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I am really scratching my head at this. I don't see anything wrong with my circuit but my regulator gets EXTREMELY hot while in operation.

What I'm building is a USB phone charger for my e-bike as part of my DIY e-bike project. I want to support a battery as high as 48 V, so I selected a regulator that can handle this and then some. The regulator is set for a maximum of 3 A, which is plenty for even an iPad. My inductor is rated to handle more than 150% of that.

With no load (reads as a few mA at the PSU of course due to losses) the chip stays cool. With a small 5 V LED it still stays cool. As soon as I add any substantial load (even just say 400 mA) the chip immediately starts getting very toasty >50-60 degrees and climbing. (I think so anyway, my temperature probe sucks and I need to invest in a thermal camera). Plug in my phone which tries to draw around 1.8 A, and it's enough to burn my finger within 3 s.

I tried adding a moderately sized heatsink (with some thermal paste from my gaming PC build), but it's not enough. Now it just takes 30 s to start roasting instead of 3 s.

On a multimeter the chip outputs a nice, smooth 5.1 V regardless of load condition (even with a USB fan), so it's clearly working, but why's it getting so hot? It seems to get scorching hot at the same rate whether my PSU is set to just a few volts above at 8 V, or all the way at 32 V (which is as high as it goes).

Is there something I don't understand here? Is it because I used a smaller inductor than the datasheet recommends? Would that lead to heat? The inductor is totally cool by the way. It's only the regulator IC that is getting hot.

The regulator is an XDS TX4138.

This is my schematic: Schematic

Edit: Here is my terrible PCB layout lol Board View Front Board View Rear

Oh, and for the folks complaining about capacitors, I'm trying not to overdo it on the input caps, because the balance board already has such massive ones that the system stays on for 10 seconds after the battery is removed. And for testing purposes I've got a 10,000uF cap I stole from a dead AC. As for output caps, I've got a 470uF one soldered into the cable going to the USB plug.

Update #2: Someone asked for the specific inductor I used, so here it is. I've ordered a new 33uH inductor to replace it. It has a higher resistance, but it's all I could find to fit into a similar footprint. I will also try to add a small ceramic between BATT and GND pins directly on the chip.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Aug 7, 2022 at 1:13

2 Answers 2

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The 10 μH inductor is unsuitably low in inductance.

Since you know output current, output voltage, input voltage, and switching frequency, you can plug the values into the formula given in the datasheet to find the minimum inductance needed for proper operation.

The output capacitor has an unreasonably low value compared to the datasheet suggestions as well, which may play a role as well.

The parts can't be randomly replaced with other values, as they need to be within limits of the operating point of the regulator chip.

This includes the switching frequency, max output current, output voltage, max input voltage, and ripple current.

The output inductor and capacitor need to have certain inductance and capacitance to meet the requirements and operate properly within limits of the operating range.

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  • \$\begingroup\$ Okay, so you think that buying a bigger inductor should fix it? I do remember having tried the formula before and the math worked fine for 10uH, though when I do it now I get: IL = 4 + 1.4/2 = 4.7; L = 5*(24 - 5)/(5*IL*200000) = 95/4700000 = 20.2uH \$\endgroup\$
    – Pecacheu
    Aug 6, 2022 at 0:22
  • \$\begingroup\$ In that case, should I select the lowest recommendation of 33uH? Or is a higher inductance better? \$\endgroup\$
    – Pecacheu
    Aug 6, 2022 at 0:23
  • \$\begingroup\$ Also, if I do the calculation for 48V (the largest battery size I want to be compatible with), I get 45.7uH \$\endgroup\$
    – Pecacheu
    Aug 6, 2022 at 0:24
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    \$\begingroup\$ With a 200 kHz switching converter, you absolutely need a proper bypass capacitor on the input, close to the device, and it must have low ESR at the switching frequency. A 470 uF electrolytic capacitor located externally won't do anything. As the circuit tries to supply higher current, the input will try to draw from the external connections, which will appear as inductors, and the input will be "starved" for voltage and collapse, causing all sorts of nasty problems. \$\endgroup\$
    – PStechPaul
    Aug 6, 2022 at 4:30
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    \$\begingroup\$ @Pecacheu At 200kHz, rise and fall times for the switch inside that device will be expressed in nanoseconds. Even tiny parasitic L's become problematic at that point. So "close" means millimeters here. And also mind the loop area. If the cap is physically close but the current has to follow a large loop due to poor layout then it's still useless... \$\endgroup\$ Aug 6, 2022 at 6:44
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Where is your input rail decoupling cap .This is important .Put one in like say 10microfarad ceramic as a minimum.Without the cap your chip is breaking inductive current which means energy .This energy will end up heating the chip .

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  • \$\begingroup\$ Yeah, I mentioned that I will try soldering one on the pins in update #2. I was told that having an input rail capacitor (currently 1uF, not counting the HUGE caps on the BATT rail on other boards) over a centimeter away like I do now is potentially too far away, so I need another one within millimeters of the chip. \$\endgroup\$
    – Pecacheu
    Aug 7, 2022 at 10:14
  • \$\begingroup\$ @Pecacheu You keep stressing you already have "massive" and "HUGE" input capacitors. The issue isn't the amount of capacitance, it's their ability to respond to high frequencies. "Massive", "HUGE" capacitors tend to be electrolytics and those are completely worthless for high frequencies, the circuit will act asif they're not even there... Think of electrolytics asif an oil tanker. Sure, it's "Massive", but worthless for competing in a F1 race. \$\endgroup\$ Aug 8, 2022 at 14:41

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