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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)

schematic

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.

enter image description here

enter image description here

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    \$\begingroup\$ More information is needed. Can you use the schematic editor to show us your circuit, along with datasheets of the parts being used? \$\endgroup\$ – MCG May 16 at 13:00
  • \$\begingroup\$ What is your energy storage capacitor? \$\endgroup\$ – analogsystemsrf May 16 at 13:04
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    \$\begingroup\$ You don't have a capacitor at the output of the regulator?? I'd recheck the datasheet... \$\endgroup\$ – aschipfl May 16 at 13:19
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    \$\begingroup\$ That's why I ask here. That's not helpful. \$\endgroup\$ – Mufasa May 16 at 15:48
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    \$\begingroup\$ Do you understand how rectifiers work? You need to put the cap after the diode not before. Your just basically shorting the supply through the cap and function generator internal impedance if you put it before the diode. Put also another cap after the linear regulator output, this will ensure you'll get smoother DC, thus more power at the LED. \$\endgroup\$ – Unknown123 May 16 at 17:06
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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.

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  • \$\begingroup\$ Ok,then the capacitance should match the input frequency to remove the imaginary part, is it? \$\endgroup\$ – Mufasa May 16 at 13:25
  • \$\begingroup\$ What do you mean capacitance to match the input frequency? perhaps this will help: learningaboutelectronics.com/Articles/… \$\endgroup\$ – theguitarfreq May 16 at 13:33
  • \$\begingroup\$ That's why I used the smoothing capacitor at the entry of the circuit. I guess that it should be after the diode \$\endgroup\$ – Mufasa May 16 at 14:12
  • \$\begingroup\$ Just think about that for a second, how can you smooth AC? Your guess is correct, it should be after the diode especially if its an electrolytic \$\endgroup\$ – theguitarfreq May 16 at 15:24
  • \$\begingroup\$ Perfect. That was really explanative. \$\endgroup\$ – Mufasa May 16 at 17:40
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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.

enter image description here The magnetic power source is shown below

Theory of Operation of charger resonant circuit.

enter image description here

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.

enter image description here

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.

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  • \$\begingroup\$ That's awesome. I will think about your tips and I will redesign the circuit to solve it. I guess that the smooth capacitance should be bigger. I will calculate it. However, John Rogers uses that capacitance. That;s the difference? I am really thankful for you time. \$\endgroup\$ – Mufasa May 16 at 17:41
  • \$\begingroup\$ @Mufasa Look at the source R and make it same or smaller perhaps using tiny stepup to step down coupling transformer. look at LC values on graph . make sense? Compute L or measure it. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 at 18:10
  • \$\begingroup\$ too bad the critics do not appreciate it \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 at 21:01
  • \$\begingroup\$ Perfect @sunnykyguy EE75. Thanks for the help. I will measure the L and then I will try to calculate the coupling capacitance for that. Very helpful! \$\endgroup\$ – Mufasa May 16 at 23:02
  • \$\begingroup\$ The graph gives you an instant ballpark figure from the intersection of LC&f also measure Rs or design an emitter follower buffer for driver or equiv \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 at 23:20

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