With your help on stack exchange, I finally finished assembling my first LM5143QRHARQ1-based buck converter! And it works! It does output 7V (well 7.03V) The only problem is that it consumes around 0.18A with the following setup (and no load): 15V lab power supply --> Step up converter to 43V (my regulator is designed to work with voltages from 40 to 55V) --> Buck converter to 7V.

This is my Buck's schematic and this is its datasheet: Buck schematic

What is weird is that the current draw changes with the bench power supply voltage. With an input voltage to the boost of 13V, it consumes 0.2A, with 15V, it consumes 0.16A...

The boost converter The above is the boost converter I'm using. It's based on the TL494C from TI. With only the boost converter attached, the displayed current draw is 0A. Only when I attach my buck it rises.

I measured the temperatures and all components on the board stay below 31C (with 24C room temperature). It's also making a weird buzzing sound when it's on. I can't test it with any load today since I don't have an oscilloscope to actually see what's happening right now. So is it normal to consume so much with no load attached? Could there be a small short somewhere and I didn't notice it while assembling? 0.18A seems outrageously high to me...

Also, this is how the board looks like. It's been soldered with an MHP30 mini hot plate for the most part and just a few manual adjustments. My own made board


So I removed a lot of the solder blobs on the PCB and resoldered a bunch of connections. I removed the intermediary boost converter from my adjustable power supply and went with the following setup: 48V fixed AC/DC power supply --> Multimeter in series to measure current --> My buck

Measuring current, **it draws around 50mA, so power loss is 5010^(-3)48= 2.4W , a slight improvement from the 2.7W previously, but not much...

When I first soldered the board, most components would stay below 31C and it would make a noticeable buzzing sound.

After another resolder, one side (since the buck is 2-phase) of the converter had its MOSFETs higher than the other side (38C hot pair, 33C other pair). The buzzing sound was still there

After yet another resolder, now the opposite side of the buck (the other 2 mosfets) are hot now and the others at 33C. Interestingly, the buzzing sound became less noticeable. I'm sticking with this configuration

The hottest parts seems to be the schematic left side diode, nearly hitting 39C with no load attached, which is really weird and I can't quite explain why. I haven't yet tested this with a heavy load or any oscilloscope. I did test it with one of those small Arduino LEDs and it brightened it without any problems enter image description here The encircled red area is slightly above the 30-33C of the rest of the circuit. The temperature there ranges from 34 to 39C, with the diode getting the hottest. The other identical diode is at 31C on the other side of the circuit. The main IC stays at 33C

I'm not sure what can be improved at this point or if I'm doing something wrong... 2.4W is a lot of power dissipation, and you would expect the reverse polarity protection N-channel MOSFET to be the hot guy (since power dissipation would be its RDS(on)Vin= 1.110^(-3)*48= 0.48W, but it looks like the diode is dissipating the most power? I hope I didn't damage it while soldering at high temperatures for too long, since it wasn't so hot in the previous attempts)

I also tried measuring the efficiency of the converter with that small Arduino light bulb (which consumed about 4mA) and a resistor using this tutorial page form TI and it resulted in 0.704 (70.4%), which is in the range of the IC graph?

  • 1
    \$\begingroup\$ I'm seeing a lot of solder balls - I would get as many of them as I could off of the board and test again. \$\endgroup\$
    – vir
    Aug 11, 2023 at 21:13
  • \$\begingroup\$ @vir Yes there is a bunch of tiny ones on the board. Should I just scrape them off with tweezers or what is the best approach to remove them? Can I wipe the board with some isopropyl alcohol? Will it damage it? \$\endgroup\$
    – Mito
    Aug 11, 2023 at 21:19
  • \$\begingroup\$ I usually use a nylon brush and tweezers. \$\endgroup\$
    – vir
    Aug 11, 2023 at 21:21
  • 2
    \$\begingroup\$ 2.7 Watts does sound excessive, BTW. Unless your switching speed is stuck in 1980, a buzzing or singing sound usually means that you have an oscillation (with some capacitors or your coils acting as the speaker). Looking at your output voltage with an oscilloscope would be a good thing. \$\endgroup\$
    – TimWescott
    Aug 11, 2023 at 22:32
  • 1
    \$\begingroup\$ @TimWescott I have added the schematic and a link to the IC's datasheet. And yes, I've spent enough to get an actual cheap scope. I'm waiting on my high school to lend me one for now as the time it takes to receive one from them or buy one from Amazon is roughly the same \$\endgroup\$
    – Mito
    Aug 13, 2023 at 8:22

2 Answers 2


First off, it is perfectly normal that if you lower the input voltage of a step up converter that the input current increase. Just think about it that way, let's say your output is at 100V and output 1A, the output power is 100W. The input voltage is at 50V, considering an ideal efficiency, the input current need to be 2A. If you lower the voltage to 10V, the current would be 10A (if your load is always 10V 1A). So, that's normal!

The second is that efficiency isn't constant over all load. Boost and buck usually are way less efficient at lower load. It is very possible that the up conversion is in the 50% efficient at those currents. If so, it wouldn't require a lot of power on it's output to create a significant load.

What I would recommend you to do is to do a proper efficiency measurement. Set the DC power supply at a constant voltage and measure as accurately as you can the current (ideally without modification to the setup during measurement. Then place known loads on it. With that, you can evaluate the efficiency curve. This will tell you if it makes sense or not. You can do those for either all your system or each board individually (I'd do the later). The simpler is your system the easier it is to debug it!

Third the noise you are earing is most likely due to an insufficient amount of current being outputted by your converter. Some buck converter will not operate properly without a minimum load. It is design and component dependent, so it might be your case.

EDIT: To PSU measure efficiency, here is what I would do if I only had a DMM:

  1. Measure n resistors (if you are using low ohms, be careful on wire resistance), You know your voltage, therefore, compute the approximate power dissipation and ensure you have a lot of headroom for them not to heat up too much.
  2. Connect the DC power supply in series with your DMM in current mode (use 10amp not 400mA due to burden voltage). Ensure that the current is less then the fuse rating. If accuracy is an issue, you will need 3 DMM and monitor input voltage as well as output voltage)
  3. Place the first resistor, apply voltage, note the current and voltage on the input side.
  4. Efficiency can be calculated as follow: Eff=Pout/Pin Pout=Vout^2/Rload Pin=Vin*Iin
  5. Repeat for all resistors
  6. Repeat for different input voltages (if needed)
  • \$\begingroup\$ I was suspecting the non-existing load to be the problem as well. So how can I do such a load test? What do you mean by "known loads"? (Known constant current draw/resistance?) I got a bunch of Arduino Light bulbs and DS3235/5160 servos whose stall torque and current is known. Should I place one of those? Do you recommend any tutorial for measuring the efficiency? \$\endgroup\$
    – Mito
    Aug 12, 2023 at 7:58
  • \$\begingroup\$ The cheapest way, by fair is to place resistors! You measure them first then place them. If they don't heat up, the resistance won't vary much. If you have a DC load, you could use that too. I wouldn't use any active device since their current is dependent of the voltage and they might get damaged. Noise could also play a factor. \$\endgroup\$
    – Julien
    Aug 14, 2023 at 13:40
  • \$\begingroup\$ @Mito I Edited my previous answer with instructions on how to measure efficiency. \$\endgroup\$
    – Julien
    Aug 14, 2023 at 13:51

You really need to poke around in the circuit with a decent oscilloscope. In the absence of that, however, put your multimeter on the "AC" range and take some readings in places where the voltage should be solid DC.

Specifically, measure the AC voltage at the output of the supply overall, and at the feedback pin to the chip (which, in a "normal" circuit, should be connected to the final output via a resistor).

If the AC voltage is only a few millivolts more than measuring the AC voltage of ground, then you're not seeing serious oscillation at those points. If it's present then the measurement is less informative.

If you looked at those points with an oscilloscope, oscillation would show up as a periodic variation in the output voltage that's not equal to the switching frequency. With an AC voltmeter you can't really distinguish between switching-frequency noise and lower frequency oscillation.


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