a dumb question about boost comverter

[sean415]

Hi,

Recently I am learning boost converters. Here I have a "dumb" question, can anyone help me?

From textbooks, it is known that if other loss is ignored, the Vout = Vin /(1-D), where D is the duty cycle of the switching signal.

Usually there is a feedback network that senses the part of the Vout and then compares it with a reference voltage Vref. The output of this comparison controls the D, therefore make sure the Vout = Vref.

My confusion is, since the feed back network lets Vout = Vref, then what is the importance of Vout = Vin/(1-D)? It seems to me this can be done even with a constant D.
Because when Vout is less than Vref, then feedback starts the switching. This then keeps boosting the Vout up. When Vout is more than Vref, then feedback shuts off the switching. This lets Vout decrease.

With a bigger D, the boosting takes less time to reach Vref. With a smaller D, it takes longer time to reach Vref.

Is my understanding correct?

Thanks a lot!

 

[grizedale]

No you cannot use a constant D.

At heart of control loop is opamp with negative feedback.

I hope u understand it.

Opamp acts to make its + and - inputs the same.....................and thus you get your regulated vout.