# Charging stage of Capacitor with AC source

First question: Can someone explain how a AC power source together with this topology can charge the capacitor? How does the process works?

how does the resonant charging behave in this topology? How does the pulsing works via a transmission line transformer TLT?

Charging C1 (red, resonant charging C2 (blue) and transmision line transformer TLT (green)

• Do you understand how a half wave rectifier converts an AC signal into a rectified DC signal and how a capacitor smooths that rectified waveform into a constant DC value? It has nothing to do with the transformers and, additionally, why have you shown a 3:1 step-down transformer followed by a 360:1 step-up transformer? I asked this on your last question and you didn't answer. Apr 16 '20 at 10:00
• @Andyaka its just the given topology, I would like to know how it works so I can understand better the circuit, how does the voltage signal of the capacitor will look like? as a sine wave or a constant value? Apr 16 '20 at 10:45
• What is the diode type and what is the AC operating frequency? What are the transformer part numbers? Apr 16 '20 at 10:49
• @Andyaka frequency 50 Hz, for the transformer i only know the turn ratio which is depicted in the figure, and the diode is RFU02VS6S from ltspice. But with this diode i get a sine wave at the capacitor, which i presume is incorrect Apr 16 '20 at 11:03

The capacitor isn't charged by AC. The cascaded transformers aren't anything unusual: the first transformer steps up the 50Hz AC to 3 times the voltage, then that voltage is further stepped up again by the second transformer by 360 times. This effectively steps up the input voltage $$\ V_{1} \$$ by $$\ 3\times360=1080 \$$.

This appears as your same 50Hz AC sine wave across the secondary of the final transformer as measured at the TRANS_OUT node, only with the voltage stepped up by a factor of 1080:

However, while the voltage across the secondary winding is AC, the diode converts the AC to DC. This is somewhat simplified of course (how diodes work is a whole topic unto itself), but the gist is this: regular PN junction diodes can be thought of as directional switches, or one-way valves. Once a diode is sufficiently forward biased (the tail or anode of the arrow is at a higher voltage than the tip of the arrow with the line, or the cathode), this turns on the diode/opens the valve, and current can flow through it minus a small voltage drop across the diode (this is the voltage that was required to forward bias it, usually called the forward voltage or $$\ V_{f} \$$). However, if the diode stops being forward biased, and instead becomes reverse biased (the cathode, or tip of the arrow is at a higher voltage potential than the anode, or tail of the arrow), this turns the diode off hard and it will not conduct.

I mental image I've always liked for this is a toggle switch, with the toggle lever at a right angle to the diode symbol. The flow of current also pushes the toggle switch in the direction of the flow, and when the toggle switch is leaning in the direction of the arrow, the switch is closed. When pointing away from the arrow, it is open circuit. The end result is, of course, you get current flow in only one direction, the direction that closes or turns on the switch. The moment current begins to flow in the opposite direction, the switch is turned off, preventing any further current from flowing.

Let's get rid of the capacitor for a second and replace it with a resistor. This will let is see what the voltage looks like on the other side of the diode, as measured at the DIODE_MADNESS node:

The DIODE_MADNESS node in yellow, and I've overlaid the TRANS_OUT node in red.

Finally, just the DIODE_MADNESS node:

See, it is no longer AC, but rather pulsed DC. And just so we're clear, AC current reverses direction, while DC flows only in one direction, but doesn't necessarily need to be constant. The voltage no longer goes below 0, current only flows in one direction as a result even if it isn't a constant flow. Hence the usage of the term pulsed DC - it is DC that is, well, pulsating.

Let's put the capacitor back:

This results in a fairly straight-forward RC charging curve, though with some jaggedness that occurs each time the AC is in the opposite polarity.

During the negative half of the AC wave where it would normally reverse direction, the diode simply does not conduct and the capacitor is also unable to discharge through the diode because that would also be in the reverse biased direction and turn off the diode.

Note: the resistor in series with the diode can be ignored, it simply limits the charging current for the capacitor.

As for the second topology, it is not possible to give an answer with the little information you've provided. S1 is not a valid schematic symbol, making the schematic unreadable. If you can post a valid electrical schematic, or maybe provide the source of those two screen shots plus the entire schematic rather that the absolute minimum amount of information, it may be possible to infer what S1 is. Normally S would denote switch, but that is not a valid schematic symbol for any kind of switch. My only other guess is that it is a spark gap, but that wouldn't make much sense and the symbol would still be incorrect for a spark gap.

If you're able to provide a valid schematic or enough information to read the incorrect one you've provided, I'll try to update my answer to explain the rest of your question.

Final note: In the comments, you mention seeing a sine wave at the diode. If the voltage exceeds the diode's reverse voltage blocking capability, depending on the spice model for the specific diode being used, it can (quite correctly) simulate the reverse breakdown of the diode, rendering it ineffective. Make sure you're not shoving 1kV across a 600V diode in your simulation.

• Thank you so much for such a detailed explanation, I updated the question where you can find the complete circuit. I hope you can explain each stage of the circuit: charging state, resonant charging and tranmssion line transformer. @metacollin Apr 18 '20 at 10:28
• if you need more information realted to the circuit please do not hesitate to ask. thank you again Apr 18 '20 at 16:05

The transformers increase the voltage, the diode rectifies the output into the capacitor, and the capacitor charges. Perhaps look up diodes or transformers on wikipedia to understand what they do if that's still not clear enough.

What may be worth expanding is how transformers deal with a DC load like this. In practice, the DC output current pushes the magnetisation of the transformer core away from zero. This produces an offset in the currents flowing, which produces an offset in the voltage drops due to winding resistance. It's these asymmetric voltages that eventually counter the flux offset, and allow the average voltages across the inductors to stay at zero.

• The signal of the voltage at the capacitor will still have a wave form? @Neil_UK Apr 16 '20 at 10:41
• The voltage on the capacitor will start at zero. Each half cycle of the AC input, when the diode is conducting, the voltage on the capacitor will rise. Each half cycle that the diode is blocking, the capacitor voltage will stay substantially constant. Eventually, the capacitor voltage will be constant and equal to the peak of the T2's output, less a diode drop. If C2 is supplying a current to an external load, then the waveform will be a sawtooth, as it alternately charges from T2+D1, and discharges to the load. Apr 16 '20 at 11:30
• thank you now i am starting to understand this. Can you also share how a resonant charging behave within this circuti (second picture) why c1 has to be greater than C2? and the function of D2 and R2. Apr 16 '20 at 13:27
• besides those new question above, how do i know whihc is the max current in the first diode D1? and the dissipation in R1? Apr 16 '20 at 13:38
• If you had to ask the first question, you are nowhere near to understanding an answer to the second. Get a simulator. LTSpice has a naff UI, but is free and more or less the de facto standard for amateurs and pros alike. Simulate an AC source for T2, add D1, R1 and your C, and do a time simulation to see what happens. Monitor the current in the diode. Observe the voltage on the capacitor, with and without a load on the capacitor. You'll learn far more than just asking question here. Then you can ask questions like 'here is a trace from my simulation, why does it do that?' Apr 16 '20 at 14:56