3016Views18Replies

# Inductor Heating Problem

I'm having a bit of trouble with an inductor implemented as part of a boost converter. The circuit works fine except for the inductor gets very hot within a few seconds after turning on. There are also noticeable high pitched noises coming from it associated with varying current loads.

I'm not really sure what the issues are as the circuit is within the maximum DC current rating (circuit draws around 120mA, inductor is rated for 250mA) and resonant frequency is also quite low (around 1.5kHz). Datasheet for the inductor I used can be found here.

Some help would be appreciated in determining what the problem is and in locating a suitable replacement inductor.

Thanks.

## Discussions

Best Answer 3 years ago

When I open that pdf, I see nothing but grey boxes. Although, selecting those boxes and copy-pasting out the text gives this:

So. Yeah. Wow! It's a tiny little thing! (3.2 x 2.5 x 1.7 mm)

If you have an happen to be in possession of a similar inductor with smaller DC resistance, and with bigger saturation current, you should try that, and see if it performs better.

BTW, the DC resistance of an inductor is one of those things you can measure with cheap tools. If you have a multimeter, you can use the ohmmeter (resistance measuring) setting for this.

Also if you want to try to guesstimate the actual peak current in the inductor, for your application, I think there are some formulas for that, here:

http://en.wikipedia.org/wiki/Boost_converter#Opera...

Answer 3 years ago

Hmm, I've just tried the PDF in a few other browsers and it seems to be working for me. Those values are all correct though, it is a pretty tiny inductor.

I did actually measure the DC voltage at about 3 ohms which seems to be in line with what the datasheet suggests. I don't have any others the same size in my possession but I could purchase some if necessary. It was quite hard to find any the same size but with higher inductance and DC current but I found a few with higher current, lower resistance at the expense of lower inductance, linked in my last post in the comment thread above.

I had a look at the formulas and I'm having a bit of trouble calculating the peak current. I get the duty cycle D to be 0.383 as 1-(3.7/6) (input over output voltage) so then ΔILon=((D*(1/f))/L)*3.7 where the frequency is 33kHz and L is 47µH (best case from one of the inductors above) and I get a value for ΔILon to be 0.00914. I'm not sure how to calculate the peak current from that though as it discharges between cycles.

Answer 3 years ago

Hey, I thought I'd try to explain how to estimate the peak inductor current, and it turns out this easy explanation only took about five hand-written pages, which is actually kinda typical for the verbose way I explain things.

I starting to think Steve's got the right idea: a bigger (than 100 uH) inductor, or faster (than 33 KHz) switching speed, or both are required.

Note the ripple current, measured peak-to-peak, (the AC part), is,

(Vin/L)*D*T = (Vin/Vout)*(Vout-Vin)*(T/L)

independent of the load current, Iout.

The time-averaged value (the DC part) of the inductor current is directly proportional to the load current, Iout.

Iavg = (1/(1-D))*Iout = (Vout/Vin)*Iout

Moar details are in the pictures-notes attached.

Answer 3 years ago

Thanks for the help, I ended up getting a 47µH inductor with an Imax of 390mA and Isat of 1.18A in a 1212 package (only 0.5mm wider so it fits on the same pads) and it seems to be working nicely in the circuit. Output is stable through the range of battery voltages with very little ripple and it doesn't get hot as before. There is still a small amount of hum but it's almost inaudible and it's not causing any problems with the audio output.

Answer 3 years ago

I'm glad to hear you found something that works for that.

Answer 3 years ago

You said your frequency was 1.5KkHz though, not 33.

Answer 3 years ago

The 1.5kHz is the inductor's resonant frequency analysed just from the LC circuit and the 33kHz is the oscillator frequency of the boost converter.

I just noticed actually the datasheet also contains a calculation for Ipk, must have missed that before.

Ton/Toff=(|6|+0.36)/3.7-1

Ipk=2Iout/((Ton/Toff)+1))

Ipk=805mA

That seems reasonable and would explain the heating. I can also plug that back in to find the minimum inductance value:

Lmin((3.2-1)/0.805)*Ton

Ton=D*1/33kHz

Lmin=32µH

So assuming I've done the maths correctly there I can get away with a slightly smaller value inductor but the peak current needs to be higher. In which case this looks like my best option for a replacement (am I correct in assuming that Isat of the inductor is the maximum peak current it can accept?

3 years ago

What ripple voltage have you permitted yourself ?

Answer 3 years ago

The value I used to calculate Co was 10mV which gives a value of 125µF. I used a 100µF ceramic in the circuit mainly due to space restrictions. It doesn't matter if the ripple voltage is higher, as it is passed through a regulator and further smoothing circuitry before being used by anything.

Answer 3 years ago

Then I'm still struck the inductor value is off.

Answer 3 years ago

Could you tell me how you're calculating the inductance? I'm using the equation from the datasheet:

Lmin((3.2-1)/0.805)*Ton

Ton=D*1/33kHz

Lmin=32µH

3 years ago

Peak current in the inductor will be VERY much higher than the load current. Increasing the frequency might make the thing place less stress on the inductor, but otherwise, you need to source something more like 2A than 250mA.

What is the input and output voltage supposed to be ?

Answer 3 years ago

Input is supposed to be 3.7v from the LiPo and output 6v.

Answer 3 years ago

My reckoning is your inductor value is way off. It should be more 2000uH, and it needs to be at least 600mA max current.

Answer 3 years ago

Thanks, will try with a beefier inductor and see how it goes.

Answer 3 years ago

Not just beefier ! A lot more inductance too.

Answer 3 years ago

I'm having trouble locating a suitable replacement inductor that will fit in the existing pad, best I can do is about half the existing value but higher DC currents and lower resistance. These seem to be the best options I could find:

10µH, 870mA, 0.312Ω

33µH, 400/620mA, 0.96Ω

47µH, 390/1180mA, 2.3Ω

33µH, 500/1300mA, 1.33Ω

10µH, 760mA, 0.1729Ω

Do you think any of those would be suitable for the higher DC current but at the loss of some inductance? It's quite a small device so I can't really fit anything much larger than about 3x3mm on the board so the only other options would be to get some more PCBs made (expensive) or solder on a larger inductor on wires (messy).

Answer 3 years ago

+1