Hi Aking and Graham,
Indeed, using semiconductors for switching involves some loss in power. But the loss can be kept at a relatively low level. Consider the equivalent series DC
resistance of supercapacitors (500 F range), it ranges from about 2 to 5 milliOhm, depending on the make and the manufacturer.
By a random search, this type has 3.1 milliOhm equivalent DC resistance:
https://handsontec.com/index.php/product/500f2-7v-super-capacitor-solder-tab-terminal/ When the switch is on, the DC resistances of the two caps in Joel's circuit are added up from the current flow point of view and consider to add the winding resistance of the 2 V transformer coil,
it may be in a similar range.
So roughly speaking, the total DC resistance in the closed circuit may amount to say an average of 10 milliOhms or so, not counting the ON resistance of the switch yet.
Using a power MOSFET like this
https://www.mouser.co.uk/datasheet/2/196/Infineon_IRF1324_DataSheet_v01_01_EN-3362809.pdf it has a typical ON resistance of 1.2 milliOhm,
max 1.5 milliOhm. Here for instance it can be purchased:
https://www.mouser.co.uk/ProductDetail/Infineon-Technologies/IRF1324PBF?qs=2r01AXMCG3Nmd46Swuw5wA%3D%3D Now consider the total DC resistance involved in Joel's closed circuit: it is around 11.2 milliOhm and now you know this depends much on the components used.
If you charge up supercapacitor C1 to say 2 VDC and C2 and the other caps are fully empty, the initial peak current at the switch-on moment would be 2 V / 11.2 milliOhm = 178 Amper (peak current).
This warrants the use of the MOSFET type I refer to above with heat sink like Joel used. OF course there are many similarly rated power MOSFETs to choose from.
So from this example, you can see the power loss of the suggested switch is the smallest, compared to the rest of the other circuit elements.
Will draw a circuit schema tomorrow. The cheap frequency generator can be like this:
https://www.ebay.com/itm/404347009846 A question to you both: In Joel's video
https://youtu.be/0c7llvTjlwQ?t=2 there is a black cylinder glued to his board, it looks like for me as a HV module like this
https://www.ebay.com/itm/314779472132 BUT as I can see in the video, the wires of this module are NOT connected to the rest of the circuit, hence no high voltage field
can be present as an additional energy source IF the presence of such field is a must to have the extra high output.
How you can see this I wonder. I pose this question because I do not know whether to include the HV module in the schematic and where to connect its wires?
Gyula