I'm developing a buck converter. I need to to change its supply from 9 V to 12 V. I also need to get more power in it (up to 10 A). I do it by reducing the load resistor.

Here is my circuit: enter image description here Note: the low-side mosfet does not have an active role here, but I need it for later development.

My problem is that when I try to gradually increase the voltage from 9 to 12, everything stops working at around 10 V. By stops working, I mean no PWM anymore, a lot of noise, and then no voltage at all at my probes.

Here is a picture of the voltage at the gate of the high-side mosfet (yellow) and source of the high-side mosfet (pink) at 9 V, when everything is still more or less working.

scope pic

Theoretically, all my components are able to handle a voltage of 12 V.

Here are the datasheets: Mosfets Capacitor Inductor

I tried varying the output capacitor value, but it didn't seem to have any impact at all on my problem. When I decrease the load to smaller values, on the other hand, the circuit stops working with smaller voltage already. That's the reason I think it's less an issue of too high a voltage and more a problem of too high a power.

I searched for stability checks and calculation for buck converters, but all I found is literature about the stability of the converter in a feedback control. I do not have such a thing yet, my PWM is fixed. Plus, my output capacitance is pretty high, specifically to avoid problems with

So, here is my question: Why does my buck converter stops working when I try to transfer more power through it, although all elements are supposed to be able to handle it, and I shouldn't have a stability problem per lack of feedback?

Edit: I also wasn't able to reproduce the problem in LTSpice

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    \$\begingroup\$ I would suggest that the low side device (Q2) should have some sort of drive as it currently (as drawn) does nothing. \$\endgroup\$ – Peter Smith Feb 7 '18 at 10:38
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    \$\begingroup\$ Normally the problem is getting enough high side Vgs >= 3Vgs(th) so the lower side must be used for PWM to drive a cap and Vdd clamp diode to create a charge pump bias supply called Vboost (see any SMPS design on boost reg) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 7 '18 at 10:42
  • \$\begingroup\$ @PeterSmith That's because it's not supposed to do something yet. The buck converter is to be expanded in a 2 quadrant converter and the low side mosfet will be used for the boost converter mode. \$\endgroup\$ – Murmeltier Feb 7 '18 at 10:45
  • \$\begingroup\$ Using a N-channel for the upper-side MOSFET operated as you shown is not an ideal option. Considering a fixed drain voltage, you need to drive the gate with a voltage above the drain by 10 V. This is what bootstrapped drivers such as the IR2110 do. They create a floating upper-side source referenced to the MOSFET source but driven from a ground-referenced input. If you want to deliver 10 A, you will have to consider it or use a transformer. Also, please consider adding a Schottky diode in // with the unused MOSFET, its body diode will be thankful, especially in deep CCM. \$\endgroup\$ – Verbal Kint Feb 7 '18 at 12:11

I suspect that your inductor's core is saturating.

You have an inductance of 1 mH and a switching frequency of 100 kHz and this tells me that you might be at the point when all the energy put into the inductor is not being delvered to the load per switching cycle i.e. you operate in CCM (continuous current mode). Operating in this mode is fine but as you increase your supply voltage the average current in your inductor can rapidly increase and saturation of the core follows. When this happens the inductance falls to a very low value and creates the likely problems you see.

Another point; operating the MOSFET as a source follower DOES NOT make a very efficient buck regulator and that MOSFET will get hot and the whole point of a buck regulator is that it is an efficient design on load. Look at the pink trace - you are only getting a peak of 5 volts coming from the source - a decent design would be close to Vin or 9 volts as per the scope picture.

  • \$\begingroup\$ I tried reducing the frequency, and it did allow me to increase the voltage. Thank you for your other advice as well (and thanks to the other commentators for theirs too), I'll try improving the design. (Bootstrap driver, schottky) \$\endgroup\$ – Murmeltier Feb 7 '18 at 13:28

In order to fully turn the MOSFET on, a Vgs of 10V must be applied according to the datasheet. This means gate voltage must rise above input voltage.

You need a proper MOSFET driver with bootstrap, not an opamp wired as comparator. You can try ADP3120 for example. Read the datasheet, everything you need is in there.

Also the MOSFET choice isn't very good. You can get much lower RdsON these days. For 10 amps output current, you can use modern SMD MOSFETs. This will make the circuit more compact, and reduce parasitic inductance and EMI.

The inductor value is way too high. It should be a couple tens of µH, at most. Calculate it using the acceptable inductor current ripple. Large inductors like you use have large interwinding capacitance, which increases losses; also a shielded ferrite inductor will work better here and radiate less.

The output capacitor is inadequate also. Its ripple current rating is too low (check datasheet). For 10A you will need at least one low-ESR polymer capacitor with a few MLCCs in parallel. Don't skimp on the MLCCs, I'd put a few 10µF there.

Choice of caps depends on inductor ripple current, which depends on inductor value, frequency, and in/out voltages. If, for example, you use an average current of 10A with 30% ripple, you need the caps to withstand 3A ripple, plus an adequate safety margin.

  • \$\begingroup\$ The current ripple has to be under 1%, which is why I have such a big inductor and high frequency. Thank you for your other advice, it will greatly help for the redesigning. I don't understand why the MLCC parallel with the polymere capacitor? \$\endgroup\$ – Murmeltier Feb 7 '18 at 13:42
  • \$\begingroup\$ Why must the current ripple be this low? In this case, you are not building a buck converter, rather a precision current source which uses a switching architecture. For example, you need an inductor with good HF behavior at the output of your buck, and this is not compatible with a value of 1mH which will be huge and have lots of parasitics. It would be better to put a 2-stage LC-LC filter after your buck. You should edit your question and clarify what you want to do exactly, so we know what we're talking about and can help better. \$\endgroup\$ – peufeu Feb 7 '18 at 13:50
  • \$\begingroup\$ At the output you need a capacitor with low inductance to filter out the switching spikes. Surface mount MLCC has very low inductance. Aluminium can capacitor has much more, so is much less suitable for filtering out HF. \$\endgroup\$ – peufeu Feb 7 '18 at 13:51

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