How do I convert a buck regulator to constant current source?

JWHassler

Joined Sep 25, 2013
275
That seems like a challenging choice for constant-current supply.
You'll have to tell the voltage-control pin to sink a current that causes the module's output voltage to be the Vf of your LED that gives the current you want.
 

DickCappels

Joined Aug 21, 2008
6,091
The the most common approach to using a voltage output buck converter as a constant current buck is to put a current sense resistor between the load and ground and then feed-back the voltage across the sense resistor instead of the voltage across the load.
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
Ok, so this chip has a 0.596V voltage reference on the FB pin with a 10K pull up resistor.
How would one use a 0.01R as feedback to FB?
How is the feedback of this sense resistor actually limiting current and not voltage?
 

Lestraveled

Joined May 19, 2014
1,946
Yes, you can configure this chip as a constant current source. As Dick said, you would put a current sense resistor in series with the Gnd pin and the load. The Feed back pin would connect to the load side of the resistor. You would size the resistor to drop .596 volts at the current you wanted. For instance, a .596 ohm resistor would cause the chip to regulate at 1 amp. The 10K internal resistor would have little effect on the very low resistance sense resistor. The down side to this design is that the gnd of the load would have to be isolated.
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
I'll be driving it at 6A.
That looks to me very inefficient and too much heat to dissipate. 21W is a lot of loss. It's almost twice as much what my load requirement is.
I only require about 11W in the load. So, I would have to generate 32W, lose 21W to gain 11W?
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
Duh, a guess it was too early to think.
Rethinking this, I would actually have .1 sense resistor with a loss of about 3.6W, which is getting better.
Now, How would I go about using as .01 sense resistors?
Do I just amplify by 10?
I swear I tried that on the simulator and it didn't work.
 

DickCappels

Joined Aug 21, 2008
6,091
Yes, you can add a non-inverting amplifier to amplify the voltage drop across a low value resistor. You can also put a series resistor between the feedback input and the current sense resistor such that the current from the 10K resistor causes an IR drop in the added series resistor so that you get to 0.596 volts when there is much less across the sense resistor.

Your idea of using a noninverting amplifier would most likely be the best solution.
 

Thread Starter

newhobby

Joined Mar 26, 2014
24
Ok, so I simulated and adjusting a few things I was able to get rid of the horrible ripple I was getting before.
I came up with a .05 sense resistor and a op amp with gain of 2, which gives me the 0.6V I need.
It regulates fine, so let's add another component.
I would like to use this source to power 3 different LEDs.
How would I go about doing this?
Do you think if I use triple high side FETs, it would work?
 
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