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I use this MOSFET:

https://cdn-reichelt.de/documents/datenblatt/A200/ZETEX_DMG1012UW_ENG_TDS.pdf

in a basic boost converter topology in DCM mode (L=100µH). The gate is driven by an ATtiny402 through a 100 Ohm resistor. My oscilloscope shows some odd behaviour: the MOSFET turns off more than 500ns after the gate has reached 0V. Turn-on is practically instantaneous.

Vds Vgs

I also measured the current into the gate. It shows some constant-ish current coming out of the gate in the delay phase.

Ig -Vgs Rg=100 Ohm

As a test I used a different MOSFET, where the waveforms look exactly as expected:

same as first picture, but different MOSFET

Unfortunately this well-behaved MOSFET is not suitable for the final application.

Now to my question(s):

What is happening here? At Vgs=0.7V Rds>30kOhm so it isn't just a low threshold voltage. I tried soldering in a new one of the same type, but the problem is the same. If possible, how can I improve turn-off performance? And no, shorting the gate directly to the pin of the ATtiny did not make a difference.

Update with some clarifications and some additional experiments:

I reduced the gate resistor to 10 Ohm, it did not change the turn-off delay. It did, however, slightly change the gate current waveform:

Ig -Vgs Rg=10 Ohm

For those that requested a schematic, it really is just your standard textbook boost converter:

Boost converter schematic

other tests I did:

  1. added a 1.5nF capacitor across the 100 Ohm gate resistor, no effect on turn off delay.
  2. reduced VCC of the circuit from 3.8V to 2.5V, the delay was shortened by 25%.
  3. reduced Rg to 10 Ohm, no effect. (I/O pin resistance ~ 30 Ohm)

clarifications:

  1. yes, I did measure directly at the gate, not at the MCU pin.
  2. yes, the MOSFET is supposed to be able to be driven from a MCU pin (700pC)
  3. where I bought the weird behaving MOSFET: reichelt.com
  4. the second MOSFET I tested (that did not behave weird) was a BSS138 https://www.onsemi.com/pdf/datasheet/bss138-d.pdf

clarification edits in the original section above:

  1. changed the Ig waveform picture to be more clear and show the remaining current better.
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    \$\begingroup\$ You refer to a specified turn off time but did you look at the conditions under which that was achieved? An MCU pin is nothing and a 100Ohm resistor is way too much even if you had a gate driver. Vgs_th is irrelevant for switching. That's when it starts to barely conduct but you care about what it takes to fully conduct. Use a proper gate driver circuit or IC and drive it with the Vgs to achieve the specified RDson. \$\endgroup\$
    – DKNguyen
    Commented Apr 27, 2023 at 0:00
  • \$\begingroup\$ @DKNguyen The linked transistor is under 100pF; output pin is ~70 ohm average so gate should see a time constant around 12ns, which agrees with the waveform more or less. Perhaps the bad one is failed, or counterfeit. \$\endgroup\$ Commented Apr 27, 2023 at 0:11
  • \$\begingroup\$ @DKNguyen I specifically chose this MOSFET to be able to be driven from a MCU. 700pC is nothing. And as seen in the oscilloscope picture, Vgs reaches the required values in the required times, but switching does not occur until much later \$\endgroup\$
    – HL65536
    Commented Apr 27, 2023 at 0:13
  • \$\begingroup\$ What is the MOSFET driving? \$\endgroup\$
    – RussellH
    Commented Apr 27, 2023 at 0:13
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    \$\begingroup\$ Could you use a 1nF capacitor in parallel to the 100 Ohm resistor and test it? I see in your first picture that the Ugs-line is not going down so fast like in your test with the second (better working) Transistor. Was there an error in the wiring. \$\endgroup\$
    – MikroPower
    Commented Apr 27, 2023 at 7:47

1 Answer 1

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According to measurements, internal gate resistance is around 400 Ω. Datasheet doesn't specify RG but from the switching times, it can't be much more than 100 Ω. You got dud parts, I'm afraid; buy new elsewhere.

This is a good demonstration by the way -- and kudos for taking complete measurements! -- of how to determine device parameters. Once you know input (gate) voltage and current, given a suitable waveform (like, a sharp step), you can solve for RG and Qg.

To clarify / for reference of those popping in, the GS and GD charges are evident here:

Gate current waveform markup

It's not obvious why Vds drops ~instantaneously (implying Crss has little effect at turn-on), but in any case it does, and this is accommodated by the positive pulse being a little taller/longer.

The negative pulse shows a mostly exponential RC decay as well, but with the Miller blip more evident. Therefore we can reasonably separate the D-G charge as the hatched area.

...Could Crss itself have considerable resistance, say? Perhaps that explains the asymmetry. I have no physical explanation for that, not based on any normal MOSFET fabrication process anyway, let alone failure mode.

There is quite a lot of ringing in the circuit (the layout is probably poor, long wires, breadboarded?), but that's evident as the squiggly lines, and all at a much shorter time constant (the ringing is over within ~100ns).

Also, the asymmetry is definitely not something like a rectification effect (say if the internal resistance had a parallel diode), because the exponential tails are fairly symmetrical (similar peak current and time constant), but for the Miller blip.

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  • \$\begingroup\$ Wouldn't an internal gate resistance of 400 Ω also affect turn-on? \$\endgroup\$
    – HL65536
    Commented Apr 28, 2023 at 0:42
  • \$\begingroup\$ It seems there's enough transconductance at low voltages (that is, at the internal gate node itself) to pull drain voltage down. So, the asymmetry is probably due to the inductive load. Still looks funny, but I don't know in what way it's failed, or otherwise different. \$\endgroup\$ Commented Apr 28, 2023 at 4:05

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