Experiments with LC circuits   part 14

 << to part 13    to part 15 >>
Back to the index

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 m 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. 0.1185 = 0.01103 m.

For set 3 and set 10, the area is simply length x height of the coil.
 

Receiver Q factor
at 1008 kHz
Coil area
(m)
Max. power
(nW)
Max. power / coil area
(nW / m)
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.


 << to part 13    to part 15 >>
Back to the index