I really appreciate your response. Infact this response has triggered questions from how the original circuit works.The diode symbols are wrong in several places!
Basically, the AC feedback path is through the lower 4n7 capacitor and 220R resistor.
The voltage from the feedback winding is rectified by the 1N4148 and charges C2. Note it is wired to produce a negative voltage. When the voltage reaches about 5.6V, the Zener diode starts to conduct and the bias is drawn away from the transistor, reducing the current in the transformer primary and hence the output voltage.
This kind of regulation isn't very precise because it can't compensate for the actual output voltage, only an approximately proportional version of it in the feedback winding. Losses in the transformer and output rectifier and changes in the load are only partially compensated.
Brian.
So will the base winding have to reverse it polarity, or will the transistor have to be forced out of saturation?To me it looks like a blocking oscillator. Similar to how the joule thief operates.
1) 330k resistor slightly biases the transistor On.
2) Admitting current through winding #1.
3) Inducing current in winding 2.
4) Sending increased bias to transistor.
5) Current builds in winding 1, and plateaus.
6) Flux stops changing in transformer. Bias ceases from winding 2.
6) Shutting off the transistor, or nearly off.
Please does the feedback rectifier conduct when the transistor is off?The diode symbols are wrong in several places!
Basically, the AC feedback path is through the lower 4n7 capacitor and 220R resistor.
The voltage from the feedback winding is rectified by the 1N4148 and charges C2. Note it is wired to produce a negative voltage. When the voltage reaches about 5.6V, the Zener diode starts to conduct and the bias is drawn away from the transistor, reducing the current in the transformer primary and hence the output voltage.
This kind of regulation isn't very precise because it can't compensate for the actual output voltage, only an approximately proportional version of it in the feedback winding. Losses in the transformer and output rectifier and changes in the load are only partially compensated.
Brian.
So will the base winding have to reverse it polarity, or will the transistor have to be forced out of saturation?
Ok thanks.the xtor turns off when it can no longer support the peaking current in the Tx, thus the voltage on the bias wdg collapses ( and then reverses ) helping turn of xtor T1, with the aid of C1 (when it is fully charged ) - it is not a flash ckt ...
Which ones are wrong ... ?The diode symbols are wrong in several places!
Brian.
Ok, I get that.When C2, the 10uF 16V cap gets to about 5v6 plus a wee bit it tends to hold the main xtor off, the coils supplies 5v6 + 0.6 volts on the 4148 to charge the cap, this is about 6.2V on this wdg, for the 7;15 ratio on the Tx the output will see 15/7 * 6.2 = 13.28 less the o/p diode drop, say 0.8 volt under load = 12.5 volt ( higher at no load ).
There is no current limiting and not very good regulation of Vout with load - it is a ckt that can fail easily - esp at start up due to high currents in the main transistor ...
Ok I kinda get you, but please you need to understand that am quite a novice, in this area.the voltage in any wdg on a flyback, goes up, at fet turn off, until limited by the clamping action of connected circuits - in this case there a re two - the o/p diode and cap ( and load ) and the feedback diode and cap ( and its 330 ohm load ) ...
where current can flow in the load(LED). So in this case of rcc flyback where there is no led, it really making me confused.
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