# Buck Current Regulator, help needed

Discussion in 'The Projects Forum' started by Eudoxus, Jan 31, 2013.

1. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
I'm designing an LED driver for a set of very high-intensity LEDs, three 3.2A LEDs with a voltage drop of 3.2V, in parallel, powered by a 10.8V or 14.4V array of batteries. So I need an LED driver >90% efficient in a 32mm diameter circular form factor that can push 9.6A at 3.2V. My work is cut out for me. I've designed a buck regulator based on a Texas Instruments TPS53319 controller with built-in switch. Problem is I've never designed a buck current regulator, only voltage regulators, and so I'm not sure how to convert the design.

Mainly I'm not sure how to determine the value of R1 (feedback sense) (in the lower left).

I'm pretty happy with the design if I can get this to work, simulation (for voltage regulation mode) shows efficiency of 94-95% and a 380mm^2 footprint, which should be very doable on the specified PCB size.

Thanks!

2. ### takao21203 Distinguished Member

Apr 28, 2012
3,577
463
0.1 Ohm or 0.01 Ohm are pretty standard.
Commonly used for these resistors.

You may consider a capacitor bank, not one large capacitor.
4.7uF are available for economic price, while 100uF cost more (as ceramic).

For such a high frequency you need special SMD electrolytic capacitors, which also can be used.

From my experience, the right coil is most important. They will work with many coils, but if you want max. efficiency, you need a matching coil.

3. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
Thanks! So just a simple 0.1ohm resistor should result in the circuit supplying just as much current as required to feed the LEDs? I'm still struggling slightly with the theory on this one.

This IC is spec'd to be able to work with all ceramic capacitors, so at 500kHz I think it should be ok. The output cap is actually low-ESR Tantalum and is fairly reasonable in price.

I spent some time looking over coils. By simulation I've picked a Wurth 744314150 (http://www.we-online.com/eisos/pdf/744314150.pdf) as the second-best one. There was one better, but it was twice the size. I decided to accept a 1% loss in efficiency for a 17% savings in footprint.

4. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
here is one way. Build one for each LED so they get proper current. The one shown is for 2.6A, you can tweak the .05 Ohm sense resistor to adjust that. Leave off RTR and the 1N4001 diode. The LM2576 is rated for 3A, it can probably do 3.2A if that's required and there are 5A versions available.

File size:
68.1 KB
Views:
107
5. ### crutschow Expert

Mar 14, 2008
13,001
3,229
For the TPS53319, the regulation voltage at VFB is 0.6V so you select R1 to give 0.6V at the desired current. For 9.6A the value would be 0.6/9.6 = 62.5mΩ with a resistor dissipation of 5.76W.

I have a concern about operating the LEDs in parallel since the one with the lowest forward drop will tend to hog the current. It would be better to operate them in series to avoid that potential problem. That also reduces the regulator current to 3.2A and will change R1 to 0.6/3.2 = 187.5mΩ with dissipation reduced to 1.92W.

6. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Very true. In parallel they need their own current source for each LED, in series you can use just one. For the circuit shown above, it means you would have to have a VIN available of at least 15V to power the 3 LEDs.

7. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
Some good points. Having the LEDs in series would pretty much require a completely new design from scratch though, the TPS53318/9 series can't deal with as small a drop as 10.8V to ~9.2V

I'm looking through some alternatives. A few might be more efficient...

8. ### spinnaker AAC Fanatic!

Oct 29, 2009
4,884
1,000
. With the configuration of leds you should consider the pt4115 instead. A whole bunch less components and pins. They only come in smd but the sot package is not all that bad.

9. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
The one I posted is the cheapest and easiest.

10. ### crutschow Expert

Mar 14, 2008
13,001
3,229
Yes, I see that the TPS53319 has a maximum output of only 5.5V so it would not work for the three LEDs in series. For that he may need a buck-boost type of regulator to operate with a minimum input of 10.8V.

11. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Nope, just a basic buck converter with a sense resistor to force it to operate in constant current mode (schematic posted above).

12. ### takao21203 Distinguished Member

Apr 28, 2012
3,577
463
It's a bit weird, I have built large 2576 regulators with large coils, but I don't get out more than 0.7 Amps.

Input is 43V, output is 32 volts. Long cables, some 3m to 5m. 200W flow through 4 cables. It is not more than 3 or 4A for each cable. 2 ground cables. I guess the DMM just can not display it properly if the amplitude of the input storage cap increases. As the brightness and temp. do well increase.

The main filter capacitor heats up to about 35C.

13. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
Actually a TPS5450 is even simpler, and I can snag a free sample. Working on the schematic now.

14. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
Updated, using a TPS5450 regulator with the LEDs in series.

Going to have to be 14V or above I'm afraid; there's no way to get >90% efficiency with the 10.8V input. I can work with 14.4V though, and I'll be getting around 94.5% efficiency, which I'm happy with.

Last edited: Jan 31, 2013
15. ### crutschow Expert

Mar 14, 2008
13,001
3,229
To avoid the problem of getting a precise, non-standard resistance for R1, you should be able to use a slightly larger standard value for R1 and add a 1kΩ pot across it with the wiper going to pin 4 to allow tweaking of the output current to the desired value.

Note that R1 will be dissipating over 3.89W and is due to the large 1.2V reference voltage at the VFB input of the TPS5450. This will significantly reduce your efficiency. That's the big advantage of the PT4115 which is designed for constant-current output and therefore has only a 100mV reference voltage, giving a R1 dissipation of just 0.32W.

16. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
The PT4115 only does 1.2A; I need at least 3.2A to power the LEDs. Are you suggesting multiple of these in parallel? At that I'm worried about PCB space.

17. ### crutschow Expert

Mar 14, 2008
13,001
3,229
My mistake, I didn't pay attention to the current limit.

But you probably could add an external power P-MOSFET to the circuit to increase its current capability.

18. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
That sounds like a good idea. Hmmm. I could drive the FET from the SW output with a ballast resistor, correct? Also, shouldn't it be an N-channel? I need to sleep on this.

19. ### crutschow Expert

Mar 14, 2008
13,001
3,229
An N-Channel would give a switching signal polarity inversion so you would have to add an addition inverter stage driver.

A P-Channel would have no inversion but you would need to slightly change the circuit design.

The connection would be V+ $\rightarrow$ P-MOSFET source, P-MOSFET drain $\rightarrow$ LEDs$\rightarrow$ inductor$\rightarrow$ common. The position of the LEDs and inductor can be interchanged if desired.

The diode anode would be connected to common with the cathode to the MOSFET drain.

The IC SW output would go to the P-MOSFET gate with a resistor from the gate to V+ (source). The value of the resistor has to be small enough to rapidly charge the gate capacitance when the MOSFET turns off. This depends upon the gate capacitance of the P-MOSFET you select. Pick one that has some reasonable margin to carry the required current with the smallest gate capacitance.

20. ### Eudoxus Thread Starter New Member

Jul 18, 2012
11
0
This is about what you mean?

Last edited: Feb 1, 2013