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Transistor Based, Switched Capacitor Power Supply
Transistor Based, Switched Capacitor Power Supply
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Old 3rd December 2020, 07:09 AM   #21
StevenStanleyBayes is offline StevenStanleyBayes  Canada
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Join Date: Jan 2012
Location: Ottawa
A big mish mash with mathematics. Sorry. Here are quick corrections.

1. RMS of a pulse signal with an amplitude of Aa and a pulse width of t and a period of T ( t < T ) :

Arms = Aa * SQRT ( t / T )

2. Power

Urms = Ua * SQRT ( t / T )
Irms = Ia * SQRT ( t / T )
P = Urms * Irms = Ua * Ia * t / T

The power is linear with the duty cycle and not with the square of the duty cycle.

3. The capacitor voltage, Uc, current, Ic, and charge period, tc, are linear :

Ic = C * Uc / tc

This means, when a capacitor provides current Ic1 to a load for a period of tc1, the capacitor has to be charged with a current Ic2 for a period of tc2 and Ic2 = Ic1 * tc1 / tc2.

Therefore, IanHegglun is right : power dissipation of the transistor cannot be saved by switching the buffer. This is why a buffer is a buffer.

The mish mash was caused by two problems :

1. RMS mathematics : Because RMS of a square wave with amplitude Aa ( Aa and - Aa ) is Aa, an automatic consideration is the RMS of a pulse with amplitude Aa ( Aa and 0 ) and a duty cycle of 1 /2 should be Aa / 2. This is not true, because the RMS is Aa / SQRT ( 2 ). Thus, the dissipated power by the transistor is the same when compared a buffer and a buffer which opens and closes.

This is a big mistake and, thanks to IanHegglun, I have realised this.

Physically, since the energy dissipated by the transistor is Uce * Ice and Uce is constant, and the capacitor charge is linear with Ice, there is no way to save current by switching. The current out for tc1 must be compensated by linearly higher current for tc2.

2. Thought of a push pull buffer, where each of the transistor consumes half of the power ( as per the formula Urms * Irms * t / T, where t = T / 2 => each transistor consumes half of the output power.

This can still be achieved with switching two transistors instead of paralleling them. There is no any advantage, better to be paralleled, although this requires additional resistors for bipolar transistors. MOSFET’s can be paralleled directly, without additional resistors. However, the buffer does not need to be switched at all. There would not be any switching noise. But this is not the point. The point is to, theoretically eliminate the power consumption of the transistors, which is impossible with an ideal voltage source. This can, theoretically, be done with real voltage sources, which have an output impedance. Then, the collector and the emitter would have the same voltage and the voltage drop is over the output impedance of the voltage source.

3. This applies to an arrangement, which was the biggest contributor to the mistake. I thought of the voltage ripple of the rectifier capacitors. The voltage ripple can, theoretically, be 150V. This way, the transistor will see Uce between 0V and 150V with close to triangular shape. This lowers the RMS of the Uce.

4. An interesting theoretical mistake. An ideal transistor will have Rce of 0 Ohms and will allow infinite current. Thus, a given Uce will make an infinitely high power consumption of the transistor, but for a zero period. But, then, the capacitor will, also be charged to the voltage source voltage for zero period by the ideal voltage source. In other words, the capacitor will be charged to 170V and not to 18V.

Anyway, the only way for the suggested idea to work is because of the ripple voltage after the rectifier. Thus, Uc is not 180V and not 170V and not 160V. Uce is between 0V and 180V. The ripple can be calculated to go from, say, 30V to 180V. This lowers Ucerms. The transistor still consumes, but, not as much. The higher the output current, the higher the ripple. The lower the output current, the lower the ripple, but, the lower the charge current too.

Anyway, big mistakes, big mess, big trouble. This is what happens when power is not considered and not calculated. Although the document reflects the high ripple voltage, the document will be removed and redone.

Regardless of how the document is redone, to rely on the ripple voltage is not a good idea and switching is not, theoretically, necessary in such a case. Synchronised switching at low Uc may not help either, because, the current has to be higher. I cannot find a way and, probably, there is no physical way, regardless of whether the transistor is high end or buffer, MOSFET or bipolar. The only way is to reply on high ripple and the output impedance of the source, which, can be increased by a resistor, but, this defies the purpose.

Again, the document is to be removed and, in case republished, the document is only for theoretical discussions and has no any practical application, maybe after a transformer which gives some voltage and another is needed, yet, there are very many, better, circuits, such as buck converters, capacitor pumps, etcetera.

Anyway, sorry for this mistake. The document is to be removed. Thanks IanHegglun for the help and advise.
 
Old 3rd December 2020, 07:27 AM   #22
StevenStanleyBayes is offline StevenStanleyBayes  Canada
diyAudio Member
 
Join Date: Jan 2012
Location: Ottawa
You did not answer the questions I have asked. In case these are allowed, similar is to be allowed too, regardless of how the standard is spun. Voltage source is a voltage source, regardless whether this comes from a power distribution transformer or an additional transformer. The voltage source is characterised by output voltage and output impedance. When you put a high power transformer with high voltage and low output impedance, you do the same.

The problem is not whether this is allowed or not, because this is and must be. Or no higher voltage than 60V maximum difference must be allowed ( or no high current, no high power ).

The problem is not whether standards allow or not. The problem is theoretical as outlined by IanHegglun with the theoretical concept and design.

However, in case the theoretical concept and design had been right, there is no problem with safety for the integrated system. Just as safe as a stove.

In case you are interested in transformerless power supplies, search with Google. There are, even, application notes by component manufacturers, such as Microchip and other companies, where, they suggest a transformerless power supply with ( simplified ) a resistor and a zener. True for low power, but, still. Some may have for a higher power. These companies are allowed in Australia, I am sure.

Had the concept been true, the concept could have been used for low power too and not only for high, just like the other publications.

In case you are interested in the theoretical mistakes and mish mash, which, I have created, please, read the previous post as well as the post of IanHegglun.

There are non electrical problems, such as fire, but, these can be countered by the case arrangements. Fire can happen when a stove is forgotten on too.

Again, the concept is not safe, but, is just as unsafe as the other concepts, as you call them, the " approved " ones too, which use a powerful transformer. These do not include the tube amplifier and the CRT TV's power supplies only. They include the power supplies of powerful, solid state, amplifiers too. The question is not what components an amplifier contains, the question is the voltage and the current, power of the transformer.

Imagine you have a 1KW concert amplifier with a 4 Ohm speaker.
 
Old 3rd December 2020, 07:28 AM   #23
StevenStanleyBayes is offline StevenStanleyBayes  Canada
diyAudio Member
 
Join Date: Jan 2012
Location: Ottawa
In other words, the problem is systematic and not related to safety and standards.
 

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