discrete buck converter understand

Can you please explain to me this discrete buck converter, i didn't understand the role of schottky diode,
and how it is forward biased ?
**broken link removed**

It conducts when the transistors turns off, collapsing the magnetic field around the inductor. The polarity reverses, making the diode conduct and passing stored energy from the inductor into the load.

Brian.

i didn't understand how it will conduct ? i know that when TR1 is switched off, but how the schottky will be forward biased ?

The inductor 'wants' the current it holds at the moment the switch goes of to continue. So, a new current path is created, through the diode which is then forwardly biased.

1.

I have a Youtube video showing animated simulations of switched-coil converters.

They portray flux fields building and collapsing, changing EMF, current bundles travelling through wires, capacitors charging and discharging.

www.youtube.com/watch?v=FT_sLF5Etm4

2.

Falstad's interactive animated simulator allows you to explore the operation of these converters...

Go to the links in my posts in the thread:

https://www.edaboard.com/threads/268178/

Can you please explain to me this discrete buck converter, i didn't understand the role of schottky diode,
and how it is forward biased ?

Click to expand...

Hi Yassin
It can be easily described as :
At start time , just consider that all of the transistors are turned off . R2 will allow some low current go through the base of TR3 . thus TR3 will try to be turn on ( as a key with limited current ) so it's collector will provide a low current path for base of TR1 till ground . so it will start the conduction too ( TR1 ) . an inductor at transient time is open circuit . so the current is low . so the voltage across the R1 is still low . but when it starts to store the energy , current will try to be increased . so when voltage across the R1 exceeds from 0.6 volt ( VBE ) then TR2 will start the conduction and thus TR1 will be turned off .
So with this trick , kind of PWM will be provided . So overall efficiency is good too .
And when Tr1 is turned off L1 will create an inertia ( according to the Lenz law ) and thus polarity of voltage across it , will be reversed . so thus D1 will conduct because it is forwarded now and then inductor will be discharged into the C2 and C3 .
I hope now you got the idea .
Best Wishes
Goldsmith

Hi goldsmith, thank you, i didn't understand how the voltage on the collector of TR1 will be 0 ? it is wriiten in the application note, and how the voltage across the inductor will be reversed ?

1. Understand that an inductor that is carrying current can not just stop doing that. It will attempt to maintain that current.
2. When TR1 is turned off, you can remove it from the schematic for better understanding.
3. So, now you have a coil that wishes to sustain the current it was carrying just a fraction of time before shutting off TR1.
4. What is the only "route" the current can take? D1!
5. So, now we have current running through D1. The anode of D1 is connected to GND (0V) and by it's very nature the cathode of a current carrying diode is about 0.7V below the anode. So in fact the anode of TR1 will not be at 0V but rather a bit lower than that ;p

Hi goldsmith, thank you, i didn't understand how the voltage on the collector of TR1 will be 0 ? it is wriiten in the application note, and how the voltage across the inductor will be reversed ?

Click to expand...

Hi yassin
When tr1 will be turned off thus it's collector voltage will be zero . furthermore , D1 will be in conduction situation so it will be the cause too . but before D1 starts the conduction TR1 is turned off so there is no path from VDC till it's collector .

About why voltage across the inductor will be reversed , it is simple ! Lenz law ! VL=-Ldi/dt did you know about this formula ? or did you see that till now ?

Best Wishes
Goldsmith

I think the problem is understanding the polarity across the coil. When the transistor is turned on, the collector side has a higher voltage than the output side so the polarity across it is + on the left and - on the right.
When the transistor turns off, the magnetic field collapses and produces a voltage in the opposite polarity so the collector end is now negative and the output end is positive. The diode is then in forward conduction mode and it holds the left side at around 0.3V above ground. Consequently the right side is 0.3V + voltage from the inductor which is enough to partially fill the gap while the transistor isn't conducting. So you can see the LED gets current both when the trasnsistor is conducting and when it isn't. This makes better use of the available energy and increases the circuits efficiency.

As goldsmith advises, look up Lenz's law, it expalns the way electrical energy is converted to magnetism and back again.

Brian.

Hi Betwixt, thank you for the explanation i will take a look at law lenz, but i have a question why you say that we have 0.3V above ground ? from where this value came ?

but i have a question why you say that we have 0.3V above ground ? from where this value came ?

Click to expand...

Hi Yassin
I think Brian's meaning was about forward voltage of the schottky diode .

Good Luck
Goldsmith

Correct! - sorry, I should have explained better.
Brian.

Correct me if I'm wrong but the cathode of the diode is 0.3 or whatever below ground.

You are right! Mind you, I also build things using transistors not trasnsistors - my spelling is even worse than my math :smile:

Brian.

betwixt said:

...trasnsistors...

Click to expand...

I noticed Digi-Key doesn't carry them