Experiments with LC circuits part 14
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In this article, we will compare the output power of three receivers with
loop antenna.
The LC circuit is loaded with resistors of various values.
There is no diode connected to the LC circuit.
The first receiver under test is
detector unit 1 .
Figure 1: detector unit 1
In part 12 of this series, you can read
how the measurement of power goes.
The measurement is done with the receiver tuned to a local station on 1008 kHz.
The result is already shown on page 12, but here it is again.
Figure 2: output power of detector unit 1.
The maximum power delivered to a resistor is 11 nW.
The Q factor of detector unit 1 has the following values:
Frequency (kHz) |
Q factor |
600 | 1276 |
900 | 1204 |
1200 | 1111 |
1500 | 912 |
At the test frequency 1008 kHz, the Q factor is about 1170.
The next receiver is set10
The loop (coil) measures 38 x 26 cm.
In this receiver, not only the diode is removed.
But also the transformer was not in the receiver, removing the transformer gives
a higher Q factor.
Figure 3: set 10
Figure 4: measurement 98, output power of set 10.
The maximum output power is 40 nW at 470 kΩ load.
The Q factor of set 10 is:
Frequency (kHz) |
Q factor |
600 | 566 |
900 | 474 |
1200 | 359 |
1500 | 266 |
At 1008 kHz, the Q factor of set 10 is about 430.
On the page describing set10, you will find lower values mentioned for the Q
factor (in measurement 3 there).
Now the Q factor is higher, because I now use better equipment, which gives
lower losses during measuring.
The third receiver in this test is set3
The loop size is 88 x 60 cm.
Figure 5: set 3
The coil has 8 turns.
The load can be connected across these full 8 turns (tap is at 100 %).
Or across 6 turns (tap is at 75 %)
Or across 4 turns (tap is at 50 %)
Or across 2 turns (tap is at 25 %)
In this test, I measured the output power at the 100% and the 50% tap.
Figure 6: measurement 99: output power of set 3.
If you connect the load across the full coil (100%), or at the tap at 50%, the
maximum power is in both cases the same (172 nW).
But the maximum power occurs at the 100% tap when the load is 100 kΩ.
And at the 50% tap the maximum power occurs when the load is about 30 kΩ.
In a real crystal receiver, you need a diode between the LC circuit and the
load.
Diodes work better (with less power loss) at higher input voltages, so the diode
at 100 % of the coil is preferable.
Maximum demodulated audio power behind the diode occurs when both diode and
load have an impedance of 3 times the impedance of the LC circuit.
So in this case, for making a sensitive receiver with this LC circuit, we need a
diode and audio transformer with 300 kΩ impedance, and connecting the diode to
the top of the LC circuit.
But for now, we go back to the measurement without any diode in the circuit.
And then there is no difference in output power for the 100% and 50% tap.
The Q factor for set 3 is:
Frequency (kHz) |
Q factor |
600 | 128 |
900 | 146 |
1200 | 144 |
1500 | 124 |
Q factor of set 3.
The Q factor of set 3 is 145 at 1008 kHz.
Set 3 receives the most power, but has also the largest coil.
Let's now see how much power each receiver gives per m²
coil area.
For this, we need for every receiver the area of the coil.
For the spider web coil in detector unit 1, the area is based on the average coil
diameter.
The inside diameter of the coil is 0.062 m, the outside is 0.175 m.
The average diameter is 0.1185 m.
And the area is: 1/4. pi. 0.1185² = 0.01103 m².
For set 3 and set 10, the area is simply length x height of the coil.
Receiver | Q factor at 1008 kHz |
Coil area (m²) |
Max. power (nW) |
Max. power / coil area (nW / m²) |
Detector unit 1 | 1170 | 0.01103 | 11 | 997 |
Set 10 | 430 | 0.0988 | 40 | 405 |
Set 3 | 145 | 0.528 | 172 | 326 |
Now we find Set 3 to have the lowest power per square meter coil area.
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