The effective area of the dipole is given by where G is the gain of the dipole, so that the needed power density S is Let us consider a numerical example. The voltage drop Vo is usually about 0.4 V. The gain of a half-wave dipole is about 1.5, and its impedance is a few tens of Q. Let us take 50 Q as a reference impedance. If we consider a frequency of 2.45 GHz, the minimum power density required is If we consider a frequency of 35 GHz, we find These low limits are very rough estimates of the power that only allow the diode to turn on without delivering any power to the system. The actual power densities should then be about one order of magnitude higher to get any DC power. Nevertheless, the reader should keep in mind that these figures assume that each antenna feeds its own diode. Several antennae could be combined to feed a common diode to increase the collection surface per diode. Another limitation exists because the parasitic capacitance of the diode degrades the conversion efficiency. This can be estimated from an energy point of view. To drive the diode, each half cycle of incoming wave must deliver enough energy to the dipole to activate the diode. The energy delivered is the product of the power density by the effective area and by the half cycle time. Thus: where E0 is the diode activation energy. This is proportional to the capacitance Cd of the diode Hence, we find and since with c the speed of light, we can write Again as an example, let us assume that Cd = 1 pF, V0 = 0.4 V and the frequency f = 2.45 GHz, then we find that the power density must be higher than
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