buck-boost converter question

[janice]

1. what is the importance of the output current?
2. what is switching frequency?
3. importance of duty cycle?

 

[simonbramble]

the output current is the current available to power the load. If your available output current is less than that required by the load, the output voltage will start to fall. The switching frequency is the frequency at which the circuit oscillates. Generally the faster the frequency, the smaller the components. The duty cycle is largely dictated by the input voltage and the output voltage and is independent of switching frequency. See my website for a more detailed look into all of the above. I hope this helps

 

[janice]

so if the frequency is fast, then is it always good thing? and if the duty cycle is fast, then what is the advantage?

 

[throbscottle]

Hi Janice, welcome to the forum

If the frequency is "fast" (I assume English is not your first language, and you mean "high"), you can use an inductor with fewer turns of wire, and smaller capacitors, BUT you have to be more careful about how the circuit is laid out, because it is more sensitive to inductances and capacitances you did not intend to be there. So it is a good thing, but not always.
I don't quite know what you mean by the duty cycle being fast? It is the mark/space ratio of the pulses which drive the inductor. You don't want the pulse of current through the inductor to last too long or it will saturate, but you do need it to be long enough so you can get the power you need back out of the inductor. So it has to be "just right".

Hope this helps

 

[Flyback]

I got you a simulation of very basic buckboost behaviour, for freely downloadable ltspice from linear.com
Just rename it with .asc
then open it in ltspice,
hit running man,
watch at your leisure...

(woops, I didn't realise you could post the .asc directly, ignore the .txt file then)