After the BOOST topology switch tube is turned on, the current flow problem

我本将心向明月

After the BOOST topology switch tube is turned on, the current flow problem [Copy link]

 

Today I saw an explanation of Boost topology on the Internet. As shown in the figure below, when K is turned on, why is the C terminal charged to form a stable voltage?

Why isn't the turned-on K short-circuited, so that the voltage across segment C is zero?

I just started to look at the switch power supply, please help me~

呜呼哀哉

You see, you are wrong. When the switch is closed, L is charged. The inductor is also an energy storage element, 0.5L*I*I.

Another question, why is the voltage across C not zero? Energy does not disappear, it must be consumed. The energy stored in a capacitor is 0.5*C*V*V. After closing, the diode is reverse biased, and the capacitor can only discharge to the load. Unless your load is very heavy and short-circuited directly, how can it be discharged immediately?

maychang

"When K is connected, why is the C terminal charged?"

When the switch K in the first post is closed, the positive end of the diode D is short-circuited to the negative end of the input power supply, and D will not conduct. How can the capacitor C be charged?

Capacitor C is charged before K is closed (of course, K is in the off state). The self-inductance electromotive force in inductor L is superimposed on the input power supply, and capacitor C is charged through diode D.

maychang

The blue words below your PPT clearly say “When K is connected…C discharges to R”.

我本将心向明月

Alas, published on 2021-5-31 17:30 You see, you are wrong. When the switch is closed, L is charged. The inductor is also an energy storage element, 0.5L*I*I. Another question, why is the charge at both ends of C...

Thanks for your answer

我本将心向明月

maychang published on 2021-5-31 17:33 "When K is turned on, why is the C terminal charged?" In the first post, when switch K is closed, the positive end of diode D is short-circuited to the negative end of the input power supply, and D does not...

Yes, the voltage across C is the voltage accumulated when K is turned off.

呜呼哀哉

I originally turned my heart to the bright moon and published it on 2021-5-31 17:51 Thank you for your answer

Power supply is not simple. If you want to read a book on this subject, I think DCDC's Sangayamani Takla's "Learn Switching Power Supply AZ from 0" is very thorough.

Jacktang

What the switching power supply needs to pay attention to is the self-inductance electromotive force in the inductor L.

okhxyyo

Alas, published on 2021-5-31 18:46 Power supply is not simple. If you want to read a book on this subject, I think DCDC's Sangayama Nitakla's "Learning Switching Power Supply AZ from 0" is very thorough...

See if it is this book:

Switching Power Supplies A to Z - Power Technology Related Data Download - EEWORLD Download Center

呜呼哀哉

okhxyyo posted on 2021-6-11 15:49 Check if it is this book: Switching Power Supplies A to Z - Power Technology Related Materials Download - EEWORLD ...

That's it. I personally feel that this DCDC theory is easy to understand. It's a pity that I haven't entered the power supply industry, otherwise I would have used it all.