SNVA829 Simple Constant Current/Constant Voltage Buck Converter

[hioyo]

The below is from TI application note SNVA829.Using buck converter as a constant current source.

May I know how the equation is obtained for CC.I tried with nodal analysis but failed.

Can someone please help me.

 

[KlausST]

Hi,

An example.
Let's say there is a voltage drop across the shunt of 0.1V
Then the Opamp regulates in a way that there is the same voktage drop (0.1V) across RIn.
This causes an emitter current of 0.1V/ Rin. Let's assume RIn to be 1k, then the emitter current is 0.1V/ 1k = 100 uA.
Let's assume the hFE of the bjt to be very high, then the collector current is the same as the emitter current = 100uA.
Since Vfb is considered constant, the RFb current is constant, too. Let's assume Vfb = 1.5V and Rfb = 2500.
Then I_Rfb = 600uA.
You have voltage regulation as long as there is current through the zener.
But zener current is 600uA - collector current. Now 600uA-100uA = 500uA.
If you increase the current: Vshunt becomes bigger, ... Icollector becomes bigger. Until collector current is 600uA..
Then you get into current limit mode.

Klaus

 

[cupoftea]

Yes Klaus is correct....its the typical hi side current monitor.....
The way to look at it is..
In regulation, the + and - inputs of the opamp will be at the same voltage. (law of opamps with neg feedback, and since current into opamp input is zero)
...then V(Rsen) = V(RIN) (kirchoffs law, taking + and - to be at same potential)
...so I(RIN) = V(Rsen)/RIN (ohms law)
...also, I(RIN) = I(RFB) ('law' of NPNs when ON)

So then you can calculate VFB like that.

But to be honest, you do not show the error amplifier in the controller.......but presume reference on error amp = 2.5V....then it woudl act to regulate VFB to 2.5v......So there you have your starting point...from which you can back_calculate I(rsen)