I'm basically trying to understand how this Dc to Dc step-up converter work. please help me.[![dc-dc step-up voltage converter 1v-9v]] I need to design a DC DC step up converter using bjts. Except the dc voltage of 1v, I'm not supposed to use any kind of external voltage. I need to get an output voltage of 5v. Please don't suggest me to use mosfets as I'm restricted to use them in this case. Explain me how the switching takes place and how the inductor charges and discharges. I think switching takes place at Q2 but not sure how. I don't have a perfect explanation for that. Please explain me the entire switching process.
closed as too broad by pipe, Bimpelrekkie, Bruce Abbott, winny, Edgar Brown Mar 23 at 16:10
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The Q1 PNP reversed gives unity current gain so the 10nF doesn't overdrive Q2 when it switches ON.
It is a positive AC feedback loop from two inversions of base to collector current and thus can become an astable oscillator with sufficient gain of Q2.
A sawtooth voltage on Q1b is expected until V2b conducts with pulses Q1 off only to repeat the cycle.
The Q2c output current from L1 is instantly switched thru the Schottky diode to continue the inductor current pulling up the voltage in typical boost fashion.
Here is an improvement