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[SOLVED] L4971 circuit explaination

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nikhillife11

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Hello friends,

I need some clarification as to why a Schottky diode is used in revese bais in the attached document. I am not understanding why they have used this diode instead of zener diode

thanks and Regards

Nikhil
 

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emontllo

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Hello Nikhil,

The L4971 is a buck (step down) switching regulator with integrated switch.
Typical signals in a buck converter:
http://i.cmpnet.com/powermanagementdesignline/2006/03/NatSemi_BuckBoost_Fig1.gif
Buck regulators have a high side switch (in this case is a MOSFET integrated in the device) and a low side switch. In your schematic the low side switch is this schottky diode. The inductor forces the current accross D1 when the MOSFET is turned off.

Ernest
 

nikhillife11

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Hey Emontllo,

Can you please explain in details plz..... How does current flow through D1 when its reverse baised even if MOSFET is turned OFF...still it remains in reverse state. So please if u can elaborate it please.

Thanks and regards

Nikhil
 

ea.arun

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when internal High side mosfet is turned of inductor tends to keep thje direction of the current in the same direction throgh the capacitor and schottky diode.Inductor acts as the source during the MOSFET off time.Schottky is used due to its reduced forward drop and fast reverse recovery.

try these....
DC-DC Converter Tutorial - Maxim
Switching-Mode Power Supply Design Tutorial
 

emontllo

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Hello Nikhil,

When the internal L4971 MOSFET is turned on, the inductor current rises. When this switch is turned off the inductor "tends to mantain the current" it and the inductor voltage becomes negative. This is why the D1 starts conducting current.
Following is a explanation of the buck converter operation:
SMPS Basics - The Buck Converter

Ernest
 

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