# coupling factor??.....

Discussion in 'Electronic Design' started by [email protected], Apr 27, 2005.

1. ### Guest

Hi,

I would like to know why coupling factor for IF transformer affects the
centre frequency. I hv a circuit in PSpice with two IF transformers...
and when I reduce the coupling factor I get the required centre
frequency which is 455KHz...

for both IF transformers

The inductor values are

L1 = 680uH
L2 = 97uH

and original coupling factor is 0.9

but after simulations the centre frequency obtained is 451KHz

when the coupling factor is reduced to 0.75 I get the required centre
frequency....

John

2. ### Tony WilliamsGuest

The k for an IF-type of coupled transformer is normally
taken as << 1. With a coupling factor of 0.9 the two
tuned circuits are probably going to merge into looking
like one circuit, which could have a resonant frequency
that is slightly offset by the leakage inductance... but
exactly how far offset depends on the definition of the
centre frequency. Possibilities could be, the frequency
at which the circuit looks resistive to the generator, or
where the maximum voltage occurs on one of the windings,
and so on.

What values of capacitor across the pri and sec?
Did you assign an estimated Q for each resonant circuit?
How is the circuit stimulated?
What did you measure in the circuit to decide on the
centre frequency?

3. ### John - kd5yiGuest

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

4. ### Jim BackusGuest

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.

5. ### Tony WilliamsGuest

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.

6. ### John WoodgateGuest

I read in sci.electronics.design that Tony Williams
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.

7. ### Tony WilliamsGuest

Sorry for the delay in replying to this post.

Ok, thank you. Presumably the Q to use is the single
value calc'd at Fres...... which reminded me of a remark
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

8. ### John WoodgateGuest

I read in sci.electronics.design that Tony Williams
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.

9. ### Tony WilliamsGuest

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

/|\
0.010812

11. ### John WoodgateGuest

I read in sci.electronics.design that Tony Williams
The textbooks generally assume that the two circuits have approximately
the same component values. Here the asymmetry is quite pronounced.

12. ### Tony WilliamsGuest

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

13. ### John WoodgateGuest

I read in sci.electronics.design that Tony Williams
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.

14. ### Fred AbseGuest

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