Transformer-based (coupled coils) output circuit (revised)

Hi

I spent some time last evening thinking about the output circuit in the transmitter that Carl launched a recent thread on.

All of my prior thinking was based on pi-network output circuits... and a lot of my intuition was, therefore, not applicable to the coupled-coil (transformer) design.

I have revised this posting several times... because the calculations are a little trickier than I thought they would be.

Attached is the equivalent circuit of a coupled-coil (transformer) output circuit... showing only those portions of the coupled coil model that are (I believe) relevant in this application. [I.e., I have left out parasitic capacitances... which, I believe, can be ignored when properly designed air core inductors are used]

In attachment 1, the number "k" is the mutual coupling between the coils... which is approximately (in this case) equal to a cross-sectional area of the smaller diameter coil / the cross sectional area of the larger diameter coil. Since the diameter of L2 is 0.75", and the diameter of L3 is 1", the number "k" is 0.75 x 0.75 / (1 x 1) = .5625.

In attachment 2, I have redrawn the the equivalent circuit by moving the output-side components to the input side of the transformer.

The equivalent circuit in attachment 2 looks like the equivalent circuit of a pi-network.

It then follows that

A) Using the pi network analogy... C3 acts as a tuning capacitor, in combination with p x C4

Note that, in the QST article that Carl was referring to...

The coupling factor, k, isn't all that small... since the inner coil has a diameter of 0.75" and the outer coil has a diameter of 1".

For the 80 meter case (using the table in the QST article), the turns ratio is: 80 turns/3 turns. It follows that p = 711

B) If the external load resistor, Rext, is 50 ohms, then: p x 50 ohms = (for the 80 meter case) 35,556 ohms.

Suppose we use C4 to resonate [(1-k) x L3] + Lext...

Then the circuit simplifies to what is shown in attachment 3

What is the associated rf load on the tube?

To calculate that one would have to redraw attachment 3 with the series equivalent circuit of: the combination of: p x Rext in parallel with k x L2 substituted for the existing parallel combination

The net result is going to be C3 in parallel with the series combination of:

A) An inductance of: [(1-k) x L2] + [k x L2 / g]
B) A resistance of p x Rext / h

where:

g = [1 + (2 x pi x f x k x L2)(2 x pi x f x k x L2)/(p x Rext)(p x Rext)]

f = the frequency of operation

h= [1 + (p x Rext)(p x Rext)/(2 x pi x f x k x L2)(2 x pi x f x k x L2)]

Note that in the present example (on 80 meters, with Rext = 50 ohms) h is a number that is significantly larger than 1, and g is a number that is only slightly larger than 1.

Thus, the output circuit looks like a pi network with C3 as the tuning capacitor, L2 as the inductor [i.e. (1-k) x L2 + (k x L2 /g) is approximately equal to L2], and a load resistance of Rext x (p/h)

Therefore, what is very important is the value of p/h

Since h>>1, we can approximate h as

h~ (p x Rext)(p x Rext)/[(2 x pi x f x k x L2)(2 x pi x f x k x L2)]

Therefore p/h is approximately:
(1) p/h ~ [(2 x pi x f x k x L2)/ Rext)] x [(2 x pi x f x k x L2)/(Rext)] / p

[Remember... p is the square of the turns ratio: n2/n3]

Ok...

Now, from pi network theory ,the r.f. load (at resonance) on the tube increases as the value the load resistor (in this case Rext x p/h] decreases.

In this case, making p larger (a larger turns ratio) makes the value of the load resistor smaller... because of equation (1)... and this, in turn, increases the r.f. load (at resonance) on the tube! [Very confusing!]

So... the "bottom line" is that if we make the turns ratio (p) too large, we will place too high a load impedance on the tube (at resonance).

The solution, in this case, is to increase the number of turns on the output winding in order to reduce the turns ratio!

Best regards
Stu