1
\$\begingroup\$

Power IGBT and MOSFET modules give some datasheet values that look like they could be used to calculate a maximum frequency, but I'm not sure how to interpret them to do so.

The datasheets generally have four timing parameters: turn-on delay time \$T_{D,on}\$, rise time \$T_r\$, turn-off delay time \$T_{D,off}\$, and fall time \$T_f\$. My first guess is that the minimum switching period would be simply \$T_{D,on} + T_r + T_{D,off} + T_f\$.

Just adding all these together can't be right, though, since for an IGBT module that I'm considering using for a project it produces a max switching frequency of over 2 MHz, far above anything I at least would consider feasible. Running this module at 2MHz would result in switching losses that exceed its maximum power dissipation by two orders of magnitude! Is the switching frequency instead limited solely by power and thermal considerations?

\$\endgroup\$
7
  • \$\begingroup\$ I think the turn on and turn off delays shouldn't be simply added. If there were only a turn on delay and turn off delay, both of equal length, the input period of the driving signal would equal the switching period. The switching period would be only delayed in time with respect to the driving signal \$\endgroup\$
    – Huisman
    Mar 3, 2019 at 14:13
  • \$\begingroup\$ @Huisman Right, but as doing the computation that way gives results that are just plain absurd (25kW power dissipation in a 250W module!), I would assume that adding up times like that isn't the proper way of doing it anyway. \$\endgroup\$
    – Hearth
    Mar 3, 2019 at 14:17
  • \$\begingroup\$ Not want to spoil it, but this IGBT has Features - "Switching frequency up to 20 kHz". (I've been taught IGBT's are suitable for high voltage, but low frequencies wrt mosfets lower max voltage, higher frequency. But IGBT's have improved much lately) \$\endgroup\$
    – Huisman
    Mar 3, 2019 at 14:21
  • \$\begingroup\$ But I'm still curious for the answer in general \$\endgroup\$
    – Huisman
    Mar 3, 2019 at 14:22
  • 1
    \$\begingroup\$ @Huisman Well, I suppose I chose the wrong datasheet to use as an example! I didn't realize this particular one specified a frequency. Still, in general they don't seem to, so I feel the question is still worthwhile. Though... With the switching energy this module lists (32mJ per cycle), 20kHz would still be dissipating 640 watts just in the switching losses, which is itself more than twice the rated power.... Huh. \$\endgroup\$
    – Hearth
    Mar 3, 2019 at 14:49

2 Answers 2

Reset to default
1
\$\begingroup\$

It comes down to how energy/power efficient you want your design to be and how much heat your design can get rid of. So, create (say) a spreadsheet that allows you to make calculations and use switching frequency as an input variable and calculate temperature (given some adequate heatsinking) as your output.

You may also want to make load current (and duty cycle) as an input variable too. Just work through it using a spreadsheet is probably the only advice I can give you. If you use the device you specify in your question at least you will be able to check that things start to go pear-shaped at 20 kHz.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you; speed being limited by thermal considerations was what I wanted to know. I don't think we'll be using the device I mentioned in the question, as our target operating frequency is at least 25kHz. We're probably going to be looking at a SiC MOSFET module. \$\endgroup\$
    – Hearth
    Mar 3, 2019 at 20:39
1
\$\begingroup\$

The maximum switching frequency will be determined by thermal considerations. The linked sheet appears to be lacking good application information. Look at the curve at the bottom of the last page. That tells you that although the device is rated for 75 amps, 600 volts and 20 kHz, 20 kHz is the limit for 50 amps and 300 volts. That operating point is valid only at 50% duty cycle and with adequate cooling to maintaining the case temperature at 85C.

There should be additional curves to show the relationship between current and switching losses and other application constraints.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.