Boost converter turn-off time

Thread Starter

anhnha

Joined Apr 19, 2012
880
For the boost converter below operating in DCM mode, the voltage conversion ratio is calculated as follows:

M = V/Vg = 1 + D1/D2
where
V, Vg are output and input voltage respectively.
From the formula, it is clear that the conversion ratio doesn't depend on D3. So, is there any requirement for D3 here?


Boost converter - intuitive.PNG
 

ronv

Joined Nov 12, 2008
3,770
From the formula, it is clear that the conversion ratio doesn't depend on D3. So, is there any requirement for D3 here?
Just that the lower the voltage drop the more efficient the conversion.
And of course the voltage and current ratings must be high enough.
 

dl324

Joined Mar 30, 2015
9,620
How does it depend on load current?
When the voltage at the cap drops below the target voltage, the switch needs to turn on. A higher load will discharge the cap faster, making it necessary for the switch duty cycle to be higher.
 

Thread Starter

anhnha

Joined Apr 19, 2012
880
When the voltage at the cap drops below the target voltage, the switch needs to turn on. A higher load will discharge the cap faster, making it necessary for the switch duty cycle to be higher.
Can I reduce switching frequency instead of increase duty cycle? I think it is the same here.
 
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