The project is about designing a transformer that converts 22V AC to 1V AC and connect it to voltage doubler as V_in

So here's my schematic and simulation result.

I'm having trouble with analyzing the result.

The D1N914 diode that is recommended for this project, had Vd,on of about 930mV when I simulated with VSIN - D1N914 - Resistor in series.

Thus I predicted the V_out(voltage across the vertical capacitor) would not be increases that much since on first positive cycle, V_out starts to increases after D1 is ON but it needs about 930mV to turn on where V_in is about 900~1000mV at best.

However, the output graph quite increased during first positive cycle about 200mV.

Did I have done something wrong? Can you explain what's going on here?

 

Everything looks fine. It takes a number of pulses for the output to charge all the way up. If the forward drop on the diodes is a problem you can reduce that by switching to a Schottky diode, but without a much more complicated circuit those diode drops will still be there.

 

Everything looks fine. It takes a number of pulses for the output to charge all the way up. If the forward drop on the diodes is a problem you can reduce that by switching to a Schottky diode, but without a much more complicated circuit those diode drops will still be there.

Thank you for answer. But what I'm wondering is the magnitude of the increase since I have to analyze it.

I cannot understand why V_out increases in that manner. As I learned, it must increases as the same amount of that of V_in.
Moreover, I have no idea why changing of capacitance affects V_out. When I increased the capacitance, V_out decreased.

can you give me some sources so that I could analyze?
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Your circuit is a half-wave voltage doubler. There must be a good analysis somewhere on the internet but it eludes me. I assume that you have already studied this one and the simple analysis is not sufficient to describe the phenomenon to which you refer.

As you know, the negative portions of the pulse are used to change C1 up to the peak voltage -some voltage is lost in the diode's forward voltage drop and a little is lost in the impedance of the driving circuit. During the positive peaks C1 is dumped into C2.

Pause for a second. If I have two capacitors of equal value, one charged and the other not charged, then place them in parallel, the voltage across both capacitors will only be 0.707 of that which was across one capacitor because you have the same amount of energy across twice the capacitance. In other words some voltage is lost transferring from C1 to C2.

When you make C2 larger it takes more cycles to "Pump up" the output to the same voltage as before, not counting losses from switching and the impedance of the source of pulses.

That get you started and help you find the analysis that covers details like this.