What dimensions to choose when winding your own inductor

Question

Suppose you want to make a wilkinson 3-way splitter with lumped components

Given the following circuit, you need a 31.5nH inductor

The inductor can be wound various ways:

1. 3 wdg, 1 cm diameter, 2.36 cm long

2. 2 wdg, 1 cm diameter, 0.8 cm long

3. 3 wdg, 0.5 cm diameter, 0.48 cm long

These are examples, you can calculate more variations for this.

My question is, given the example of building a wilkinson 3-way splitter with lumped components as described in the link above, for 435 MHz from the same example, for 50 Watts transmit power: what way should the inductor be constructed ?

thanks

Answer 1

It boils down to practical issues — you want to make it as small as possible, but no smaller.

The power-handling requirements will dictate the wire diameter (to handle the current) and the end-to-end spacing (to handle the voltage). For an air-core coil, you also want the wire to be stiff enough to hold its shape well.

You will want to wind the coil on some sort of handy form (drill bits are useful for this), so this puts some limits on your choice of inside diameter. The one variable you have the most flexibility with is the length. Indeed, it is possible to tweak the final inductance value after construction by squeezing/stretching the length of the coil a bit.

Answer 2

For 435MHz, a better way to implement that pi-network is to use a microstrip line. A short line segment thinner than the 50ohm line functions as a series inductor. For the shunt capacitors, add short segments thicker than 50ohm. I suggest this not because of 31 nH but because of stray capacitances; 4.22pF shunt capacitors will be difficult to implement accurately. Now, if you consider this approach, you could divide this pi-network into two stages to achieve lower loss and broader bandwidth.