I am trying to get a boost converter working. This is the schematic:


simulate this circuit – Schematic created using CircuitLab

The clock signal is 12V and 100KHz, the output capacitor is 330uF and I got it from the output of a PC power supply so is better than a common electrolytic capacitor. The load is a 7W resistor and the coil is a toroidal inductor I got also from a PC power supply but I don't know how much Henri's it is and I don't have an inductometer to measure it. The inductor is something like this:

enter image description here

I can't get this to boost the 12V input voltage and the MOSFET heat up a lot even with a large heat sink on it. I am having from 2A to 3A of current being drawn from the power supply. Here is the the waveform at the output of the boost (diode cathode):

enter image description here

As you can see I am getting a 24.8V peak (the zero voltage line is two divisions below the center horizontal line) but just a peak, I can't get a constant boosted voltage. In the MOSFET gate the signal is a perfect square wave of 12V so problem should not be there.

Can the MOSFET heat up problem being caused because I need a diode in anti-parallel with the MOSFET?

  • 2
    \$\begingroup\$ That waveform looks more like the one you should see on the anode of the diode. Are you sure the diode isn't shorted or otherwise damaged? \$\endgroup\$
    – Dave Tweed
    May 18, 2014 at 20:20
  • \$\begingroup\$ The waveform on the anode is exactly the same but with a lower peak voltage. I measured the diode and it isn't damaged \$\endgroup\$
    – Andres
    May 18, 2014 at 20:23
  • \$\begingroup\$ The time constant for the capacitor and resistor is 8.6ms. When the diode is off (the MOSFET is on), the capacitor should discharge through the resistor with that time constant. But, according to your scope trace, the voltage on the cathode drops almost instantly from about 12V to about 5V. So, if the capacitor is good, the diode probably isn't (assuming the schematic accurately reflects the actual circuit). \$\endgroup\$ May 18, 2014 at 20:27
  • 2
    \$\begingroup\$ @Andres How can the anode peak voltage be lower than the cathode. This can't be true. \$\endgroup\$
    – Andy aka
    May 18, 2014 at 21:00
  • \$\begingroup\$ Are the MOSFET and the diode uninsulated on the same heat sink (shorting the diode out by connecting the drain to the diode cathode)? \$\endgroup\$ May 19, 2014 at 5:00

2 Answers 2


Your toroidal coil doesn't have enough inductance for this application, and it's saturating. This limits the amount of energy it can store, and it also causes the current through the MOSFET to be way too large.

Try putting a much lighter load on the output (i.e., larger resistance for R3) and driving the MOSFET with a much lower duty cycle. See whether that produces better waveforms.

But the bottom line is that the inductor is a crucial part in any switchmode power converter, and you really need to know what you have, in terms of the exact inductance value and its peak current capability. If you can't measure what you have, then purchase parts with known specifications.

  • \$\begingroup\$ I really appreciate the straightforwardness and tone of this answer. \$\endgroup\$
    – D. Patrick
    Feb 28, 2019 at 1:40

As Dave said, your inductor is saturating. I had the same problem when I built my boost MPPT. Little did I know about saturated core.

Easy solution: use iron core powder instead. My bet is you have used ferrite core inductor.

  • \$\begingroup\$ Yes I used a ferrite core, I solved it changing topology to half-bridge but why iron core powder is better? \$\endgroup\$
    – Andres
    Aug 29, 2014 at 14:21
  • \$\begingroup\$ Because it does better when it comes to saturating, the powder makes saturating almost impossible \$\endgroup\$
    – BELSmith
    May 19, 2021 at 6:06

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