# explain the working of this circuit [closed]

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][2]][2] 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 BrownMar 23 at 16:10

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• The schematic is very badly laid out. Where did you get it? What do you understand and where are you stuck? – Transistor Mar 22 at 20:48
• what is the working of Q1 and Q2? I think Q2 acts a switch.explain me the whole circuit, – Stranger_sid Mar 22 at 21:02
• I believe they act as switches. I want to step up the dc voltage from 1V to 10v using bjts in particular without using MOSFETs. I'm not supposed to use any kind of pWm . – Stranger_sid Mar 22 at 21:07
• do you understand it may not work with an inverted PNP? stick to circuits that work and are well documented, and never make an illogical (non-conforming) drawing unless its a paper napkin – Sunnyskyguy EE75 Mar 22 at 21:08

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

• can you please explain how to calculate the frequency of switching? – Stranger_sid Mar 27 at 16:03
• I wonder if any other expert here can do this? perhaps pipe, Bimpelrekkie, Bruce Abbott, winny, Edgar Brown ?? I might answer if it is re-opened, but that's up to you to correct the question and make it more specific ( to appease the Mods ) – Sunnyskyguy EE75 Mar 27 at 16:12
• e.g. start with my image pasted into your question and ask the above question, then you will figure out how it works (OK?) – Sunnyskyguy EE75 Mar 27 at 16:36