Experiments with LC circuits   part 13

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In part 12 of this series, we determined the power which my detector unit 1 delivers to various values of load resistors.
The detector unit was tuned to a strong local station on 1008 kHz.
The voltage across the LC circuit was measured (measurement 97), and the power to each load resistor (Rload) calculated.

Now we take a closer look at the results of that measurement, and do some extra calculations.
We are adding another resistor to the story, which is the loss resistor (Rloss) of the LC circuit.
This is not a real resistor, but an imaginary one, that represents the losses in the coil (L) and tuning capacitor (C).
The loss resistor is assumed to be in parallel with the coil and capacitor.
As we want low losses in the LC circuit, the value of Rloss must be as high as possible.

The value of Rloss can be calculated with:

Rloss = 2. pi. f.L.Q

f = the frequency in Hz, in this case 1008000 Hz.
L = the inductance of the coil, for detector unit 1 this is 230 uH, or 0.00023 H.
Q = the unloaded Q factor of the LC circuit
For detector unit 1, we have at 1008 kHz a Q factor of about 1170.

Now the value of Rloss can be calculated to be about 1700 kΩ at 1008 kHz.

On other pages on this website, this same loss resistor is sometimes called: Rp, referring to the internal parallel resistance of the LC circuit.
But in both cases it is the same thing.


Figure 1.
In this circuit diagram, we have the LC circuit, with the resistors Rload and Rloss parallel to it.
Because the losses in the coil and capacitor are represented in resistor Rloss, we can consider the coil and capacitor drawn in this circuit as being lossless.

When we know the voltage across the LC circuit (V), the following currents can be calculated:

Iload = V / Rload

Iloss = V / Rloss

Itotal = Iload + Iloss.

And also the following power values can be calculated:

Pload = V / Rload

Ploss =
V / Rloss

Ptotal = Pload + Ploss

Pload is the power delivered to the load, we already determined this value in
part 12, but the next table is showing the values again.
Ploss is the part of the received power from the transmitter, that is in the LC circuit converted into heat.
Ptotal is the sum of both powers

The next table shows the calculated current (I) and power (P) values.

Received voltage V
(mV RMS)
330 44 132 26 158 5.75 1.12 6.87
680 77 114 46 160 8.82 3.53 12.35
1000 101 101 59 160 10.16 5.98 16.14
1200 112 93 66 159 10.37 7.32 17.69
1500 127 85 75 160 10.84 9.56 20.40
1800 141 78 83 161 10.99 11.64 22.63
2200 154 70 90 160 10.74 13.90 24.64
3300 174 53 103 156 9.22 17.89 27.11
4700 190 40 112 152 7.70 21.29 28.99
10000 222 22 131 153 4.93 28.98 33.91
infinite 268.7 0 158 158 0 42.47 42.47

The values in this table are based on the measured voltage values in measurement 97.
In this table the voltage is given in mV RMS (and not mV peak).

Figure 2.
In this diagram the calculated values for Iload, Iloss and Itotal are plotted as function of the load resistor value.

As we see, Itotal is quite constant around 158 nA, regardless of the value of the load resistor.
When we disconnect the load resistor at all, the voltage V is:  268.7 mV RMS (this is 380 mV peak as measured in part 12 ).
Because we then still have Rloss in the circuit, Iloss is then equal to Itotal at a value of: 0.2687 V / 1700 kΩ = 158 nA.

I think we can conclude that in this test the LC circuit (without Rloss) acts as a current source, with value Itotal determined by the strength of the received station.

And the voltage across the load, is determined by the value of
Rload and Rloss in parallel, multiplied by Itotal.

Figure 3.
The LC circuit can be seen as a current source (in this case 158 nA) with Rloss (1700 kΩ) parallel to it, as in the left circuit in figure 3.
But you can also see the LC circuit as a voltage source (in this case 0.2687 Volt), with Rloss in series connected, as in the right circuit of figure 3.
In both cases load resistor
Rload, gets exactly the same voltage, current and power.

As long as Rloss is higher then zero and lower then infinite Ohm, you can replace a current source with a resistor in parallel, by a voltage source with the same resistor in series.
The value of the voltage must then be: V = I . Rloss
Both circuits will behave the same at the load.
And as Rloss is not a real resistor, you can either assume it to be in parallel (to a current source), or in series (with a voltage source).

At the beginning of this page, I assumed Rloss to be in parallel, and that leads inevitably in finding the source to be a current source.

Figure 4: the power in Rload, in Rloss, and the sum of these two.

The power in the load resistor is maximum when the load resistor is equal in value to the loss resistor (1700 kΩ).
The maximum power to the load is about 11 nW.

With the load resistor disconnected: Ploss = Ptotal = (0.2687V) / 1700 kΩ = 42.47 nW.

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