LaPlace Transforms for this Circuit?

I would think this would apply to a working circuit, but no expert here...

http://www.dummies.com/how-to/content/laplace-transforms-and-sdomain-circuit-analysis.html

Thanks! I did check the web for "How To" articles, and it brought a lot of memories back, although those neural networks were really dusty. The proof, of course, is in the final analysis and it can be verified graphically. Critical thinking used to be so much easier than it is now. Being critical is still too easy, for me that is.

By the way, I am always amazed at the pictures you find and post. What a repertoire you must have!

I figured I'd throw it out there and give it a day before I dug into the problem, just in case I had started off with the wrong assumptions. It is rare that I've had to do this kind of work. I know that this approach will require "partial fraction expansion" which is really easier than it sounds, but I don't remember how I used to do that either.

I hear retirement calling me! I just have to ignore it a few more years.

I have the feeling that something is not OK. The Darlington pulls the voltage to zero so that the coil is between full voltage and zero. The capacitance actuates as a supplementary source of small internal resistance. The coil has a quite high inductance due to the iron core. this means that the current will follow an exponential growth even if the voltage makes a step. It follows that the secondary voltage which depend on the magnetic field variation thus on the current cannot be a step. It should be a fucntion of the type Ue=Ui/R*(1-e^-t/T). "T" being the time constant depending on the inductance of the primary.

As for the Laplace transform it would be better if you first write the differential equation and then jumps to the transform. This gives you a better feeling about what happens physically.

I believe I can neglect the 220uF capacitor as just localized filtering. I'm also going to ignore the fact that my Darlington pair can not really reach zero volts. The energy available then is quite small due to the high impedance of the transistors when turned off.

What seems really odd to me is thinking of the secondary as an ideal voltage source in series with its internal inductance. LaPlace transforms really don't have a conversion for square waves. So, I'm treating this as a periodic step function which does have a transform and it includes initial conditions. So, I think it will be OK, but that is why I'm trying to make sure I get started correctly.

A square wave is the difference of two step functions with a time delay (e^-T).

You can define it as a series of such differences.

Yes, you are correct, thank you. That is what I think I will have to do as well. It is the assumptions that I want to make sure that I have right. Energy is stored in the transformer core via the primary coil while the driver is ON or current inflow is highest. Then it is off for a short while with only leakage current on the primary.

On the secondary, without a load, the output is stepped up and it follows the input very closely, except that with magnetic isolation, it is now equally split (both positive and negative) about the zero crossover. When a load is added, the waveform is dramatically changed because the core runs out of energy too soon to follow the input so you end up with something that can no longer be described as a square wave. And to describe what is really going on requires some integration over a fixed period of time. Getting to that point is what is hurting my brain.

This one should be in the text books but it requires a lot of skills to work through the whole thing. I'm trying to avoid starting off in the wrong direction.

Laplace transform of a rectangular voltage pulse of duration T is (1- exp(-sT)/s.

This is the "Variable Q Ripost"post again. I do not think you are helping yourself by going into Laplace transforms. You are assuming that the energy going into the transformer core is significant.

Rather, I think it is negligible compared to the energy going into the capacitor across the transfo secondary and the variable inductor connected to the secondary.

Looking at your scope diagrams of secondary voltage, I see no sign of overshoot at the vertical transitions due to the transfo leakage inductance discharging into the 0.1 mmF capacitor. Nor is there any "rounded edge" indication that the leakage inductance of the transfo is slowing the charge of the 0.1 mmF when the drive transistor turns on.

In circuit terms, the transformer is a "perfect" transfo with a winding inductance many times that of the "variable inductor" load, so it can be ignored for analysis purposes - its impedance is also likely to be many times that of the 0.1 mmF capacitor @ 330 Hz.

Referring to your scope diagram, I am now wondering how you got the "secondary voltage" since neither side of secondary is earthed.

Did you use a two channel oscilloscope in differential mode??

Thanks for the reply. I may need to examine my scope settings again because I see what you mean. This circuit seems to be taking full advantage of the storage in the transformer core. When I tested it without capacitors I noted the following:

1) With an open circuit on the secondary, the output looks a lot like the input, only at a higher voltage. That is to say, there were nice sharp corners and vertical lines during the change from negative to positive. Being magnetically isolated, it was centered about the zero cross over line.

2) With a resistive only load (about 2000 ohms) I notices some slope in the vertical change as well as an abbreviated pulse in the negative direction, returning to zero for the remainder of the cycle. That tells me that all of the magnetic energy was consumed by the load before the cycle ended. I would expect this since the input was at zero volts during this portion of the cycle and because without a load, the output had no choice but to follow the input.