The following circuit converts high-frequency electromagnetic fields to direct current (DC) power.

enter image description here

I'm trying to understand how it works. The diodes are used to rectify the signal, however it's not clear to me why they are put in that particular configuration and why we need four of them (and not only one). It this sort of voltage doubler ?

Also, AFAIK, C3 and C4 are used to store the electrical energy. But what is the purpose of C1 and C2. Does it act as a filter ?

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    \$\begingroup\$ Where did you get the too-simplistic-and-not-representative-of-reality picture from? \$\endgroup\$ – Andy aka Oct 28 '16 at 20:51
  • \$\begingroup\$ From here : youtube.com/watch?v=XpLCK88nVgU \$\endgroup\$ – tigrou Oct 28 '16 at 21:00
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    \$\begingroup\$ Engineering by YouTube, love it! Going to trade in my degree for a cup of coffee. \$\endgroup\$ – Tyler Oct 28 '16 at 21:19
  • \$\begingroup\$ @tigrou .. its bogus as shown. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 29 '16 at 1:10

That circuit is overly complicated, and thus, burns most of the energy input in losses from the much too many unneeded components.

The following is the much simpler circuit of a crystal radio, which makes you hear AM radio from the nearby station, without a battery.


simulate this circuit – Schematic created using CircuitLab

It's so well-known it even was on Hal Roach’s Rascals and a gimmick in some comic magazine

Well, how does this work? Energy is taken from the EM field, rectified/demodulated by the low drop, soft I/U characteristic, hence Germanium, diode, then fed into the high-ohms earphones. Coil and capacitator are only there for selection of a specific station. When you leave them out, you hear AM-Babble.

And that's what your circuit does. It leaves out the frequency selection, feeding all received energy to the consuming circuit on the right. The capacitators and diodes indeed work as a voltage doubler, but that's nonsense because they have so much loss you are better without them.


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