MoIL manipulate with times when is T1 open and when is closed. When T1 is opened through its collectors weight (primar of transofrmer) current Ic flows.



T = T1 + T2 = const.
T = 1/f = const.
f = 40 kHz
T = 25 µs






1. T1 = T2 = T/2 (stacionarno stanje)
2. T1 > T2
3. T1 < T2

Case 1:
In some stationated condition with constant weight, T1=T2=T/2. This means that transistor is 50% closed and 50% opened.

Case 2:
If becomes rise of use, consumers will take more Ampers then before. (from 8 to 12). Output voltage +5 V will start to fall and will fall to 4,99 V and supervisor block will detect that and "inform" MoIL about that. MoIL will react with extending time of opened or closed T1. Longer T1 means longer openess of transistor or longer current Ic flew through transformer primar. This means that through primar flews more energy so more energy is transformed to secundar of transformer so it is kompenzated higher use of energy and output voltage is now 5 V.

Case 3:
If begins reduced use of energy and output current instead 10 A falls to 8A that will influence to output voltage that will start to easily grow and it will grow from 5 V to 5,01 V which will detect supervisor block and "inform" MoIL about it.

MoIL reacts with shorting T1 and lenghtenin T2 and thats why transistor is shorter time opened so collector current Ic flews for short time period and reaches minor value. That means less flew of energy through primar and time and through secundar of transformer and less energy comes to output and output voltage is back to 5V.

Reaction of power unit supply is very fast for duty load because everything is done within one period T (25µs) which is enough fast, for even the most sensitive components (RAM) not to notice it.