On Sep 27, 10:28*pm, K7ITM wrote:
On Sep 26, 8:25*pm, brian whatcott wrote:



rtfm wrote:
On 2009-09-24, Tomylavitesse wrote:
HI,

Often I can see coils where some turns are shorted in order to modify there
value.
I wonder whether these shorted turns would not be seen as a short cicuit and
could decrease the quality of that coil.
How can I have an idea of the lost of quality without complex equipment ?
I plan to build this kind of coil for a shortened antenna...

The circuit to shorten turns in such environments has to be very low
loss. Otherwise you will have lots of power lost in this short
circiuted parts of the coil. And in PA / antenna environments you
always have power. This is why switches are always "heavy duty" in PA
and antenna switching units.

If you do so, the current in this shortened turns is a complex
current. And as you know, complex currents do not consume power. Only
the real parts of current do.

This is totally different from the situation e.g. in a power
transformator. If you would shorten turns there, you would have power
loss, because the resistant of the turns in power transformators has a
much higher real part compared with rf coils in PA / antenna units.

OK?

This seemed like an insightful response - but not one with universal appeal.

Let me try embroidering on this theme a little more....

The leakage path for current in an insulated wire is strikingly
different from the leakage path in a magnetic conductor - usually called
a core or stamping.
The leakage resistance can easily be 1000 megohms compared to the
conductor's resistance of (say) 1 ohm. A ratio of a billion to one.
The leakage path for cores and stampings is lucky to be a thousand times
more "resistive" than the path through the core - if it's an iron
stamping, an iron dust core, or a ferrite core. A ratio of a thousand to
one.
For air cored coils, the leakage path is lower still, so that the
magnetic path does not couple all turns together at the best of times.

If an end turn or two is shorted, the reactive current in the shorted
turn pinches off most of the magnetic coupling from the remaining coil
altogether, so the effect is not as dramatic as we might expect.
Waddaya think of that?

Brian W

It's actually not too difficult to come up with a reasonably accurate
model that you can then analyze...for example, you can put it into
Spice and look at the losses, the lowering of Q, etc.

What I've done to build the model is to consider that the shorted
turns are one coil and the remainder of the turns are another. *Before
those turns are shorted, each of those two coils (that happen to form
one coil) has a self-inductance, and when they are put end-to-end to
form the one coil, they then also have mutual inductance. *You can use
your favorite coil inductance calculator to find the inductance of
each of the two individual coils, and the net single coil composed of
those two. *Those three inductances let you calculate the mutual
inductance (or the coupling coefficient), and that's the core of the
model. *You can add in the RF resistance of each coil: *there are
various ways to get a good estimate of the Qu, and from that you can
get the RF resistance at the operating frequency.

If this is too muddy, I could offer a specific example...

Cheers,
Tom

So, just for fun I ran an analysis this way...
I asked the Hamwaves on-line inductance calculator about three coils,
all 2.5mm wire, 5mm pitch, 75mm mean coil diameter, default copper
wire. You can put effective shunt capacitances into the model too,
but I didn't in this case. The test frequency is 10.0MHz. These are
the "lumped equivalent" values.

L1: 13 turns 65mm long -- 9.272uH, 0.491 ohms, Qu=1187
L2: 17 turns 85mm long -- 13.191uH, 0.646 ohms, Qu=1283
L3 (L1 and L2 end to end): 30t, 150mm -- 26.600uH, 0.880 ohms,
Qu=1899
(My impression is that the RF resistance and Qu calculation yields Qu
that's a bit too high, but that's what the calculator says...)

From these, I calculate the mutual inductance. L1+L2+2*M = L3, so
M=2.0685uH. Coupling coefficient equals M/sqrt(L1*L2), or 0.187.

I put L1 and L2 and their respective RF resistances in series in
Spice, with a resistance representing a shorting switch across L2.
Note that this model, with the switch open, will give an RF resistance
simply equal to the sum of the RF resistances of the two coils, which
isn't the same as the Hamwaves calculator gives for L3.

The results are kind of interesting. It doesn't take much shunt
resistance in the switch to lower the Q (increase the net RF
resistance). You want to be sure your switch is really OPEN when the
coil isn't shorted. You can actually stand quite a bit of resistance
(a good fraction of an ohm) when it's shorted without really bad
effects. Here are the results I got, again at 10.0MHz:

Effective net
R(switch) L(net) series resistance
ohms uH ohms Qu
-------- -------- -------- --------
1e9 26.598 1.138 1469
1e8 26.598 1.146 1458
1e7 26.598 1.229 1360
1e6 26.598 2.056 813
1e5 26.597 10.32 162
1e4 26.476 92.4 18.0
1e3 19.405 545 2.24
1e2 9.201 132.4 4.37
10 8.950 13.886 40.5
1.0 8.947 1.845 305
0.5 8.948 1.175 478
0.2 8.947 0.774 726
0.1 8.947 0.640 878
0.05 8.947 0.573 981
0.02 8.947 0.533 1054
0.01 8.947 0.520 1081

Cheers,
Tom