Buck converter - controller and diode getting hot (~40°C) - Page 1
Forum section: electronics / beginners

author: doru.cazan
author_rank: Contributor
timestamp: April 17, 2023, 03:34:39 pm
content: Hello Everyone!I am a beginner so any thoughts on this issue would be much appreciated!Built a buck convertor with LMR14030 (pictures attached) and run it with a 24Vac transformer (there is a bridge rectifier and a couple of capacitors in the test circuit) with output 3.3V. The output on oscilloscope shows less than 60mV ripple (need to mention that I used an esp32 as load because I do not have yet a programmable resistive load). I do not see any reason not to be happy with the output (maybe I do not see the reason because I do not know enough). The current should be less than 500mA (it works from USB) while the controller is rated for 3.5A. All components values are the ones recommended by Webench Power Designer but I arranged the parts on the PCB myself, the best I could.Getting to the point: because it felt a bit warm, used a thermocouple and measured the controller and diode temperature: about 40°C after 2 minutes and remained stable. All other components, except controller and diode, remain at the room temperature (e.g. inductor, capacitors).Questions: Is this temperature increase normal? What more should I check? I wanted to use this in a thermostat but the convertor will be less than 7cm apart from the temperature sensor so, in case this is normal, will need to find other solutionMany thanks!

author: Benta
author_rank: Super Contributor
timestamp: April 17, 2023, 03:55:15 pm
content: That'd not hot, it's normal. Hot is when you get blisters touching it.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 17, 2023, 04:05:35 pm
content: Those SMD type electrolytics on the output of the bridge rectifier may not be a good choice.

author: doru.cazan
author_rank: Contributor
timestamp: April 17, 2023, 04:15:46 pm
content: Thank you @Benta, @wasedadoc!@wasedadoc: could those electrolytic capacitors lead to increased temperature of the controller and diode? There is a third input capacitor that it seemed a bit small to me but it was recommended by Webench Power Designer. What do you think, regarding the input capacitors, could improve the design and lower the temperature?

author: mariush
author_rank: Super Contributor
timestamp: April 17, 2023, 04:24:09 pm
content: Yeah, I was just about to write that....If you have a 24v AC transformer, the bridge rectifier converts that to a peak DC voltage of Vpeak = sqrt(2) x Vac - 2 x voltage drop on rectifier diode = 1.414 x 24 - ~1.5v = 34-1.5 = 32.5vBut if you're running this at night you may have higher AC input which in turn will cause higher AC output .Also, lots of transformers no-load voltage is 10-15% higher than load voltage.Basically, 35v is risky, 50v would be safer.As for the layout, personally, I would have moved L1 more to the right or move the IC more to the left. The diode could also be rotated 90 degrees.If all was more to the left, there would be more copper fill on the right side to act as heatsink for the diode to radiate some heat away, if needed.From the datasheet :9.2.2.6 Input Capacitor SelectionThe LMR14030 device requires high frequency input decoupling capacitor(s) and a bulk input capacitor,depending on the application. The typical recommended value for the high frequency decoupling capacitor is 4.7 μF to 10 μF. A high-quality ceramic capacitor type X5R or X7R with sufficiency voltage rating is recommended.To compensate the derating of ceramic capacitors, a voltage rating of twice the maximum input voltage is recommended. Additionally, some bulk capacitance can be required, especially if the LMR14030 circuit is not located within approximately 5 cm from the input voltage source. This capacitor is used to provide damping to the voltage spike due to the lead inductance of the cable or the trace. For this design, two 2.2 μF, X7R ceramic capacitors rated for 100 V are used. A 0.1 μF for high-frequency filtering and place it as close as possible to the device pins.The ceramic capacitor closest to chip should be 0.1uF 50v or higher, x7r or np0/c0g if available (this 0.1uF is not really critical, could be a bit less), it wouldn't hurt to have a second 2.2uF - 4.7uF ceramic capacitor, x7r preferable, x5r acceptable, again rated for much more than peak dc voltage (maybe 100v would be best) - ceramic capacitors capacitance will drop with input voltage.Ideally, use a polymer capacitor but 50v and higher polymer capacitors are probably expensive.

author: Siwastaja
author_rank: Super Contributor
timestamp: April 17, 2023, 04:25:32 pm
content: Forum's attachment system being broken again, I can't look at the schematic as it returns a different image when downloaded/enlarged.Building a 3.3V switcher in non-synchronous buck topology is going to have limited efficiency, because the diode is an (approximated) constant voltage drop. For example, 0.4V is pretty big % of 3.3V. Not so much of 12V, for example. So the diode is usually the most lossy component of the whole switcher. If you want to do better, use asynchronous buckswitcher IC. If you use one with integrated MOSFETs (just like yours, but two instead of one), then you have one external part (the diode) less, so eases your work, too.Otherwise than that, 40degC surface temperature is not bad at all. Such tiny board, especially if just 2 layers, does not provide much heatsinking. Even a little bit of wasted power makes the temperature increase. Note that temperature rise itself makes the thing capable of dissipating more power; at very least linearly, but usually somewhat more as the natural convection starts to work (air begins to move as heated air rises and gets replaced by cool air).If you want to measure efficiency, add enough capacitance to input and output to get rid of AC measuring artifacts (you probably have enough already, though), then measure input current, input voltage, output current and output voltage. You would need 4 multimeters to do all of that simultaneously, but if you trust your voltages are stable, that reduces to two, and if your load is constant, you can do two different measurements with one multimeter. Then you can calculate efficiency Pout/Pin and compare to what Webench said. Probably something around 75-80% efficiency is to be expected; 15% of the input power goes to diode loss, 5-10% to other losses, and 75-80% to the actual load.Quote from: wasedadoc on April 17, 2023, 04:05:35 pmThey are fine and you need a lot of capacitance when you rectify 50/60Hz. But you may want toaddlarger MLCCs there in parallel because low-ESR, low-ESL input cap is critical for a buck converter. Try to supply the thing with DC instead of AC to remove any heating of the electrolytic capacitors caused by the 100/120Hz ripple. Then if they heat up, that is all due to buck converter input current, and it would be a good sign to increase the MLCC input capacitor size.

author: doru.cazan
author_rank: Contributor
timestamp: April 17, 2023, 04:37:56 pm
content: Thank you @mariush, @Siwastaja!This is a test PCB for the buck converter, in the final circuit 50V capacitors are used (more expensive so I thought I can use 35V for test board).It is indeed a 2 layers board with back plane as GND except one single trace for the feedback. Only the controller and diode are getting hot / warm. All other parts, including electrolytic capacitors, remain at the room temperature.@Siwastaja, so much to learn from your comment! Will solder a larger MLCC input capacitor, see if I can use a better suited diode and will definitely buy a programmable load, it seems that it could help a lot with measurements.Another thing I did not think to measure, but I should, is the temperature in the enclosure (have a couple of prototypes). I think it will raise a bit more

author: mariush
author_rank: Super Contributor
timestamp: April 17, 2023, 04:42:28 pm
content: I edited my post and added some stuff... I see you thanked me before the edit and just in case you won't re-read it I'm posting again.To add something, see this application note about capacitance of ceramic capacitors variation with voltage and temperature :https://pdfserv.maximintegrated.com/en/an/TUT5527.pdfand some suggestions for synchronous buck switchers :AP64500https://www.digikey.com/en/products/detail/diodes-incorporated/AP64500SP-13/10419714AP64350https://www.digikey.com/en/products/detail/diodes-incorporated/AP64350SP-13/10420257LMR50410 (max 36v input voltage) :https://www.digikey.com/en/products/detail/texas-instruments/LMR50410YFQDBVRQ1/13563000RT7272 (max 36v input voltage) :https://www.digikey.com/en/products/detail/richtek-usa-inc/RT7272BGSP/5724397

author: AnalogTodd
author_rank: Regular Contributor
timestamp: April 17, 2023, 04:48:00 pm
content: Quote from: doru.cazan on April 17, 2023, 04:15:46 pmThe big electrolytics are there as energy storage elements for when the rectified input voltage drops. They are needed to keep the controller alive every half cycle of the AC. They are used because they give larger amounts of capacitance in a small form factor and ESR is non-critical. The third capacitor is because this is a buck regulator that switches on and off at high frequency. When the switch turns on, it pulls a large slug of current out of the input capacitors; if you relied solely on the electrolytic capacitors the ESR is high enough you would have a significant transient on the input every time the switch turned on.The temperatures of the controller and diodes are actually in a normal operating range and should not degrade operation over time. What is happening is that the energy to run the buck regulator and supply the load is enough that it causes the temperature rise.You are already using a synchronous buck converter, so I am assuming the diodes you are measuring the temperature rise on are the input rectifier diodes.(Sorry, somehow pulled up the wrong data sheet and didn't notice)Now, you said this is supposed to be in a thermostat and that the temperature rise is unacceptable. The question from here is what do you actually need for your supply in terms of voltage and current? If you require too much power, then you're going to have issues. Efficiency can never be 100%. Right now, with a 24V supply your efficiency at light loads is already only about 80% based on the data sheet. If your load is 1W, you're dissipating 125mW in the rest of the circuitry which gives the temperature rise.The rectifier diodes are a ~1.4V drop at the net input current (1.125W/24V=47mA) and thus dissipate 65mW.Options are to find circuits that operate at higher efficiencies and possibly switch toideal diode circuits instead of a basic rectifiera synchronous buck converter for better power dissipation there.Others are to move the transformer and rectifier circuitry out and away (to get the heat of the diodes separated from the rest of the circuit) and just feed the circuit with the rectified DC, then move to a better switcher circuit to remedy that heat dissipation.Sucks that trying to get to the circuit schematic gives the board picture, would have noticed better that I somehow pulled the wrong data sheet if I had been able to view that better.

author: Siwastaja
author_rank: Super Contributor
timestamp: April 17, 2023, 04:48:07 pm
content: Quote from: doru.cazan on April 17, 2023, 04:37:56 pmThere is often not much you can do. Schottky it has to be, which it probably already is. Smaller C means less switching loss, but it's usually not that much to begin with. You can try to find one with smaller Vf which will directly translate into less loss, but be careful since the very lowest Vf diodes have surprisingly high off-state leakage current; something which we usually approximate as zero. Be careful: leakage increases with increasing temperature, possibly leading to thermal runaway (chain reaction). Check if unsure.Temperature clearly isn't a problem currently (unless you are going to add significantly more load, then you should evaluate again), so for that reason alone no need to go for synchronous switcher. Clearly you are not running on battery either (so that battery life would be a concern). Also I don't think saving another 100mW would save the planet. Therefore, the non-synchronous design is probably fine as is.

author: doru.cazan
author_rank: Contributor
timestamp: April 17, 2023, 04:56:08 pm
content: This is the 4.7uF input capacitor I used:https://www.digikey.co.uk/en/products/detail/murata-electronics/GRM21BC81H475KE11L/4905516Will have study all provided documentation and solder a larger one with better characteristics to see if temperature decreases.The thermostat will be powered from the 24Vac valve control line and will work internally with 3.3V with all components using about 350mA (WiFi connected, a dedicated/separate MCU for the thermostat, ...)And will definitely have to learn about synchronous buck convertorsSince ~40°C it is normal, it seems that different type of buck converter is the right answer for a lower temperature.Or I can design the enclosure with a separator wall and some holes to redirect the warm air away (the sensor is already at the bottom but there are no holes on top for the air flow).

author: AnalogTodd
author_rank: Regular Contributor
timestamp: April 17, 2023, 05:33:52 pm
content: Quote from: doru.cazan on April 17, 2023, 04:56:08 pmThat capacitor is not likely a temperature issue. What your temperature issue stems from is the overall efficiency of your circuit. Look at the curves in the data sheet--you're going to be running 70-75% efficiency on this circuit. With 1W of power going out, that's over 300mW of power between the controller and diode. The reason the diode gets hotter at higher input voltages is because it is on for a larger portion of the duty cycle at higher input voltages. You're talking about 84+% of the duty cycle at 350mA, so a schottky with a 0.5V forward drop will run 0.5V*350mA*0.84 or close to 150mW. If you were to go to a synchronous buck where the Rds(on) of the bottom side is 0.1ohm, that would be 350mA^2*0.1ohm*0.84 or 10mW.Next is the overall operation of the switcher itself. Looking at the data sheet, the 40uA operating current is specifiedNON-SWITCHING! To get the higher frequency switch edges, it requires higher internal operating currents and that ups the power dissipation. To get 15deg. C rise with ~25deg.C/W thermal resistance, that's 0.6W of power dissipated. Top switch will run the 0.09ohm resistance times 350mA squared * 16% duty cycle or 1.8mW, so that says the chip overall burns 0.6W when switching. Just some rough calculations based on circuit efficiency in the data sheet suggest this may be close to the mark.Best bet would be a synchronous switching regulator (kills the power dissipation), and then find one that doesn't necessarily go to as high a switching frequency so it isn't driving the daylights out of the switches and burning so much power. Additional idea is to add a bunch of ground plane on the back side to lower overall thermal resistance and help dissipate the power better.

author: doru.cazan
author_rank: Contributor
timestamp: April 17, 2023, 05:44:01 pm
content: Thank you @AnalogTodd, will definitely look at synchronous switching regulators. @mariush helped with few suggestions so will start with those.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 17, 2023, 07:08:27 pm
content: My concern about the SMD electrolytics is the ripple current they experience. Do not overlook that the narrow conduction angle of the bridge means a spikey current into those caps and a relatively high peak current.The above will not affect the temperature of chip and diode but could lead to less than expected lifetime of those caps.

author: Siwastaja
author_rank: Super Contributor
timestamp: April 17, 2023, 07:22:56 pm
content: Quote from: wasedadoc on April 17, 2023, 07:08:27 pmSMD and through hole parts are roughly similar in ratings. The size is of course small, but there are two of them and the input current to the buck stage is so small I don't think that's a problem. Easiest way to test your hypothesis is to touch the capacitors by hand and feel if they warm up. If they are near room temperature even under operation, no problem. If they heat up, then one should confirm whether it's because of the 100/120Hz ripple or the f_sw ripple by the procedure I described above (running the input on DC to see f_sw related heating alone).