Energy Received By
|Fig. 1 A capacitive antenna drives a resistor|| ||Fig. 2 An inductor is added, forming a tuned circuit|
Above we have two variations on a theme: a pair of LARGE parallel metal
plates creates a strong AC e-field between themselves, and a pair of SMALL
metal plates intercepts a bit of energy from this field. The received
energy heats up a load resistor. Assume that all this takes place in the
nearfield, where all plate-sizes and distances are far smaller than
c/500KHz = 600 meters. Imagine the big plates to be a couple of feet
In Fig. 2 on the right, an inductor has been added, and its value is chosen
in order to create a tuned circuit with a center frequency the same as
the frequency of the voltage-source which drives the large metal plates.
In both diagrams the load resistor has been adjusted for maximum received
energy. Assume that any stray capacitance of the resistor and inductor
are included in other capacitances.
Analysis is very straightforward. Let's add the parasitic capacitances
and plug in some real numbers. I'm arbitrarily making the capacitances
form a 100:1 voltage divider. If you object, feel free to choose some
|Fig. 3 Comparing the two circuits|
First let's choose a value for R1 on the left. Note that the parasitic
capacitances "Ca" happen to form a
capacitive voltage divider with C1. To adjust R1 for maximum power, we
set R1 equal to the Thevinin series reactance created by the Ca-C1 voltage
divider, which equals 1/(2*pi*f*100pF) or approximately 3.2K ohms. The voltage
which appears across R1 will be .707 times 100V, times the 1:100 voltage division,
so V(r1) = 0.7v. Knowing that power = V^2/R, the power received by R1 will be
around 160 microwatts. It's fairly small, as you probably expected. After all,
the voltage across the C1 plates obviously should be far smaller than
the 100V applied to the large plates.
OK, on the right we've added L1, and adjusted it to resonance. Something
interesting occurs: the resonant circuit acts like a pure resistor, C2
essentially "vanishes," and the voltage across R2
becomes very large. C2 becomes part of the reactance of the LC circuit, and since I'm
assuming lossless components, its reactance is infinite. To adjust R2 for
maximum energy, we set it equal to the series reactance of the voltage
source, which is two Ca capacitors (2pf) in series, or 1/(2*pi*f*1pf) = 320K ohms.
Since C2 and L1 "vanish", the voltage across R2 is simply .707 times the
100v drive, or 71v, and the received power is around 16 milliwatts.
We've found that the maximum power intercepted by each of the two circuits
is very different. The circuit with the resonator works better by a
factor of 100!
By adding a tuned circuit, we've eliminated the 100:1 capacitive voltage
divider. And if the large plates were even further apart (but still in
the 500KHz nearfield), then the 100x difference between the two circuits
would be even greater.
To check my numbers, I divide the power equation for one circuit by the
power equation for the other. I find that everything cancels out except
the 100:1 from the capacitive voltage divider. The circuit with the
resonator receives 100x more power.
Also note that the voltage across C2 is higher than the voltage across C1
by a factor of 100.
This implies something interesting. If you held a little bitty electrostatic
field meter near C1, it would indicate a low value; a value just about as low
as if the C1 plates were not present at all. On the other hand, if you
used the same meter to measure the field
strength near C2 on the right, you would find it to be very large, almost as
if the small plates forming C2 were directly connected to the large, distant
driver plates! All this is a consequence of high-Q resonance. Detune the
circuit, and the large voltage across C2 goes away.
Pd without tuned circuit: 0.16 mW
Pd with tuned circuit: 16 mW
Increase in received energy: 100x
Increase in e-field near the small plates: 100x
| New info:3/18/2000
Engineers on SCI.ELECTRONICS.DESIGN have pointed out that my above analysis of FIG.2 is not quite correct: the R2 power is NOT limited to only 16 milliwatts as I have calculated. This is true because the voltage across L2 is NOT limited to 71V like I thought. Instead, it is limited only by the Q factor of the resonator, and is proportional to the value of R2. If the resistance of R2 is made very large, then the Q becomes large. (The resonant frequency of L2-C2 must be retuned slightly to maximize the power in R2.) By making the value of R2 larger, the Q becomes larger, and the voltage across R2 will rise proportionally. Because the power dissipation for R2 depends upon the *square* of the voltage across it, increasing the value of R2 doesn't result in the same received Pd, instead it lets the antenna receive a larger Pd.
If lossless components are used (particularly for the coil L2,) the voltage
across R2 can be FAR higher than 71V, and the Pd received by resistor R2
can be FAR higher than 16 milliwatts.
As a result, my calculation of 100:1 power difference between Fig. 1 and
Fig. 2 is perfectly correct as it stands. However, if the above changes are
made, then the 100:1 difference could be immensely larger, even far higher
than 100V. And because the voltage across R2 and C2 rises much higher, the
e-field near the C2 plates is much higher too.
This change does not alter my conclusion: R2 dissipates far more power than R1, and the e-field adjacent to the small C2 plates in Fig. 2 has a far larger value than the e-field adjacent to the small C1 plates in Fig 1.
OK sci.physics people, all interpretations of the "Energy Sucking" phenomenon aside, do you agree with the following?:
Feel free to check my calculations. I'm
a digital designer and my analog math is rusty.
SIDE NOTE: This is the electrostatic case. The entire article can be
rewritten so that the "transmitter" is a large hoop-shaped coil driven by
a 500KHz current, and the two receivers are smaller coils nearby, one of
which features a tuning capacitor connected across it. Rather than
coupled capacitors, we can form loosely- coupled transformers, and the
results will be similar. The tuned "antenna coil" will intercept more
energy than the RL circuit. And the b-field near the tuned "antenna coil"
will be much larger than the b-field near the other.
ANOTHER NOTE: this might explain why crystal radios work much better if a
tuned circuit is used. A tuned circuit is not just a filter. Instead it
creates a higher signal voltage in the radio circuit. This is not magic,
because in order to help to exceed the Vf of the crystal diode, we could
instead place a step-up transformer between the antenna/ground section and
the rest of the circuit. But there is some magic to be had: if we place
a parallel LC circuit between the antenna and ground, we can eliminate the
antenna/ground capacitance, and we can actually increase the net amount of
energy received by the antenna as if the antenna was electrically larger.
The tuned circuit in a crystal radio is *not* simply a bandpass filter.
Instead its oscillations grow as it takes in energy, which *drives* the
receiving antenna and creates a strong EM field. And this EM field then
"funnels" the incoming EM waves to the antenna which would otherwise pass
right by. (Or from another viewpoint, the resonator drives the receiving
antenna, causing it to emit an EM sphere-wave which superposes with the
incoming waves to form a diffraction pattern... and this diffraction
pattern takes the form of an "EM shadow" which appears downstream from the
receiving antenna: a shadow-region where some EM energy is missing. The
missing energy has gone inside the crystal radio.)
In Fig 5 I've sketched in the profile of an "absorber disk" which has an
effective aperture equal to that of the small resonator. Lines of
Poynting flux which pass through this disk are deflected so as to strike
the tiny resonant absorber. For high-Q absorbers at very long EM
wavelengths, the effective aperature can be enormous. In theory a loop
antenna a few inches across could gather the same amount of energy as a
long-wire antenna hundreds of feet wide. It's just a matter of achiving a
high enough "Q" rating.
Fig 5. Circle shows the approximate size of an
"absorber disk" which has an area equal to the
Effective Aperature (EA) of the resonant antenna.
Chu, J.Appl.Phys. Dec. 1948
Hansen, Proc.IEEE Feb. 1981.