coils : turns shorted = quality ?
K7ITM wrote:
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
Good post!
Brian W
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