Inductor in buck converter

Question

I have a couple of questions regarding the switching inductor in a buck converter.

1. Is it normal for the switching inductor in a buck converter to heat up? In my circuit, it reaches around 65 Celsius. I let it run for an hour and everything seems fine.

2. Does a higher switching frequency result in a warmer inductor? While the switching frequency is 15kHz, the inductor runs pretty cool, I cannot feel any heat at all when touching it. When the frequency changes to 22kHz, it heats up to 65 Celsius.

3. How can you tell if the inductor is saturated? My output current is 50mA, the inductor I use, Wurth 7447709821 is rated for 1.1A. Does this mean the inductor is not saturated in my case, and all the heat is due to the resistive component? The resistance of this inductor is 0.815 ohm, which results in about 40mW based on the output current. I don't think it is possible to get from room temperature to 65 Celsius with only 40mW.

Edit: Adding the circuit and some measurements as requested.

Schematic and PCB: Vin 110Vdc, Vout 32Vdc, Iout 52mA. The entire bottom layer is the ground plane.

Measurements:

• Channel 1: switch node, 22kHz.
• Channel 2: Inductor current measured with a clamp meter, ratio 1mV/10mA.

Answer 1

Is it normal for the switching inductor in a buck converter to heat up?

Yes. This is due to the winding resistance, which varies depending on the average DC current. In general, smaller-value inductors will have lower DC resistance, which is why you try to select the smallest value you can while getting an acceptable ripple.

That said, heating to 65C is a lot for such a light load. Something is broken.

Does a higher switching frequency result in a warmer inductor?

It can, as parasitic losses are increased. A higher switching frequency however allows selecting a lower-valued inductor, which reduces loss, so it should be a wash.

But the range of change you mentioned (shifting up from 15 to 22 kHz) should not be significant here. Again, something else is going on I think.

How can you tell if the inductor is saturated?

By using a current probe with the inductor is wired in series, or seeing if the switch node is ON for close to 100%. (By 'saturated', meaning that the high-side switch is on all the time, rather than achieving an RMS ripple of 30% of DC. This should never happen.) If you see that, again, something is wrong.

My output current is 50mA, the inductor I use, Wurth 7447709821 is rated for 1.1A. Does this mean the inductor is not saturated in my case, and all the heat is due to resistive component? The resistance of this inductor is 0.815 ohm, which results in about 40 mW based on the output current. I don't think it is possible to get from room temperature to 65 Celsius with only 40 mW.

You're correct, the inductor should not be heating up that much at such a low current.

Can you simulate it?

Answer 2

1. It's normal if the losses in the inductor due to DCR, AC resistance and/or core losses are enough to raise the temperature significantly.

2. Yes, a higher switching frequency can result in higher AC and core losses, and can increase losses. However, for a constant inductance, a higher switching frequency will also result in lower current ripple.

3. You calculate your inductor current ripple and the peak inductor current at max load. If that's less than the inductor saturation current limit then you should be OK.

Having significantly different core and AC losses when going from 15 kHz to 22 kHz with your inductor doesn't sound likely, especially with 50 mA of output current, so you should look for some kind of abnormal operation like subharmonic or high frequency oscillations. If you add your schematic, key operating parameters (Vin,Vout,Fsw) and some waveforms (switch node, output, inductor current if you can) we may be able to pinpoint the issue.

Answer 3

Losses in an application can be checked on the manufacturer's website:
https://redexpert.we-online.com/we-redexpert/en/#/home
You will have to register an account to use REDEXPERT. There, you will find a design tool including AC and DC loss calculations. Entering the relevant parameters and selecting the part in question shows:

AC (core) losses are wildly dominant for this application. Even at twice the DC output (100mA here). Calculated total is still 94mW at 50mA.

They don't provide a power dissipation rating, but we can convert the DC rating to one. For R_DC = 0.815Ω, and 0.95A at 40K temp rise, P = (0.95A)²(0.815Ω) = 0.73W. Thus we can expect a thermal resistance around (40K)/(0.73W) = 54 K/W.

If ambient temperature is 22°C, 65°C is 43K rise, or slightly above ratings. You must be dissipating about 0.8W, at least if it's entirely in the inductor itself; the regulator and catch diode will heat up some as well, basically raising the ambient for the inductor.

This still doesn't seem to account for the overall temp rise, but it's at least a strong contributor. It's also possible that Würth's calculations are naive, not well suited to the large peak current in this condition; core loss calculations tend to be based on many assumptions, which core materials and typical geometries don't necessarily satisfy very closely.

In any case, it seems like the inductance value is much too low for the size of core, core material used, and switching frequency. Losses start to come down by 300kHz or so, but I'm guessing this is far too high for most any LinkSwitch product.

Nice layout by the way.