, a practical value for L2 is 7µH.

The impedance transformation design depends on the load, in this case near infinite impedance of the grid circuit save some capacitance, and the losses in the input coils. Lets assume the coils both have a Q of 200 for the purposes of this exercise. The function of the input circuit is to provide some selectivity, whilst capturing the maximum power available from a 50Ω source, most of which in this case will be lost in coil resistance, but at the same time delivering the highest signal voltage to the grid of the input tube.

Fig 3 above shows on the left hand side, two coupled coils L1 and L2 with some mutual inductance M. On the right hand side, it shows a Tee equivalent circuit of the coupled coils.

In this application with L2=7µH, it is the combination of L1 and M that sets the impedance transformation. Taking L1 to be one fifth of L2, or 1.4µH, and assuming coil Q to be 200 at 3.6MHz, coupling such that M=0.33µH provides a solution. Different combinations of L1 and M would work, those chosen for this explanation are practical.

Fig 4 above shows the Tee equivalent circuit without showing the coil resistances. If this network is solved (with coil resistances), the input impedance is 50+j0Ω. This network was designed by finding the value of M and C given L1, L2 and Q that gave an input impedance of 50+j0Ω, M implies the flux coupling factor k is 0.104 which is practical for a coil assembly of this type.

In practice, the coupling between the coils, and / or the inductance of L1 would be varied until maximum output was obtained at resonance from a source with a Thevenin equivalent source impedance of 50+j0Ω. The turns ratio does not imply the impedance transformation ratio.