I have one doubt in this schematic. It has one diode D1. Why it is there won’t it will get short and all the current will drain through it.I have attached the schematic.

 

What do you mean "it will get short"? Why would it? It's a flyback diode for the inductor current.

 

Its diode D1 I am talking about. Its D3 diode that will work as flyback diode.

 

Hello,

I have taken the schematic from the PDF, so others can view it directly.

The D1 diode is for the reference voltage for the current setting.

Bertus

 

Ok I got it.Thanks Bertus.

 

So the answer to the original post, based on Bertus' reply is "Yes, is has current going to ground; that's the way its supposed to work."

My question (not intending to hijack) is whether the 2 mv/°C drift is compensated for somewhere.

 

What about D2? Why is this D2 used in reverse polarity ?

 

What about D2? Why is this D2 used in reverse polarity ?

IC1 is a LM393 comparator http://en.wikipedia.org/wiki/Comparator

 

Its diode D1 I am talking about. Its D3 diode that will work as flyback diode.

The circuit comes from this site ttp://www.instructables.com/id/Poormans-Buck/
R1 will limit the current through D1 to a safe value for the diode. Without R1, D1 would have been acted as short. And very quickly gone up in flames

 

D1 is acting as a crude voltage reference ~ 600 mv forward conduction voltage.

 

IC1 is a LM393 comparator http://en.wikipedia.org/wiki/Comparator

Yes, I know the IC is a LM393 comparator. but why do we obstruct the output voltage of comparator?

 

Yes, I know the IC is a LM393 comparator. but why do we obstruct the output voltage of comparator?

Because the LM393 has what is named a open collector output http://en.wikipedia.org/wiki/Open_collector

 

D1 is acting as a crude voltage reference ~ 600 mv forward conduction voltage.

Ahan yeah I see that! THanks a lot!