Why Is Adjusting the Duty Cycle Necessary in Boost Converters?

[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 will make sure that Vout = Vref, then what is the importance of Vout = Vin/(1-D)? Why do we need to adjust 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 let's Vout decrease.

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

Is my understanding correct? Can boost converter works with a constant D?

Thanks a lot!

 

[fleem]

The maximum power a supply can deliver is at some output voltage less than the no-load output voltage (assuming no negative feedback). That's why you design a switching supply such that its maximum no-load output voltage (without regulation) is something notably more than the working output voltage, then use negative feedback to regulate it to the working voltage.

 

[sean415]

Thanks fleem.

I guess I know how it works. The output power that a boost converter delivers Pout=(Vin*Vin*Ton*Ton)*f/(2*L)

So if Vin, f and L are all fixed, then you need a certain Ton to get certain Pout. Ton needs to be adjusted if Vin or Iout changes.

Alternatively, if both Ton and f can be adjusted, then one can fix Ton and adjust f.