In this article I continue my project where I needed a constant current source. In the previous article I talked about the first approach the LM317 constant current source which didn’t worked out that well but could of been improved as suggested by some of my readers in the comments. So I decided to use a dc-dc converter and turn it into a constant current source. Since I’m using the TPS54232 in another project and I have it around I decided to use it. This technique can also be applied to other dc-dc converters with a bit of reading on the subject.
I’ve started by reading SLVA374 app report from TI which is a Step-Down LED Driver Design Guide based on the TPS54160 dc-dc converter, normally used as a buck voltage regulator. The schematic from the app report can be easily adapted to our device with a few changes like adding a sense resistor and an output capacitor:
If you remember from my last post I needed 1.12 A to power the two led packs that I’ve build connected in parallel. We have a simple equation that we use to find out the feedback resistor value: R=Vref/Io In our case Vref is 0.8V taken from TPS54232 datasheet, Io is our desired current of 1.12 A. Doing the math we get the R value = 0.71 ohms. This is not a standard value, but we can use two 1.5 ohms resistors in parallel(R1 and R2) to get 0.75 ohms which is pretty close. With 0.75 ohms feedback resistor I have 1.06A at the output which divided by 56(the number of white LED’s) means 18.9 mA for each white led.
We must consider the power dissipation for this resistor and we can calculate it : Pdis = Vref^2/R In our case we have Pdis = 0.8^2/0.75 = 0.85W. The only 1.5 ohms resistors that I could find locally were 5 W so I had to work with a bigger package. The EN pin is used to enable and adjust the Undervoltage Lockout but I’m not going to use that feature since my input voltage is always gonna be high enough not to cause any problems. The resistors are placed on the pcb anyways so you could use those pads and soldere the necessary resistors.
We also must consider the power dissipation in the low side diode. During the converter on time, the output current is provided by the internal switching FET. During the off time, the output current flows through the catch diode. The average power in the diode is given by: Pdiode = (1-Voled/Vin) * Vfd * Io , where: Vfd is the led forward voltage, Voled is the supplied output voltage and is approximated by: Voled = Nled * VLed + Vref where : Nled = number of LEDs, Vled = forward voltage drop of each LED. In my case Pdiode = (1-(4.2/12))*0.75*1.06=0.51W so I used a 2A 40V schottky diode part number CDBA240-G.
You would also have to consider the inductor and the input and output capacitors, but you can read all about that in the datasheet of the TPS54232 or in the app report mentioned above. After I had all of my circuit figured out, I routed the board and etched it using my photo etching technique. For such a small board its not worth taking out the solder paste so I soldered it using the soldering iron, as you can see its a mix of through hole and surface mount parts but they fit together quite nice.
The solder drops that you can see on the back of the board are a sort of DIY thermal vias. I’m not sure how efficient they are but I gave them a try with this board, probably because I had too much time available :-). First when I designed the board I included a area of copper in the top layer right beside the resistors. Next I drilled some 0.7 mm holes and I placed 0.7 copper wire in the wholes securing it by soldering on both sides. The trick is to make the solder joint as small as possible so it doesn’t spaces your component from the board. As I’ve said I don’t know their efficiency but I think they work, I can feel the heat transferring from one side to the other faster. If you like them you can try them.
The testing went smooth, the LED’s light up perfectly, no problems at all. During the testing I noticed one bug though: if the input wires are not firmly attached or secured and there is an imperfect contact the converter will tend to output less current than the programmed 1.02A. I’m not sure why this is happening, it might have something to do with the Undervoltage Lockout feature that I skipped on but I’m not sure. Anyway its not that much of a problem since once in place the power source will have the cables firmly attached so no worries. That’s why I used the cable connectors on the PCB in the first place, I knew it would save me some trouble later.
Now I had one last thing to do. Since this circuit is going to be operated in the outdoors the corrosion would set in pretty quickly so I improvised once again. I had this idea for quite a long time but never actually tried it. So I used my Bison universal hobby glue, which is transparent, and covered the copper traces on the PCB in glue. After it hardened, it looks like you could dip these board into water and nothing would happen to them. Well, except from the terminals which stick out of the PCB :-). Nonetheless I think these boards will run no problems even after a few years. If you’re wondering where the bubbles come from, they form when the glue dries out :-).
I’m pretty happy how this project ended up and I feel like I know more about LED’s and ways to power them. I’ll definitely need to experiment more with some high power LED’s, maybe use them to light my workbench. As for the efficiency of this circuit I don’t know if I’m calculating the right way because I’m using the equations from page 16 of TPS54232 datasheet, and those are clearly stated to be used only under continuous conduction mode. Since the circuit has been modified to act as a constant current source I’m not sure the same equations apply. But I did the math anyway and I got an efficiency of 75.36% and according to the same equation if I would connect the two led packs in series I would get an efficiency of 79.01%. Once again I’m not sure these calculations are correct and I ask the readers to comment on these.
There is more that you can do to improve the efficiency of this circuit. It turns out you can reduce the power losses in the current sense resistor by lowering the voltage across the resistor. The solution is to inject a bias voltage, but I’m not going to try this solution since I’m already happy with my design. You can read more about it in SLEA004 app report from TI.