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

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?

• Well, as far as I can see, it's you who designed this and so maybe you can explain why you chose to use this type of transformer? Your title should not be all-caps. It's seen by some as rude. Mar 26 at 16:30
• Hi, I apologise for all-caps. I have designed the circuit with hepl of books and this website. I saw some FET hartley circuit and that circuit contains this transformer. Mar 26 at 16:52
• I have edited the question and added the orijinal circuit. Mar 26 at 16:54
• It's running about 7 MHz as far as I can tell so, is there any reason to use such an ornate sinewave generator feeding a fast comparator when you could make a 7 MHz oscillator just from the comparator? Mar 26 at 17:36
• Well, a Colpitts oscillator can use a JFET and it only has one inductor. Like this: electronics.stackexchange.com/questions/419750/… Mar 27 at 9:21

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.

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.

• For an un-gapped toroid core, I'd suggest FT37-61 toroid. One turn produces less than 100nH, but a reasonable # of turns for 27uH...(about 22 turns). For a toroid like this, fractional turns are a bit of a problem Apr 8 at 15:56
• 2 turns on an FT37-61 toroid gives 220 nH so, it's reasonable. The "61" part in the toroid part number made me think of Fair-rite because it reminded me of their material classifications and, indeed that is the case. Link: fair-rite.com/product-category/inductive-components/toroids and, as it happens they do make some toroids with an $A_L$ of 25nH/1_turn in material "67". Apr 8 at 16:16
• Nice answer! I'd add maybe a small word of advice about real circuits involving transistors that rely on small capacitances (C5: 0.5 pF (to 10 pF?)): You need to really wind with a big separation between windings if you don't want to incur parasitics between the windings that are larger than these really small capacitances. Also, C5 is a 0.1 F capacitor. Such capacitances only exist in electrochemical double-layer capacitors ("supercaps"), and those won't work at these frequencies, at all. Apr 8 at 16:18
• Link to Fair-rite's equivalent toroid type 5961000201: fair-rite.com/product/toroids-5961000201 but, you'll need to check that it will take the windings in the space available. Apr 8 at 16:23
• @MarcusMüller I believe you have just done that!!! I expect that 100 nF will suffice for the 100 farad capacitor. However, that would series resonate around 1.5 MHz and might be a tad to near to the 7 MHz output to be acceptable. Apr 8 at 16:25