Experiments with LC circuits part 12
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As you can read in part 11 of this series, the voltage received by detector unit 1 was measured using this circuit:
Measuring the received voltage of an LC circuit.
No external antenna was used, only the receiver coil is used as a loop antenna.
Before starting the next measurements, I once again measured the received voltage when tuned to my local station "Groot Nieuws Radio" on 1008 kHz (100 kW 41 km distance).
Detector unit 1 received 380 mV, which is in line with measurement 91 to 94 (see part 11)
The measured voltage is the peak levels off the (un-modulated) carrier.
The next thing I wanted to measure was the power which my receiver was
receiving from the transmitter.
For doing this I made a test setup, which to my surprise gave strange and useless results.
But this made me aware of some very interesting phenomenon, of which I had never thought before.
The idea of the power measurement was as follows:
- Disconnect the diode from the LC circuit.
- Load the LC circuit with some resistor value.
- Tune the LC circuit to a strong local station.
- Measure the received voltage across the LC circuit with a high impedance measuring device.
- Calculate the received power in the load resistor.
- Repeat the measurement with other load resistor values.
The following test setup is used:
10 Resistors of 330 kΩ are in series connected to the LC circuit of my detector unit 1
With a crocodile clip, a junction between two resistors can be connected to ground.
Which means, we can adjust the load resistance of the LC circuit in 330 kΩ steps from 330kΩ (as drawn in the circuit with the solid line), to 3300 kΩ (as drawn with the dashed line).
The wires of the resistors are cut off at about 1 cm, so the string of 10 resistors was kept compact, with just enough space between two resistors to place the crocodile clip.
The voltage across the LC circuit is measured with a high impedance amplifier (version 2), followed by this diode detector .
The receiver is again tuned to the same local station on 1008 kHz.
Here are the results of the received voltages:
LC circuit (mV)
Measurement 95: LC circuit loaded with string of 10 resistors of 330 kΩ.
The strange thing about this results is, that the voltage across the LC circuit
hardly increases when the load resistor is more the 3x 330 kΩ.
Even if no point of the 10 resistor string is connected to ground, I only measure 158 mV.
And with the resistors removed (as in figure 1) I measured 380 mV, so much more.
What is the difference between these two situations?
To find out what is happening, I removed the string of 10 resistors, and
connected one single resistor to the LC circuit.
So with one lead to the top of the LC circuit, and the other lead of the resistor not connected to anything, but just sticking in the air.
The leads of this resistor are kept at their original length of 25 mm.
Figure 3: a resistor connected with one lead to the LC circuit.
The next table shows the received voltage by the LC circuit as function of resistor value R.
|Resistor value R
LC circuit (mV)
Measurement 96: Received voltage across the LC circuit with a resistor
connected with one side to the top of the circuit.
Figure 4: this diagram shows the results of measurement 96.
We see, the resistor value connected to the top of the LC circuit has
influence on the received voltage.
Despite the fact that the other side of the resistor is not connected.
With a resistor value of about 300 kΩ, the effect is the strongest.
Is this strange or not?
After some thinking I became aware that the resistor lead which is not
connected has in fact some capacitance to ground.
Although this capacitance is maybe very small, it should at least be some capacitance higher then zero.
So the LC circuit is loaded with a resistor (R), in series with a (very) small capacitor.
The actual capacitance value is probably depending on the length of the resistor lead, and it's distance to any grounded point.
Figure 5: the not connected resistor lead has some capacitance to ground.
But could this capacitance cause this effect?
To find out, I used the complex impedance calculator which I made on my other website.
In this calculator you can enter a resistor value in series with a capacitor value.
And then calculate at some frequency the equivalent parallel circuit.
The parallel circuit is a resistor in parallel with a capacitor.
The parallel capacitor is just adding to the value of the tuning capacitor, and will cause a slight detuning of the LC circuit.
This is no problem, because you can turn the tuning capacitor to bring the LC circuit to resonance again.
But the parallel resistor is a worse thing, this appears to be in parallel to the LC circuit, and will reduce circuit Q factor.
And a lower Q means, higher losses, and lower received voltage.
I let the calculator calculate the parallel resistance at 1000 kHz for series capacitances in the range 0.1 to 0.6 pF.
The next diagram shows the results.
Figure 6: parallel resistance calculated as function of series resistance and series capacitance at 1000 kHz.
We see the parallel resistance shows a dip at some combination of series
resistance and series capacitance.
To get the dip at 300 kΩ series resistance (just like in figure 4), we need a series capacitor of about 0.5 or 0.6 pF.
The calculated parallel resistance is then about 0.6 MΩ.
The LC circuit of detector unit 1 has of it's own a parallel resistance of about 1.7 MΩ, and placing 0.6 MΩ parallel to this would mean a lot of extra losses, and much lower circuit Q and receiver sensitivity.
The string of resistors in figure 2 is not useable for loading the LC circuit
in this measurement.
The capacitance from the resistors to ground causes too much extra signal loss.
We see it is a bad idea to connect a resistor to the top of the LC circuit, while the other lead of the resistor is not connected.
Maybe you (just like me) did this sometimes to create a point for easily hook on some measuring device to the LC circuit.
If you do so, the resistor must be of low resistance, let's say below 100 Ω.
Or better; use just a piece of wire for this.
Now instead of using a string of resistors (figure 2), I now placed a single
resistor parallel to the LC circuit.
Figure 7: LC circuit loaded with one single resistor (R) across it.
For several values of resistor R, the received voltage is measured
Measurement 97: received voltage (Volt peak) of detector unit 1 at 1008 kHz.
Figure 8: received voltage of detector unit 1
The red line shows the results of measurement 97, so with each time a single resistor as load.
The blue line shows the results of measurement 95, where the load consists of a string of 330 kΩ resistors.
For comparison, the voltage without load resistor connected is also shown (0.38 Volt).
We see when the load is a single resistor, much
more voltage is received then with the string of resistors.
Now we can calculate the power in the load resistor.
First convert the Volt peak value to Volt RMS by multiplying by 0.707 (= 1/√2).
Then calculate power with:
Power = (Volt RMS)² / load resistor value.
|Power in load
This table shows the calculated power values from measurement 97.
These power values are shown in the next diagram.
Figure 9: received power of detector unit 1, tuned to a strong local station at 1008 kHz.
Once again: there is no external antenna connected, only the receiver coil is receiving the power.
The maximum received power occurs with a load of about 1800 kΩ, this is very close to the calculated impedance of the unloaded LC circuit which is about 1700 kΩ.
This is in line with the theory, which says the power in a load resistor is maximum when the load resistor value is equal to the source resistor value.
This is different from the situation in a normal crystal receiver, where you have a diode between the LC circuit and the load.
A diode is a non linear device, and then diode resistance and load resistor should be 3 times the impedance of the LC circuit for maximum power in the load.
But in this measurement I am only interested in how much power the receiver can deliver to a resistor.
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