The brain fog and lack of notebook is really eating my lunch, the second figure is my attempt to redraw the first:

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would someone please proof the schematics please?

The circuit matches, but the error is that two things, (D1 and C2) are grounded instead of being connected to the line from input common to output common. YES, I AM rather picky about that sort of thing. I will be the first to admit it. That is because it matters. In this drawing we need to assume that both input and output are referenced to "ground", which is not always the case.

 

Thank you, isn't this a ground?

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Thank you, isn't this a ground?

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In this drawing it is a round connection, BUT in the drawing that W was asking us to verify there was no actual indication that the input or output were referenced to ground. It was only presumed.
And the whole thing looks a bit "off" because resistor R1 is connected between two identical ground symbols.
In addition, consider that the TS had expressed concern previously that the return side of the output was not commoned with the return side of the power input source, which was an initial requirement.

 

I'm looking forward to playing with this, I have to wonder if D5 and C5 are necessary? Time will tell.

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Be prepared for the voltage to drop if you try to draw "too much" current out of the 15V point. I'd suggest starting with a LED current of 2 mA or so.

 

In post #1 the 555 output goes to 0V (ground) and to +5V. It never goes negative.
The transformer output goes positive and negative with ground in the middle.

The literature shows 4 diodes and 4 capacitors in a two-stage multiplier then I show 6 diodes and 6 capacitors in a three stage multiplier.

 

As long as the square wave source will both source and sink current it should work. At least in theory.

 

It can work with half stages.

 

C1 charges thru D1 only when the input signal goes negative.
When the input signal goes positive, C2 charges thru D2 from the charged voltage on C1 plus the input voltage.
Minus the low forward voltages of the Schottky diodes.

 

I don't think that is a supportable position, but you're entitled to it. It does affect my opinion of your credibility on a whole range of other things.

In post #1 the 555 output goes to 0V (ground) and to +5V. It never goes negative.
The transformer output goes positive and negative with ground in the middle.

The literature shows 4 diodes and 4 capacitors in a two-stage multiplier then I show 6 diodes and 6 capacitors in a three stage multiplier.

So the highlighted schematic is flawed, I had not realized that I needed a balanced signal. Which means I will need qty2 555 circuits to drive it. So I will go with this schematic:

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now the question is where can I tap the outputs for voltages:

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would this assumption be correct?

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So now here's my next brain fart:

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So what say you all, would this work?

 

I probably screwed up on orienting the mosfets, but I am hoping folks on this site will correct me if I got it wrong funny thing is I don't even know if I have these parts I plan on ordering ordering them via Amazon if I don't (of course the mosfets are logic level devices):

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The symbols show an N-Channel and a P-Channel, but I think you have them in the wrong place and the P-Channel device is in the wrong orientation. Also, your gate resistor R2 at 10K is way too large. There is no DC current that flows in and out of the gate. The following shows the correct way to draw a complementary MOSFET push-pull driver. I tried to find logic level devices, but the p-channel device has a pretty large rds(on) at 6.25Ω

 

Thank you very much. So it should look more like this?

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that is if I didn't make another brain fart, no one's complained about the smell yet.

 

You still need to make the resistors labeled R2 smaller than 10K. You want to switch the gate as rapidly as possible, and the small resistance is to damp any ringing that might be produced.

 

OK thanks. I have updated the drawing. I need to study how a mosfet is biased.

 

OK thanks. I have updated the drawing. I need to study how a mosfet is biased.

It is much easier because a gate draws no quiescent current.

 

Studying this I realized I had got the transistors wrong yet again:

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Now to layout the protoboard.

 

That circuit will triple 2.5V, which is the RMS voltage of the input. I have been playing with this in LTSPICE.

To triple a 5V square wave, you need to build the multiplier on top of the 5V supply, like this:



Here, I have tripled the 5V input and got out about 13V at 13mA.

 

That circuit will triple 2.5V, which is the RMS voltage of the input. I have been playing with this in LTSPICE.

To triple a 5V square wave, you need to build the multiplier on top of the 5V supply, like this:

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Here, I have tripled the 5V input and got out about 13V at 13mA.

Works better if you use Schottky diodes instead of signal diodes.