**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.

R_{load}(kΩ) |
Received voltage V (mV RMS) |
I_{load}(nA) |
I_{loss}(nA) |
I_{total}(nA) |
P_{load}(nW) |
P_{loss}(nW) |
P_{total}(nW) |

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 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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