# In a single diode rectifier circuit, is capacitor offering additional energy to Load?

I am a graduate student majoring in electrical engineering. Now, I am studying about wireless power transfer and energy harvesting. While reading a paper written by Bruno Clerckx, who is senior member at IEEE, I encountered something that doesn't make sense to me (maybe makes sense for experts naturally.)

Take a look at the following circuit diagram; single diode rectifier circuit.

Surely, the authors did not consider Vin as a simply sinusoidal signal, but for simplicity and for striking what I do not understand, I assume Vin is a single sinusoidal signal.

Then Vout will be a signal $$V_{out} = \max ~\Big\{V_m\;\cos(\omega t + \theta), ~0 \Big\}$$ if the capacitor does not exist.

On the other hand, if the capacitor exists like the figure, the Vout will be smoothened like the figure.

Of course, Vout keeps higher in this case than in the case without the capacitor, so I expect to harvest energy better.

However, in the power view, Let P1 be transmit power, P2 receive power, and P3 output of rectifier's power. P2-P1 is the loss of the power due to the wireless channel, P3-P2 is the loss of the power due to the block for rectifier.

Then, I think P3 is anyway determined by the efficiency (power loss) of the channel, the one of the rectifier block, and the transmit power.

Why can we get more harvested energy if adding the capacitor, even though the capacitor is not a source but an energy consumer (anyway, one of resistances).

• Why would a diode cause the circuit to violate the laws of thermodynamics? Measure Pin and Pout! – winny Jul 19 '17 at 6:49

Why can we get more harvested energy if adding the capacitor, even though the capacitor is not a source but an energy consumer (anyway, one of resistances).

• You don't get more harvested energy - you get better managed energy.
• The capacitor accepts power in short bursts during charging.
• The capacitor supplies power when the AC < DC.
• Most likely the circuit that you are powering requires steady DC rather than half-wave rectified power. The capacitor helps supply that.

Re-run your simulation with and without the capacitor and monitor the current at the top of Vin and the current through RL. You will see a big change in both.

• Thank you for a respond. Sorry for having a more question. Why do we have to give an energy to the capacitor (for the capacitor to supply charged energy to the load) ?? Just giving all energy to the load is better reasonable thought, isn't it? – God Danny Jul 19 '17 at 7:28
• Without the capacitor you only get pulses of power when the AC signal is positive. That would switch your load on and off at high frequency. Do the simulation and you'll see the problem. – Transistor Jul 19 '17 at 7:39
• It's a bit like having an intermittent water supply - it's on for a minute, off for a minute, ... Add in a reservoir tank in your attic and you have continuous smooth supply. – Transistor Jul 19 '17 at 7:40
• I did simulation by orcad (pspice). I understand Vout become just positive signal of inputs when no capacitor, but Vout become almost constant value when having big capacitor. Why do this situation happen~? For example, we have 2 choices (1) give Jane (load) 10 apples whenever I buy 10 apples or (2) give her 5 apples and Jake (capacitor) 5 apples whenever I buy 10 apples and when I do not buy an apple Jake gives her his apples. As seeing this example, option (1) and (2) are anyway same in respect to the number of apples Jane received. – God Danny Jul 19 '17 at 7:58
• Apples are delivered every second minute. You can take as many as you want. Jane eats one apple per minute but can only hold one at a time. Without Jake she takes one apple, eats it and has none for the next minute. With Jake there when the truck arrives she takes one, he takes one (to top up his supply) she eats hers and then eats one of Jake's while waiting for the truck to come back. (This is getting ridiculous!) Without Jake Jane averages 0.5 apples per minute. With Jake she averages 1 per minute. – Transistor Jul 19 '17 at 8:09