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Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
zesla wrote ...
Hi marko,
Thanks a bunch for your post, Is the output current very much so that we need a parallel bank cap?
Actually I am not able to see the reality of pi*supply for the tank yet?!
thanks again
Hi,
the tank voltage seems to come from the condition of the DC link choke current staying constant, for which it has to see only AC voltage, or exactly as many volt-seconds in the negative cycle as in positive one. The current walks a bit up and down, the same amount every half cycle, but the frequency is so high and the inductance of the RFC is so large that this is negligible and we can think of this current as constant.
It's probably easier to imagine this with two choke configuration than the center tapped one. In operation, voltage on the mosfet drain is 0 for half a cycle and then half a sine wave with amplitude pi*supply voltage.
The choke sees this voltage shifted down for the supply voltage, resulting in a weird function which you can integrate on it's period and confirm the integral to be 0 when you take the peak of the sine thingy to be pi*supply. Probably someone here can think of more elegant ways of visualizing this.
Marko
PS. Wolfram alpha gives a nice representation of why mazzili voltage has to be pi (supply voltage taken to be 1V for simplicity).
Registered Member #3857
Joined: Sat Apr 30 2011, 10:16PM
Location:
Posts: 9
Marko wrote ...
zesla wrote ...
Hi marko,
Thanks a bunch for your post, Is the output current very much so that we need a parallel bank cap?
Actually I am not able to see the reality of pi*supply for the tank yet?!
thanks again
Hi,
the tank voltage seems to come from the condition of the DC link choke current staying constant, for which it has to see only AC voltage, or exactly as many volt-seconds in the negative cycle as in positive one. The current walks a bit up and down, the same amount every half cycle, but the frequency is so high and the inductance of the RFC is so large that this is negligible and we can think of this current as constant.
It's probably easier to imagine this with two choke configuration than the center tapped one. In operation, voltage on the mosfet drain is 0 for half a cycle and then half a sine wave with amplitude pi*supply voltage.
The choke sees this voltage shifted down for the supply voltage, resulting in a weird function which you can integrate on it's period and confirm the integral to be 0 when you take the peak of the sine thingy to be pi*supply. Probably someone here can think of more elegant ways of visualizing this.
Marko
PS. Wolfram alpha gives a nice representation of why mazzili voltage has to be pi (supply voltage taken to be 1V for simplicity).
Inductor voltage:
It's integral:
Thanks Marko, But actually I am not able to understand your last post (Maybe because I am a beginner)!
What are those links reffer to actually? Still I must say that I do not know why the voltage of the tank section goes higher than the supply by factor of Pi?! I can not understand the Wolfram alpha representation as I dont know what it does tell and how to interpret it for your circuit design.
Forthermore why do we need to a constanct current source? why don't use just 2 resistors instead of 2 chocks?
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi zesla,
What are those links reffer to actually? Still I must say that I do not know why the voltage of the tank section goes higher than the supply by factor of Pi?! I can not understand the Wolfram alpha representation as I dont know what it does tell and how to interpret it for your circuit design.
To keep it simple, the law of self-inductance says that if you have an ideal inductor carrying a DC current component, it will remain constant (or zero) only if the inductor sees only AC voltage on it (which can be 0 volts too, but that doesn't matter here). An AC waveform can be seen as one that has equal surfaces (volt-seconds) both over and under 0V, no matter what the waveform looks like.
The voltage on mosfet drain (to which the inductor is connected) is like half-wave rectified sine wave due to resonant action of the tank circuit.
You must now understand that inductor voltage is this drain voltage, minus the value of DC supply voltage. The first wolfram alpha link shows what it looks like.
We pretend for now we don't know what the peak value of the drain voltage is, and we can vary it, but our DC supply voltage (Udd) is fixed and hence inductor voltage is always shifted down for the same amount.
When the circuit is turned on the choke current rises, and higher it is it feeds more power to the tank circuit and the amplidude of the drain voltage increases. This in turn makes the current rise more slowly because the above-zero surface of inductor voltage rises, while one under zero remains constant. And when it's amplitude reaches pi*Udd - Udd, the under and above zero surfaces will become equal, and current will settle on whatever value required for this. This can be proved by integration as seen on the other wolfram alpha pic, and pi*Udd - Udd on the inductor translates back to Pi*Udd on the mosfet drain.
I think that's fairly clear and simple enough, please be more specific about what you don't understand.
Registered Member #3857
Joined: Sat Apr 30 2011, 10:16PM
Location:
Posts: 9
Thanks Marko . You actually are a kind and helpful guy.
Well, to be specific, I do not know which componetnts cause the voltage of the tanks goes to pi times supply.
surely I know that there are kinds of amplifiers called bridged amplifiers, which are able to affectively generate 2X supply output. If I suppose that your design is working in a such manner, then there actually would be just 2X supply at the ends of the tank circuit not pi X supply. apart from than bridging if I suppose that the pi X supply comes from emf or so called back emf of the inductors & chocks in your circuit then I remmebr that you told me that there would not be any back EMF (specially a greate back emf) generated in the chocks and inductors of the said circuit when normally working(that's very strange to me yet because any change in the current of an inductor will cause a back emf induced in the circuit (that's why I can not able to see why the inductor's current does not change while its related fet is turned off for some portions of time!)).
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
zesla wrote ...
Thanks Marko . You actually are a kind and helpful guy.
Well, to be specific, I do not know which componetnts cause the voltage of the tanks goes to pi times supply.
surely I know that there are kinds of amplifiers called bridged amplifiers, which are able to affectively generate 2X supply output. If I suppose that your design is working in a such manner, then there actually would be just 2X supply at the ends of the tank circuit not pi X supply. apart from than bridging if I suppose that the pi X supply comes from emf or so called back emf of the inductors & chocks in your circuit then I remmebr that you told me that there would not be any back EMF (specially a greate back emf) generated in the chocks and inductors of the said circuit when normally working(that's very strange to me yet because any change in the current of an inductor will cause a back emf induced in the circuit (that's why I can not able to see why the inductor's current does not change while its related fet is turned off for some portions of time!)).
Hi zesla,
I never said that chokes produce absolutely no induced voltage (back emf), since that is pretty much the point of any inductor and all I wanted to explain is that there are no large variations in current on them and hence no extreme voltage spikes (which would happen if both mosfets went off at same time). What you call "back EMF" is the very voltage waveform I posted on wolfram alpha. If you drew the choke current waveform on the same diagram, it would rise slowly wihle voltage is positive (mosfet on) and fall slowly while it's negative(mosfet off), because of large choke inductance, and only for naked eye it looks like it's constant.
While mosfet is off the choke feeds the energy it stored while the mosfet was on into the tank circuit, and the process repeats over and over.
Registered Member #195
Joined: Fri Feb 17 2006, 08:27PM
Location: Berkeley, ca.
Posts: 1111
hi zesla, I have added a second plausibility. RFC is perhaps 220uH it acts as a high impedance to L and C. the primary is a resonant parallel tank and the so is the secondary. If this was a flyback the voltage will be determined by the Q and turns ratio of the transformer. The calculations are complicated. When the primary and secondary are in resonance at the same frequency your voltage amplification will be greatest. When a parallel tank becomes resonant the circulating current is only restricted by the resistance in each component L & C. Because of this the type of L and C is important. I prefer a primary circuit that uses a polypropylene snubber capacitor because of the low series inductance and resistance built in the component known as ESR and a primary winding with only 4-6 turns of heavy wire say 14 AWG. This reduces resistance and increases primary circulating current which we want. The way I tune these oscillators is to add or remove capacitors to adjust the oscillator's primary frequency. When you are tuned the secondary output will reach a max. If you understand tesla's the principle is the same where you use some type of switch to get a primary to oscillate with the secondary. if you wanted to make a comparison a VTTC uses the same principles to transform voltage. I hope this is helpful and I didn't confuse you further
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
The factor of pi is because (1) the tank voltage is a sine wave, and (2) as Marko points out, the average voltage across an inductor must be zero.
By (1) the voltage at the transformer centre tap is a full rectified sine wave, whose peak amplitude is half the peak amplitude at the ends. Using integration, you can prove that the peak value of this waveform is pi/2 times the average. Therefore, the peak voltage at either end of the coil is pi times the average voltage at the centre tap.
By (2) the average voltage at the centre tap is equal to the DC supply voltage.
Therefore, we have proved that the peak voltage at either end of the transformer (which is where the FETs are connected) is pi times the supply voltage.
For the version with two chokes and no centre tap, the math is different but the answer is the same. Half a cycle goes missing, so the peak-to-average ratio is doubled and becomes simply pi.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Steve McConner wrote ...
The factor of pi is because (1) the tank voltage is a sine wave, and (2) as Marko points out, the average voltage across an inductor must be zero.
By (1) the voltage at the transformer centre tap is a full rectified sine wave, whose peak amplitude is half the peak amplitude at the ends. Using integration, you can prove that the peak value of this waveform is pi/2 times the average. Therefore, the peak voltage at either end of the coil is pi times the average voltage at the centre tap.
By (2) the average voltage at the centre tap is equal to the DC supply voltage.
Therefore, we have proved that the peak voltage at either end of the transformer (which is where the FETs are connected) is pi times the supply voltage.
For the version with two chokes and no centre tap, the math is different but the answer is the same. Half a cycle goes missing, so the peak-to-average ratio is doubled and becomes simply pi.
Oi Steve,
Here I made a diagram of the inductor voltage for the centre-tap inductor as you described:
If that is not clear, I took the supply voltage to be 1V in both cases (that's where the -1 factor comes from) and the sine waveform has a peak at pi in the first diagrams, and pi/2 in this new one. Both leads to pi*supply voltage half-pulses on mosfet's drains.
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royer Oscillator by Marko
