Quote from: hartiberlin on July 22, 2014, 07:20:45 PM
@x_name41Nice !
What is the output versus input power ratio in your device ?
Many thanks !

Regards, Stefan.

I have no idea, because i used sound card as a oscilloscope and i had no opportunity to calibrate and adjust the ratios, for the ratio between input and output power i can say that in the DC power supply i have 3,45V/0,54A.- without load, and with load have 3,68V/0,53A. Load is it incandescent bulb on a 24V/0,020A. I would like this device, to use it for laptop power supply

I am still playing with the simulator.

I came up with this  output circuit
by also using there a switched cap which makes the output pulses bigger.
The switching pulses are inverted to the main switching pulses, so this second cap
is 180 degrees out of phase switched onto the load resistor.

Well, as thze second scopeshot from the left ( the white one) shows the
actual peak values of the input power, where the minus amplitude means
POSITIVE input power and the positive wave means NEGATIVE input power (power delivered back to the grid)
you see, that this circuit draws around 826 - 610 Watts= 216 Watts...
But these are only the peak values...so no real RMS measurements.

At the load resistor the peak values are 279 Watts, but surely the curves must be integrated
to see the real RMS output power. This is unfortunately not available in this simulator...

Well you can see, if you switch the output switching OFF with the switch at the top,
the amplitude at the load resistor falls to around 96 Watts Peak.

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Quote from: x_name41 on July 22, 2014, 08:13:47 PM
I have no idea, because i used sound card as a oscilloscope and i had no opportunity to calibrate and adjust the ratios, for the ratio between input and output power i can say that in the DC power supply i have 3,45V/0,54A.- without load, and with load have 3,68V/0,53A. Load is it incandescent bulb on a 24V/0,020A. I would like this device, to use it for laptop power supply

Maybe you can try to self-loop your device. Instead of lightbulb use brige rectifier and 400V cap, and bring all power back. It will work or not .
All this mesurment with light bulbs and standard multi-meter are inconclusive anyway.

Guys, I would stick to the basics and go from there to see if it works before trying new stuff, or it will be endless...  Even just sticking to the basics we need to make assumptions. For instance, I started my thinking about the serps starting from a basic tank circuit. Visualize the cap charging and discharging with regards to voltage and current. As soon as you start inserting a load in series with the cap and coil (the secondary of our transformer) the phase shifts and the source starts to see the tank as a load. The higher the resistance the worse it gets. BUT if we force switch the tank to stay in the 90-180-270-360 phase switching in theory it should work right? If we now start inserting a load (resistance) it takes more time to charge/discharge the cap, but we need to charge and fully discharge it before each reversal. So we know that the higher the load resistance, the smaller our cap value and vice versa. Keep in mind that a non switched tank circuit works well with open core coils. Closed cores, be it toroidal or E type is terrible, so here me have to make assumptions. Do they use standard ones or modified cores..?
Jim's early demonstration shows he was using the full duty cycle (25%) per quarter.  In the last one they only use about 13-14% of the duty cycle. And the waves look different. When I went to parallel/series switching the current waves actually looked the same as the ones on the plastic box scope shots of the latest demo, except that the discharge current was about half.
We have to analyse and test what we know and work from there.


regards,
Mario


btw, what has the bitoroid stuff to do in this thread?

Quote from: x_name41 on July 22, 2014, 08:13:47 PM
I have no idea, because i used sound card as a oscilloscope and i had no opportunity to calibrate and adjust the ratios, for the ratio between input and output power i can say that in the DC power supply i have 3,45V/0,54A.- without load, and with load have 3,68V/0,53A. Load is it incandescent bulb on a 24V/0,020A. I would like this device, to use it for laptop power supply

Let me see if I understand. Your load is 24V bulb drawing 20 milliAmps, for a power of about half a Watt.
You are supplying 3.68V at 0.53 Amp, or nearly two Watts. This isn't really very good efficiency.

Are you thinking you can just take the _difference_ in the supply power, loaded and unloaded, and claim that that is your input power lighting the bulb? No, you aren't allowed to do that and here is why:

Turn on your garden hose and direct it out into your yard. Measure the flow rate at the faucet. This is your "unloaded" flow rate. Now, without changing anything, bring a bucket over and let the hose play half into the bucket and half into the yard. The bucket, your load, is filling up, right? Measure the flow rate at the faucet. Is it different than before? You are being charged for this flow rate, not just what hits the bucket.