Please consider the following circuit.
For this case, my following notion is correct?
- When v(t) is greater than diode turning-on voltage (v_d), the forwarding current (i_d) occurrs, and then energy will be harvested.
- When v(t) is less than v_d, i_d is zero, and then energy will be not harvested.
- Thus, the amount of harvested energy is depending on the i_d only when v(t) is greater than v_d.
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.