High frequency reduces power on circuit

Question

I am not expert in electronics so there must be many errors that I am not considering. I designed a simple half wave rectifier that gives signal to a LDO voltage regulator of 3V3. In the input I have a sine wave that should have a frequency of 13.56 MHz and an amplitude of 10 V. However, when I choose a 500Hz wave the LED at the output of the LDO is working but at 1KHz, the LED is off. the voltage drops to a few mV.

I am using the following schematic that is extremely simple (the purpose is to design a bioimplantable device for the brain, so that's why it must be simple)

^{simulate this circuit – Schematic created using CircuitLab}

The datasheets of the LDO that I used is:

https://www.sparkfun.com/datasheets/Components/LD1117V33.pdf

Just for leting you know, I am basing my circuit in the design of John Rogers :

https://www.nature.com/articles/s41928-018-0175-0?WT.feed_name=subjects_engineering

What is the problem and how could I solve it. I attach some pictures.

Answer 1

Why your circuit fails.

• You used a Zener instead of a Schottky diode shown. ( similar but different symbol)
• you put the cap before the Zener instead of after a Sch. Diode.
• 2.2uF is a short circuit at 13MHz but 10k at 10Hz , 1k at 100Hz
• 50 Ω gen is too high impedance for this design
• no series inductor
• no series resonant circuit.
• no Q voltage amplification
• no impedance matching.

The impedance of RLC is shown below vs frequency. Resonance occurs at the intersection of L and C at some impedance and frequency.

The magnetic power source is shown below

Theory of Operation of charger resonant circuit.

The charger here consists of a series RLC resonant loop current circuit with two Schottky series diodes to a DC storage cap. THe crossover impedance of L and C above indicates both the impedance at resonance and the frequency. For a series resonant power circuit, you need a low impedance drive and resonant circuit with a higher impedance load shunting the capacitor so that the voltage gain ~ Q= impedance ratio R/X(f).

The LDO load capacitor is bigger which affects the resonant frequency only when it conducts at peak voltage. Thus the higher the Q, and reduced ripple takes longer to charge up but affects the resonant frequency less. It is larger so that the ripple is low and thus has less dynamic frequency shifting effects from conducting at a lower duty cycle. The cap input cap affect the tuning of the series according to the duty cycle at which the storage cap is conducting thru the diodes that affect the resonant tuning. This can be analyzed as a resonant circuit with a ripple-controlled pulsed capacitance at a low duty cycle. Therefore C ratios must match ripple and total C must resonate with secondary inductance to achieve some Q of 5 to 10 without being too critical on component values.

Proof of Concept

This explanation is for electronic experts. For the non-experts, use the simulator to see how it performs when you move the sliders for LC value away from optimal resonance. As the simulation is slowed down from real-time, the response will also be slow.

Above shows about 77mW average input power and 34mW output DC power. (almost 50% = ideal)

Maximum Power Transfer occurs when overall source to load impedances are matched at resonance and then you get 50% efficiency yet with a higher reactive to real impedance ratio to get higher Q and voltage gain. This requires computing what your load impedance is then choosing your reactive parts to resonate at this. It may require a step transformer externally and a step-down transformer or tapped coil to reduce the series resonant impedance.

An RLC meter is needed to verify component values. Note it is also possible to eliminate the lower Schottky Diode with some tradeoffs on threshold and tuning sensitivity.

Answer 2

If you want a half wave rectifier, put the capacitor between diode and regulator, not before the diode. Now the incoming AC can discharge the capacitor. Also at high frequencies the average voltage over would be zero, at lower frequencies the capacitor voltage will follow the input voltage.