Buck converter

Thread Starter

Dritech

Joined Sep 21, 2011
863
Hi all,

I did the following buck converter simulation which should output 10V 2A. On simulation it worked fine, but will it work in prectice? are the values selected good?

Any help would be appreciated.



Edit: The voltage on the transistor base is 15V.
 

crutschow

Joined Mar 14, 2008
24,969
Very little increased efficiency is obtained when going from 12V to 10V with a switching regulator.
A linear regulator is 83% efficient for that conversion, so even using a 90% efficient switching regulator will only gain 7% efficiency or about 2W less dissipation.
And you'd be hard pressed to get 90% efficiency using a BJT emitter follower for the switch since it has a minimum drop of about 0.7V when conducting.
 

Thread Starter

Dritech

Joined Sep 21, 2011
863
Don't see how it could work. No circuitry to regulate to 10V.
Thanks for the reply. This will be an open loop circuit, so I will connect the output to a multimeter and vary the function generator duty cycle (at the base of the transistor) till I get the disired voltage.
 

Thread Starter

Dritech

Joined Sep 21, 2011
863
Very little increased efficiency is obtained when going from 12V to 10V with a switching regulator.
A linear regulator is 83% efficient for that conversion, so even using a 90% efficient switching regulator will only gain 7% efficiency or about 2W less dissipation.
And you'd be hard pressed to get 90% efficiency using a BJT emitter follower for the switch since it has a minimum drop of about 0.7V when conducting.
Hi crutschow. I am going to implement this circuit just for the fun of testing a switching regulator :)
But before I buy the components and assemble it, I wish to hear from others with excerience if this would work in real world, or it I need to change something.
 

DickCappels

Joined Aug 21, 2008
6,387
It should work with the right Q1.

BUT it would work better and be more efficient if you use a PNP transistor so that the collector drives the inductor. If you drive the inductor with Q1's emitter there is a good chance that the free-wheeling current will flow through Q1 and the 12 volt supply instead of through D5. That will increases Q1's dissipation and reduce overall efficiency.
 
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