Please consider the following circuit.

enter image description here

For this case, my following notion is correct?

  1. When v(t) is greater than diode turning-on voltage (v_d), the forwarding current (i_d) occurrs, and then energy will be harvested.
  2. When v(t) is less than v_d, i_d is zero, and then energy will be not harvested.
  3. Thus, the amount of harvested energy is depending on the i_d only when v(t) is greater than v_d.

For example:

  1. enter image description here

  2. enter image description here

First signal's averagy power (E[|v(t)|^2])is less than second one's. However, the first one turns on diode for a longer time than second one's. (that is, the first signal's sum of red-brakets time is longer.)

In this case, although second signal has more average power, the amount of harvested energy by first signal is much, right??

Thank you for reading my question.

  • 1
    \$\begingroup\$ It would be nice to know where all this came from. Is there a web site you've been reading? If so, link it so that those willing to spend the time can go look. If this is entirely from your own head, you should write more about your thinking. It doesn't appear to illustrate a clear division of ideas. You appear to conflate energy, power, and the area above the diode line and I worry this is going to be a difficult knot to untie in your mind. Besides that, the answer to your question will likely require some additions to your circuit, as well. \$\endgroup\$
    – jonk
    Commented Jul 4, 2017 at 5:31
  • \$\begingroup\$ @jonk This is entirely from my head. Sorry for a vague question. I think it is because I do not know well on circuit theory. Sorry. \$\endgroup\$
    – Danny_Kim
    Commented Jul 4, 2017 at 6:40
  • \$\begingroup\$ That's fine. We all start from somewhere. You just need to spend more time learning concepts, I think. Luckily, there is a lot of material on the web regarding RF energy harvesting. Texas Instruments has a bunch of stuff on the topic as they are (or were) active in it. RF antennas are not simple voltage sources as you would seem to have it, though. So there is some learning ahead of you. \$\endgroup\$
    – jonk
    Commented Jul 4, 2017 at 6:49
  • \$\begingroup\$ @jonk Thank you very much. If you don't mind, could you recommend me some materials including basic concepts? The explanation in wikipedia is too rigorous for me to understand the concept. \$\endgroup\$
    – Danny_Kim
    Commented Jul 4, 2017 at 6:55
  • \$\begingroup\$ Perhaps start here? ti.com/solution/energy_harvesting \$\endgroup\$
    – jonk
    Commented Jul 4, 2017 at 7:00

1 Answer 1


The answer to your question is more or less yes, but I'm not sure you're asking the right question.

1) Only when there's a voltage across and a current through the load is energy being delivered to the load, which you've drawn as a capacitor.

2) I_d is zero when the source voltage is too low.

3) The amount of energy delivered to the load depends on i_d and v(t)-v_d

The input signal does not 'have a power', as the power it delivers depends on the load, and the load the signal sees changes with the load's voltage, due to your diode.

The input signal has a voltage, and an output impedance, the latter complicating what happens to the voltage that appears across a load when a current flows.

  • \$\begingroup\$ I am studying of wireless power transfer field. In other words, antennas at receiver get a signal r(t) from transmitter, and it is considered as v(t) as the figure above, and I finally want the signal v(t) to become an energy for the device. (I want to make a small device for circuit class. For example, youtube.com/watch?v=HEfPgx51cas) \$\endgroup\$
    – Danny_Kim
    Commented Jul 4, 2017 at 6:43
  • \$\begingroup\$ @Danny_Kim Watched the video. In your dreams! The complex conjugate beamforming discussed between 0:50 and 1:20 needs a lot of processing, and a lot of transmitter hardware. It's integral to 4G MIMO. I'm amused that the scare over radiation from mobile phones cooking people's heads has been forgotten and now people smile while much higher RF energies are used simply to avoid a charger connection. Given your present level of understanding, you have snowball in Hades' chance of doing a Coto. Start with dipoles, then Yagis. \$\endgroup\$
    – Neil_UK
    Commented Jul 4, 2017 at 7:30

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