anditechnovire
Newbie level 6
Ok
Anywhere, it a little bit uneasy to comprehend.
Ok thanks for your answer. So the flyback voltage on the secondary winding is dependent on the level of regulation on the feedback winding, not the turns ratio.No, what is being said is the output would not stabilize unless there was a load. In a Joule Thief, the LED is the load, in your schematic, it is the 330 Ohm resistor across the output. If you remove it, the spikes in secondary voltage will make it rise much higher.
There is no output voltage stabilization at all in the circuit, it relies upon the feedback winding picking up the same signal, although not necessarily the same voltage as the output winding. If one produces more voltage, so will the other. When the rectified feedback voltage exceeds the Zener diode breakdown voltage it 'strangles' the transistor and limits the current it can pass and hence reduces the output. When running it will try to reach an equilibrium point where the transistor is conducting just enough that the Zener holds it back. Under increased load, there will be less voltage on the feedback winding so the Zebner will conduct less and the transistor will switch harder.
A much better system is not to monitor the feedback winding but to monitor the real output voltage, that way the losses in the transformer and rectifier are also taken into account. The problem that introduces is that the secondary is no longer isolated from the AC lines so it poses a risk of electric shock. That is why you will see an opto-coupler in most small power supplies of this kind. The output voltage is converted to current on the input side of the optocoupler (LED side) and its output side goes to the switching transistor. It works like the 1N4148 and Zener diode in your schematic
Anywhere, it a little bit uneasy to comprehend.