"Radio" Magazine, Issue of January 1997.

In these latter days the interest in powering up a radio with "free energy" of powerful broadcasting radio stations increasing rapidly. In some magazines there are reports about "loudspeaking" crystal radios and radio receivers with headphones, that powered up from a powerful radio station and receiving broadcast of others, less powerful stations. Because the reason of such things are vague, there are lots of incredible schematic solutions that allegedly provides even more incredible results.

The goal of this article is to help to radio amateurs, who are interested in this problem, to look into that problem and find out real capabilities of radio receivers powered with "free energy" of powerful radio stations. Crystal radio circuits will be described later.

It is known that electromotive force (emf) in the antenna of a radio receiver can be calculated with the next formula:

ε = E * h_{e},

where E - strength of field;

h_{e} - effective height of antenna. We have to maximize not the emf, but the power of the signal at the detector, its input impedance R_{in} depends on the type of circuit in which the detector is included, and also it depends in some degree on emf in antenna. Because the power at the detector is

P = U * I,

where U - voltage at the detector;

I - current through it,

the input impedance is

R_{in} = U / I,

therefore to maximize the power it takes to change the input impedance of the detector, using different circuit types of matching the detector and antenna, increasing voltage across detector and decreasing the current, and vice versa.

From the other hand it is known that the source (antenna network) provides maximum power to the load if its active impedance matches the load impedance, i.e.

R_{A} = R_{in},

and the reactive component is compensated with a reactive impedance of opposite sign. This is usual conditions of matching the source with the load. How to provide them in the real situation?

Mostly powerful broadcast radio stations are broadcasting on long wave band. Wet soil, water, and especially sea water at these frequencies are having performance of a conductor, in which conducting currents are more higher than biasing currents. As a result, horizontally polarized waves at the ground are significantly attenuated. Because of this, vertically polarized waves are used for radio broadcasting, they are radiated by vertical metal rod antennas with good grounding.

Problems about designing long and middle wave antennas were solved in 1930's, and the theory can be found in handbooks of 1940's-1950's, that why books in the reference section of this article are old.

Fig. 1

The design of a vertical antenna with grounding is shown in Figure 1. The resonant wavelength of that antenna is

λ_{0} = 4 * I_{e},

the effective height of antenna is

h_{e} = 2 * I_{A} / π.

(The resonant wavelength is that wavelength, at which frequency the impedance at the jack J1 is active and equal to impedance of 1/4 wave asymmetric dipole antenna, i.e ~37 Ω ). It's almost impossible to build 1/4 wave vertical antenna at home, because it would be too high, so they usually use inverted "L" (Fig. 1, B) and "T" (Fig. 1, C) - shaped antennas, with parameter

λ_{0}= K * I_{A},

where I_{A} = h + I_{L},

K - a coefficient - see the table below:

Antenna | K |

L-shaped with I_{L}, < h | 4.5...5 |

L-shaped with I_{L}, > h | 5...6 |

T-shaped with I_{L}, > h | 6...8 |

Umbrella | 6...10 |

It may be recommended to use an umbrella antenna with 3 or 4 radial wires are connected at the top, but its design is quite complicated, so it is used extremely rarely.

The umbrella antenna uses only vertical part to intercept radio waves, horizontal parts are used as a capacitive load to electrically lengthen the wavelength of the antenna and its effective height. The more developed horizontal parts, the more precisely is the ratio

h_{e} = h,

and more effective the antenna.

In most cases an antenna intercepts radio waves with length more than the antenna's length λ >λ_{0}, the impedance of the antenna is a complex impedance (Z_{A}) with the active (R_{Σ}) and reactive (X) components is given by the formula:

Z_{A} = R_{Σ} - jX;

R_{Σ} = 1600*(h_{e} / λ)^{2};

X = W*ctg(π*λ_{0} / λ),

where W - impedance of antenna wire, it is about 450..560 Ω.

Fig. 2

To compensate the capacitive impedance of the antenna, an inductance (a loading coil) must be included in its network, so the equivalent circuit of the antenna turns into the circuit shown in Figure 2. Now we can calculate the power transferred from the antenna into the load (the detector), without considering losses in its network. When the input impedance of the detector equal to the active impedance of the antenna R_{in}=R_{Σ} the power across the load is maximal and it's equal to:

P_{0} = (ε/2)^{2} / R_{Σ}.

Substituting in this formula expressions for ε and R_{Σ}, we get this formula:

P_{0}= E^{2} h_{e}^{2} λ^{2} / (4*1600*h_{e}^{2}) = E^{2} λ^{2} / 6400

The derived formula determines the maximal power that can be induced by a radio station in an ideal lossless antenna. It's worth to note that the power doesn't depend on antenna design and size. So here are some conclusions from the said above:

- The possibility of powering radio receivers with "Free Energy" depends only on the electromagnetic field strength of the radio station at the location of reception;
- It's better use long and very long waves;
- It takes to match active impedances of detector and antenna to make reception effective.

For example, let's calculate the maximal power at the antenna induced by the electromagnetic field of a long wave radio station, operating at a frequency of 171 kHz (λ=1753 m) with the field strength of 20 mV/m:

P_{0}= E^{2}*λ^{2}/6400 =0.02^{2} * 1753^{2} / 6400=0.19 W.

Such power is enough to provide loudspeaking operation of portable radios, because it is equivalent to U_{sup} = 9 volts at the current of 20 mA.

Unfortunately, the real situation is far from described above. The problem is that antenna network includes resistance of losses R_{l}, it's the sum of antenna wire resistance, resistance of the coil L (see Figure 2), and grounding resistance. The efficiency of such antenna can be described by the formula:

η = R_{Σ} /(R_{Σ}+R_{l}),

and the power of such antenna is:

P = P_{0}*η = E^{2} λ^{2}*η / 6400.

Let's calculate the efficiency of an antenna. Resistance per unit length of a cooper wire with the diameter of 1 mm to direct current is 22.5 Ω/km, the resistance is almost two times more at the frequency of 200 kHz [1]. For a cooper wire with the diameter of 2 mm the same parameters are 5.5 Ω/km and 3 times, respectively. Thus, an antenna wire with a length 20...50 m has a resistance R_{wa} of 0.3...3 Ω. A resistance of grounding wires P_{wg} are even higher. M. V. Shuleykin introduced the empirical formula for ground losses calculation [2]:

R_{wg} = Aλ / λ_{0},

where coefficient A is in range from 0.5...2 Ω for good grounding and till 4...7 Ω for bad grounding. The resistance of the matching coil R_{wl} depends on its quality factor Q that can be calculated using next formula:

R_{wl} = X / Q.

Let's calculate the efficiency of an inverted L-shaped antenna with vertical part of 10 m and horizontal part of 20 m, h_{e}=10 m, using values from the example above. According to the table the coefficient K=6, then the wavelength of the antenna:

λ_{0} = 6*(10+20) = 180 m,

λ / λ_{0} = 10.

For the wire 1 mm in diameter, R_{wa} = 22.5*2*0.03 = 1.3 Ω, acceptable grounding can be achieved if R_{wg} = 3*10 = 30 Ω. With the wave impedance of the antenna W = 500 Ω the antenna reactive impedance is

X = 500*ctg(π/10) = 500/0.31 = 1600 Ω = 1.6 kΩ.

For a matching coil with Q=250, its resistance is
R_{wc} = 1600/250 = 6.45 Ω. A total loss resistance of the antenna is equal to 38 Ω (this is the sum of all resistances found above), the radiation resistance is

R_{Σ} = 1600*(h_{e}/λ)^{2}=1600*(10/1753)^{2} = 0.05 Ω,

it means that the antenna efficiency η = 0.05/38 =0.14%!

Therefore, the signal power transferred from the antenna into the load is 0.19*0.0014 = 0.26 mW, it's the equivalent of power supply voltage of 1 volt at the current of 0.26 mA. This power is enough to power a radio with headphones, but it's not enough to power a loudspeaking radio.

It's worth to notice that main losses of the antenna were caused by the grounding. It takes to dig the ground to reach the groundwater and bury a metallic object to make a good grounding. A network of suspended horizontal wires buried at shallow depth, plumbing system or a metallic fence also can be used for grounding.

But here is an important question of how to achieve impedance matching between the antenna and the detector? Using reactive components will only deteriorate the efficiency because they having losses, so it's better to use only these components that shown in Figure 2. In this case the circuit of radio receiver turns into a circuit that shown in Figure 3. The variable inductance coil L1 and the capacity of the antenna are forming a resonant tank, tuned to the frequency of a powerful radio station. Reactive impedances of the antenna and the coil are equal and compensate to each other. The serial active resistance of the antenna network:

R_{A} = R_{Σ} + R_{l}

recalculates into equivalent resistance

R_{ce} = X^{2} / R_{A},

connected in parallel with the coil L1. If this resistance is too high to be matched with the input impedance of the detector, then the detector should be connected to a tap of the coil L1 in a such way that the equation is satisfied:

n^{2}*R_{ce} = R_{in},

where n - ratio of the coil winding the lower end - tap to the entire coil winding. The circuit diagram of the crystal radio is self-explanatory (see the Figure 3).

Fig. 3

In the example above R_{ce} =1600 ^{2} /38 = 67.4 kΩ. If the detector has the input impedance of about 2 kΩ (that value is correct if the detector is loaded with 4 kΩ headphones), n = (2/67)^{0.5} = 0.17, therefore the tap is at about 1/6 of the coil winding.

An important problem in a rural area is a lightning protection of antenna systems. It's better to connect the antenna to the ground. The circuit diagram of the receiver shown in Figure 3 meets this condition. Nevertheless even quite far lightning strikes induces emf in antennas of kilovolts, that is very dangerous. A gas discharge arrester or a neon lamp, connected between antenna and ground can protect the diode D1. And anyway, it's better to close the switch S1 if there is thunderstorm is nearby.

V. Polyakov, Moscow

Reference:

- G. Ginkin. Handbook of Radio Engineering. Moscow. - L:GEI, 1946.
- G. Belotserksvskiy. Antennas. - Moscow, Oborongiz, 1956.

Magazine editorial comments: A paradoxical result in the constant power, transferred from the antenna into the load, that is not depends on the antenna size (if there are no losses and the antenna is matched with a load) can be easily explained. It is known that a transmitting antenna (if there are no losses) matched with a load, transmits all the power from a transmitter. Therefore different antennas with the same radiation pattern create equal intensity of the electromagnetic field. It's important to note, it doesn't depend on the antenna size. Of course, for a real antenna with losses the last statement becomes less relevant to practice. With decreasing an antenna in size, its radiation resistance gets too low, reactive impedance increases, all these complicate the matching of the antenna and the load, losses grows, therefore the efficiency of the antenna falls sharply.

Because antennas are reciprocal, therefore with the equal strength of field, matching the load, and with no losses, receiving antennas of different types and sizes will provide the same power in ther loads. Of course, this is only a theory, because in reality there are losses and it's hard to match the antenna and the load.

It's important to note that all the calculations in this article are applicable only if antenna size is significantly less than a wavelength.

See also "Eternally speaking" radio receiver and Loudspeaking "crystal" radio receivers.