# How does the secondary coil affect the Q of the primary coil?

This is an LC tank circuit in which load resistor $$\R_L\$$ is link-coupled through the inductor:

My textbook stated that:

Link coupling is another way of coupling the signal to a small load resistance. Link coupling means using only a few turns on the secondary winding of an RF transformer. This light coupling ensures that the load resistance will not lower the Q of the tank circuit

What I don't understand is the bold context that implies that if I decrease turns of secondary winding, the Q of the primary inductor (or the LC tank as the whole) will increase.

My thoughts so far is that the reflected impedance on the primary when the secondary has fewer turns (step-down) is that:

$$R_{p} = (\frac{Np} {Ns}) ^2* R_L$$

$$\R_p\$$ will be very big and this will be in series to inductor of LC tank, but if the series resistance increases then won't Q decrease instead of increasing?

What point am I missing?

• Think in terms of the energy lost in the secondary load resistance, per cycle or half cycle.
– jonk
Aug 15 '20 at 7:29
• Think of the extreme case where \$R_L$was an open circuit i.e. infinite resistance. Would you expect an open secondary coil to reduce the Q (take away energy) of the primary coil ?
– AJN
Aug 15 '20 at 8:04
• @jonk AJN Ah yes, Power lost in secondary is also power lost in primary so if the turns of secondary is a lot lower(less secondary voltage) then I can make sure there will be less power lost in the primary, thus Q is higher. (is this right?). BTW does that mean the reflected impedance I mentioned above has no effect or whatever in the Q? Aug 15 '20 at 8:48
• @IwataniNaofumi good point. I suspect that the reflected impedance formula in the question is not applicable for loosely coupled inductors. Tightly coupled inductors in transformers use the formula you have supplied. A quick web search for loosely coupled inductor reflected impedance tells me that the formula is different. Link 1 pdf, slide 6
– AJN
Aug 15 '20 at 9:08
• continued Link 2 slide 22 Link 3 One of the links tell me that "Those circuits cannot be characterized by turns ratios; rather, they are characterized by self and mutual inductances".
– AJN
Aug 15 '20 at 9:12

## 1 Answer

This question is really about transformer theory and isn't "made special" because the primary may be part of a tuned circuit.

R_p will be very big and this will be in series to inductor of LC tank, but if series resistance increase then won't Q will decrease instead of increasing? What point am I missing?

The point you are missing is that the resistance "projected" from the secondary to the primary is not in series with the primary winding but, in parallel with it.

Imagine a regular (but ideal) transformer that steps-down 240 volts AC to 24 volts AC. Imagine also that the secondary load is a 24 ohm resistor. That would cause a secondary current of 1 amp RMS and, in the primary, will be a current of 0.1 amps RMS.

So, from the perspective of the AC applied to the primary, the load looks like 240 volts ÷ 100 mA = 2400 ohms. That 2400 ohms is 24 ohms x turns ratio squared (10²). In other words, the load is projected from the secondary to appear in parallel with the primary magnified by N².

The higher the step down ratio, the higher the projected load is by N².