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 (R_load) 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 (R_loss) 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 R_loss must be as high as possible.

The value of R_loss can be calculated with:

R_loss = 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 R_loss can be calculated to be about 1700 kΩ at 1008 kHz.

On other pages on this website, this same loss resistor is sometimes called: R_p, 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 R_load and R_loss parallel to it.
Because the losses in the coil and capacitor are represented in resistor R_loss, 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:

I_load = V / R_load

I_loss = V / R_loss

I_total = I_load + I_loss.

And also the following power values can be calculated:

P_load = V² / R_load

P_loss = V² / R_loss

P_total = P_load + P_loss

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

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

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 I_load, I_loss and I_total are plotted as function of the load resistor value.

As we see, I_total 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 R_loss in the circuit, I_loss is then equal to I_total at a value of: 0.2687 V / 1700 kΩ = 158 nA.

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

And the voltage across the load, is determined by the value of R_load and R_loss in parallel, multiplied by I_total.



Figure 3.
The LC circuit can be seen as a current source (in this case 158 nA) with R_loss (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 R_loss in series connected, as in the right circuit of figure 3.
In both cases load resistor R_load, gets exactly the same voltage, current and power.

As long as R_loss 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 . R_loss
Both circuits will behave the same at the load.
And as R_loss 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 R_loss to be in parallel, and that leads inevitably in finding the source to be a current source.


Figure 4: the power in R_load, in R_loss, 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: P_loss = P_total = (0.2687V)² / 1700 kΩ = 42.47 nW.

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