Hello,

I would like to start off by saying that there is an old thread that is similar to this one: http://forum.allaboutcircuits.com/t...-regulator-to-constant-current-source.107534/
However, the rules state that I shouldn't just hijack someone else's thread and should instead reference it in a thread of my own.
So here we are.

What I am attempting to do is convert a high current switching buck power supply to be a constant current source for driving a high powered diode. (Don't worry, I have the proper eye protection.)

In the older thread, the stated solution was to manipulate the FB pin of the IC using a calibrated resistor in-line with the load and an op amp to generate the desired voltage signal at the desired current. My question is whether this would be possible with my setup, which uses the attached schematic and has a "Sense" pin. The datasheet pdf for the controller is also attached.

For reference, my operating values are 12V IN, 2.2V, 30A OUT.

Thank you,
Obli

 

I should mention that the "VAC" on the schematic was for simulation purposes and is not present during operation of the circuit.

 

My question is whether this would be possible with my setup, which uses the attached schematic and has a "Sense" pin.

it will work.

i have done this many times over. from a topology perspective, switching mode led drivers are no different from a smps.

from an efficiency perspective, you want to pick one with the lowest Vfb.

 

Thank you for the response.

I'm guessing that Vfb is the feedback voltage.
I understand that keeping the resistor value to a minimum should minimize power dissipation, but what do you mean by minimizing Vfb?

 

the power dissipation on that resistor is proportional to the voltage on it, squared. the voltage on that resistor is chip-specific. so you should pick a chip with as low of a feedback voltage as you can. you will notice that pretty much all switching-mode led drivers have very low feedback voltage, for that reason.

 

the power dissipation on that resistor is proportional to the voltage on it, squared. the voltage on that resistor is chip-specific. so you should pick a chip with as low of a feedback voltage as you can. you will notice that pretty much all switching-mode led drivers have very low feedback voltage, for that reason.

I see. Since I already have the power supply, and the feedback voltage appears to be the actual output voltage (as far as I can tell), I guess I will just have to go with a small resistor. Since the output is 2.2V 30A, my equivalent resistor would be around 0.0733 Ohms. Since I would be amplifying that with an op amp, I should be able to reduce this further, to something like 0.018 Ohms right? Then I would just need to have a gain of 4 through the amp.

 

you probably want to read a little on how this approach works - i have a couple examples on my blog page.

essentially, the load and the current sense resistor form a divider, with the middle going to the feedback / sense pin of the converter. the load is on top of the current sense resistor.

you pick the sense resistance so that it drops Vfb at the target current levels. the converter will maintain the right output voltage so to deliver just the desired current through the load and the current sense resistor.

in your case, the power dissipation on that sense resistor would be 2.2v * 30a = 66w, assuming you want to deliver 30a into that little sense resistor.

with that much current, I think you are far better off with an off the shelf commercial product.

 

you probably want to read a little on how this approach works - i have a couple examples on my blog page.

essentially, the load and the current sense resistor form a divider, with the middle going to the feedback / sense pin of the converter. the load is on top of the current sense resistor.

you pick the sense resistance so that it drops Vfb at the target current levels. the converter will maintain the right output voltage so to deliver just the desired current through the load and the current sense resistor.

in your case, the power dissipation on that sense resistor would be 2.2v * 30a = 66w, assuming you want to deliver 30a into that little sense resistor.

with that much current, I think you are far better off with an off the shelf commercial product.

Right, just using a resistor alone would dissipate a lot of power. In the older thread I linked to however, they were able to use an op amp to amplify that voltage drop to minimize power dissipation across the resistor while still getting a controllable response.

 

Amplifying the sense voltage would help. A regular opamp or a dedicated current sense amp would work.

But you have to experiment for stability.

 

Alright. I'll give it a try and see what happens.
Thanks for the responses.

 

Hmmm... I seem to have let the magic smoke out of something.
Looks like I'll have to try another power supply solution...

 

Looks like I'll have to try another power supply solution...

there are many ways a particular implementation can fail: bad theory, bad connection, bad parts, bad luck, ...

without your providing more information, others cannot help you.

 

Specifically, what happened was that I absentmindedly connected the wrong wires and caused something to make a very audible "pop" when power was connected. The main controller board was a pain to solder in and I don't want to try desoldering it at the moment to investigate.

I do have an alternative solution in mind though. Please see the attached documents.
I plan to take several LM3409 LED drivers and combine them in parallel using level sense shifter circuits to ensure they share the load.
The attached schematic is a bit low-quality, so I apologize. I also attached the datasheet for the LM3409 and an article describing using this method to connect parallel LED drivers.
I think this would be the simplest solution if it works, and pretty easy to implement since it can be done with DIP rather than SMD IC's.

What do you think?
Also, would it be better to make a new thread for this?

 

it should work. the pnp pair is essentially a high-side current sense amplifier. the difficulty I see with the maxim implementation is that the two pnps have to be reasonably matched - you cannot use degeneration here unfortunately.

also, a minor issue is that the design needs sufficient voltage drop on the sense resistor. at high current levels, this can create a problem.

a simpler approach is to use a high-side current sense amplifier, either a dedicated chip or one built around opamps.

however, i'm not sure if it will solve your problem. the maxim solution is only useful if the load is not divisible and it needs more current than one driver can deliver: one led, or one led string. if you have multiple leds, it is much easier to just split the leds into individual strings and drive them accordingly.

this takes me back to your case. it seems to me that you already have this buck converter that can deliver sufficient current. So I'm not really sure how this approach would help you.

 

Well the idea here is that the individual components cost much less than my existing buck converter, which I may have damaged irreparably. If this new solution works, I could apply the much cheaper and modular LM3409 design to other projects as well. Then any level of current requirement (within reason) could be met by just adding or removing a few modules.

It makes sense that the voltage drop on the sense resistor would be a problem like before, but the existing LM3409 "typical application" circuit also uses a sense resistor at high loads (up to 5 amps), so I think it should be fine, right?

 

From the datasheet for the LM3409, it looks like the required voltage drop is around 0.25 V.
1.25 W of dissipation at 5 A seems acceptable.

Also, regarding the pairing of the pnp transistors, the recommended component from the Maxim implementation is this.

 

it just seems to be complicated, with an expensive p-ch mosfet + other parts. on the flip side, the external switch allows more current capability and it operates at high switching frequencies -> small coils.

i would look around for easier solutions.

 

Note that the Parallel Coupling pdf attached in post #13 contains errors: R4 and R8 shown in Fig 1 have totally wrong values, although the formula for calculating them is correct.

 

Thank you for the replies. Since the two threads seem to be overlapping, I'll just leave a link to the other one here.
I think it would be more suitable to carry on the conversation there since the focus has shifted away from converting a SMPS to current sensing.