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Having read most of the online references to this circuit and performed some simulations, I still don't have a firm grasp on how this circuit functions.

Here's a schematic to refresh (possibly painful) memories: enter image description here

and here is one of the better intuitive explanations (in my opinion) of how the circuit works.

If you simulate this circuit and make a tight coupling between the primary and secondary of the "transformer", the circuit will not work as expected. If the coupling is tight, the top of the secondary will be in-phase with the primary, as you would expect. But, as the explanation above describes it, you want the secondary windings acting like inductors so that there is a 90 degree phase shift at the top and bottom of the secondary with respect to the centre tap (at the resonant frequency).

So, if my understanding is correct and the secondary winding halves are acting like inductors to the current arriving from the top of the primary tank (via C at RF) why does this circuit need transformer coupling at all?

Anybody care to confirm or reject my speculation in the comment section? Anybody - sob!

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  • \$\begingroup\$ Signal inversion? \$\endgroup\$ Sep 1, 2017 at 3:32
  • \$\begingroup\$ My feeling is that this has to do with coupled inductors and resonant induction. I think the two tanks need to be magnetically coupled enough to produce resonant ringing in the secondary, but not enough to overpower the RF current arriving to the centre tap. The resonance in the secondary ensures no phase shift occurs to the arriving CT primary signal - but, you do get the 90 degree shift at inductor ends. Am I on the right track?? \$\endgroup\$
    – Buck8pe
    Sep 1, 2017 at 6:43
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    \$\begingroup\$ I think I understand it now. With loose coupling the secondary tank is free to act more inductively and can develop a phase shift above and below resonance. That is, currents in the secondary can be phase shifted relative to voltages in the primary. This is in contrast to a typical transformer with tight coupling where that phase shift wouldn't occur (I'm speculating!). The primary voltage arriving at the centre tap then acts as a kind of reference against which the phase shifted voltages top and bottom are compared. \$\endgroup\$
    – Buck8pe
    Sep 1, 2017 at 9:04

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You have already described that you need loose coupling in order to get a 90 degree phase shift between the primary and the secondary. The phase shift will vary with the deviation of the primary frequency from the resonance frequency of the secondary. It is only 90 degrees right at resonance.

But you are wrongly assuming that the energy that makes the secondary tank circuit oscillate, is fed to it via Cc and the center tap of the secondary. This is untrue. It is the loose coupling that provides the energy transfer from primary to secondary. The optimal amount of coupling depends on the Q of the secondary tank. The higher the Q, the looser the coupling.

The center tap, and the voltage coupled into it via Cc, acts as a phase reference, as @Buck8pe has written. If you look at just the top half of the secondary circuit, consisting of the top half of the secondary winding between center and top, and the corresponding rectifier circuit, then you realize that the anode of the rectifier sees the sum of the primary voltage (via Cc), and half the secondary voltage (via the secondary winding). This means you are adding two sinusoidal voltages with the same frequency, but a phase difference between them. The resulting signal is again a sinusoid with said frequency, but with an amplitude that depends on the phase difference. Hence you have converted a phase difference into an amplitude variation, which you can detect using the rectifier/capacitor combination, as usual.

The bottom half does the same, except that the inverted secondary voltage is used, so you get the difference rather than the sum of the two sinusoids. Combining the two has the effect of linearizing the demodulation, hence reducing distortion.

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  • \$\begingroup\$ Super stuff! It's been a while since I pondered this one, but when I get back to it (and I will) this extra insight will be very helpful. \$\endgroup\$
    – Buck8pe
    Feb 9, 2019 at 10:29

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