coupling factor??.....

The k for an IF-type of coupled transformer is normally taken as

The mutual inductance (M) between the primary and secondary is M = K * sqrt(L1 * L2). As you can see, the mutual inductance changes with the coefficient of coupling (K).

John

Have you checked the frequency response over a range of (say) +/- 25 kHz? You may find that the simulation is showing a double peaked response. Coupled tuned circuits used to be described as undercoupled, critically coupled and over coupled. When circuits are undercoupled, the centre frequency is correct but the output amplitude is reduced. With critical coupling the maximum output is obtained with a single resonant frequency. If overcoupled, the circuit has two peak responses, one below the nominal resonant frequency and the other above. The output at the nominal resonant frequency is reduced relative to the two peaks. Overcoupling can be useful in increasing the bandwidth of a coupled circuit.

Refer to one of the old textbooks for more explanation - I've just checked it in a book called "Radio Engineering Handbook" editied by K Henney and published in 1950. Terman would cover it but I don't have a copy.

In article , Jim Backus wrote: [snip]

There's a brief description in an old Samuel Seely textbook. However, at moderate Q's (20-ish) the k would be around 0.05 for optimum coupling, going even lower for higher Q's. This is at least a decade below the values of k that OP is talking about.

With a k near unity the circuit might be better imagined as a tuned power transformer, with the leakage inductance and secondary capacitor transformed over to the primary.

I read in sci.electronics.design that Tony Williams wrote (in ) about 'coupling factor??.....', on Sun, 1 May 2005:

I think caution is necessary. Some people use 'k' to mean what other people call 'Qk' (Q x k). Critical coupling is achieved at Qk = 1.

If the OP's values were not 'Qk' values, he would have seen response peaks very far away from his expected centre frequency, not just a small amount.

Ok, thank you. Presumably the Q to use is the single value calc'd at Fres...... which reminded me of a remark made by DNA the other day, about apparently rigorous final sums (for transformers) being based early assumptions that may not precisely carry through.

In this case the textbooks develop a sum for the final output voltage (against coupling), based on the assumption that Xc=Xl at Fres. But what happens to other things, like the primary current, at just off Fres (where Xc is not equal to Xl)?

So I LTspice'd it, and the .asc file is at the end of this post. The numbers were jigged slightly to ensure that Fres and Q were the same on both sides. Fres is 455249Hz, Q is 92.48, giving a critical K1 of 0.01812.

Sweep it, look at I[R2], and there is the single peak of critical coupling in the secondary current. Fiddle K1 up and down, and you can see the effect of over/under coupling.

But now look at I[R1] when K1 is 0.01812. At critical coupling the primary current is already double-peaking, and it is not until K1 is significantly reduced that it reduces to a single peak, at Fres.

Perhaps this is what the OP was seeing?

------------------------------------------------------------ Version 4 SHEET 1 880 680 WIRE -208 272 -208 224 WIRE -208 224 -64 224 WIRE 0 224 96 224 WIRE 96 304 96 336 WIRE 96 416 -208 416 WIRE -208 416 -208 352 WIRE -208 448 -208 416 WIRE 208 224 352 224 WIRE 352 224 352 288 WIRE 208 416 352 416 WIRE 352 416 352 352 WIRE 208 304 208 336 WIRE 352 416 352 448 FLAG -208 448 0 FLAG 352 448 0 SYMBOL ind2 80 208 R0 WINDOW 0 -64 45 Left 0 WINDOW 3 -66 81 Left 0 SYMATTR InstName L1 SYMATTR Value 679µ SYMATTR Type ind SYMBOL ind2 192 208 R0 SYMATTR InstName L2 SYMATTR Value 97µ SYMATTR Type ind SYMBOL res 80 320 R0 WINDOW 0 -58 39 Left 0 WINDOW 3 -59 75 Left 0 SYMATTR InstName R1 SYMATTR Value 21 SYMBOL res 192 320 R0 SYMATTR InstName R2 SYMATTR Value 3 SYMBOL cap -64 240 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName C1 SYMATTR Value 180p SYMBOL cap 336 288 R0 SYMATTR InstName C2 SYMATTR Value 1260p SYMBOL voltage -208 256 R0 WINDOW 3 37 87 Left 0 WINDOW 123 37 115 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value "" SYMATTR Value2 AC 1 TEXT -76 129 Left 0 !.ac lin 2000 450000 460000 TEXT 56 200 Left 0 !K1 L1 L2 0.010812

------------------------------------------------------

I read in sci.electronics.design that Tony Williams wrote (in ) about 'coupling factor??.....', on Tue, 3 May 2005:

Maybe. This double-peaking of the primary current is a little-known effect, mainly because it hardly ever matters. But if you thieve off a bit of energy from the primary side of a critically-coupled IFT for some purpose, you may run into the effect.

There are other apparent anomolies. For example, the theoretical value for critical coupling (K1 = 0.010812) appears too low. At this K1 there is still a single peak, shifted down slightly to 454.7KHz. K1 has to be upped by about 6% to get even the first hint of double peaking. There is another slight departure from the textbook sums when over-coupled and calculating f1 and f2.

These departures don't matter in practical terms, but it is an interesting observation on textbook sums, and the approximations that were made to reach a final answer.

/|\\ 0.010812

I read in sci.electronics.design that Tony Williams wrote (in ) about 'coupling factor??.....', on Tue, 3 May 2005:

The textbooks generally assume that the two circuits have approximately the same component values. Here the asymmetry is quite pronounced.

Try identical comps each side, it doesn't make any difference, nor should it afaics. The textbooks' initial assumptions only seem to require precisely equal L*C products and (preferably) similar Q's.

The OP does seem to have noticed an oddity in the behaviour of coupled circuits. As long as that primary current is double-peaking then the sec- -ondary passband is distorted slightly, resulting in a peak below Fres. At about 2/3 critical coupling (for my chosen Q) the primary double-peaking more or less disappears, and the secondary voltage peak is then at about Fres.

I read in sci.electronics.design that Tony Williams wrote (in ) about 'coupling factor??.....', on Wed, 4 May 2005:

But the textbooks assume *simplified* circuits, just as your model does. The inductors have shunt capacitance and shunt resistance as well as series resistance, and the capacitors have both series and shunt losses.

Considering the thousands of precision theoretical analyses and millions of actual measurements that have been made on tuned coupled circuits over the years, a claim to have found an asymmetry (other than one produced by varying either C or L to 'align' the circuits) has to be regarded as an extraordinary claim, requiring extraordinary proof.

If you actually twiddle trimmers or inductor cores to 'tune up', then you DO get asymmetries. This is exhaustively dealt with in two books by W Th H Hetterscheid, for example.

That's what I get, too. Double peaking was the first thing that came to mind here.

BTW, it might be a good idea to switch on "Greek mu conversion to lowercase u" for posted netlists. Some news readers (Pan for example) display mu correctly, but convert to a quoted string on copy-and-paste. Took me a while to figure out what the curly braces that appeared on the copy were all about.

Sorry about that...... I didn't notice that LTspice changed a 'u' typed-in on the keyboard into a mu.