How can I create my JFET Hartley oscillator's transformer in real life?

Question

I designed this FET Hartley circuit. How I can make the transformer in the circuit in real life? I have designed the PCB in Altium, but I don't know how to deal with the transformer. What should I do about this problem?

Answer 1

I don't know how to deal with the transformer. What should I do about this problem?

You need to wind this yourself because you won't find one that is off-the-shelf. Here's a close-up view of your transformer: -

I have put red lines around L2 and L3 because these set the scene on how you build the transformer yourself. They both have values of 100 nH and, to wind 100 nH with a single turn requires a core with very low permeability. So, you need to find a core that has an effective permeability that produces 100 nH with a single turn.

This means it will have an \$A_L\$ factor of 100 nH for one turn. If you are not familiar with the term \$A_L\$ you probably need to read a bit about it. But, it just tells you that 1 turn will produce an inductance of x. Because multiple turns sharing a core have shared flux, it means that two turns would produce an inductance of 4x (not 2x). 3 turns produce 9x inductance etc..

• You should also be able to see that a core with an \$A_L\$ value of 25 nH per single turn, can only produce 100 nH with 2 turns.
• So, your best bet is to find a core with an \$A_L\$ of 100 nH for a single turn.
• It can be a gapped core of course; gapping can significantly reduce the effective permeability of a core.

Then, for the 27 μH winding, 5 16 turns would produce 25 25.6 μH so, I suggest you recheck your simulation (knowing what I've just said) and, of course, good luck. Winding your own transformer isn't as hard as it appears to be but, do make sure you observe the winding directions as depicted by the dots.

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A few more words on inductance factor \$A_L\$.

It derives from the equation for the inductance of a coil thus: -

$$L = \dfrac{\mu N^2 A}{\ell}$$

Where,

• \$\mu\$ is the permeability of the core (\$\mu_0\cdot \mu_r = 4\pi\times 10^{-7}\cdot \mu_r\$)
• \$A\$ is the cross-sectional area of the core
• \$\ell\$ is the effective length of the core
• \$N\$ is the number of turns

So, \$A_L\$ equals \$\dfrac{\mu A}{\ell}\$ and, is specified by all core suppliers for any given ferrite core.

For a toroid core it translates to this: -

Images above stolen from HyperPhysics.