When it comes to motor control, I understand that we have the option of using discrete MOSFETs or IGBTs. Also, there are some products in the market where 6 IGBTs are placed in one single package, such as GB25XF120K. (Here's another example part, from Infineon: FS75R06KE3)

However, I don't know how to compare and contrast this solution to using 6 discrete MOSFETs, in terms of:

  • Switching speed
  • Power dissipation (static; what's the IGBT equivalent I2 * RDS, on?)
  • Power dissipation (switching)
  • Cooling (Why isn't there any junction-to-ambient thermal resistance published?).
  • Gate drive circuitry

Also, all the sources I've read on the subject "recommends" IGBTs for high voltages (> 200V) but they don't really get into the details. So I ask the question again, perhaps a bit differently: Why wouldn't I want to use an IGBT for -as an example- a 48V brushless DC motor?

  • \$\begingroup\$ In your Infineon link look at K/W, its a thermal resistance. Just in Kelvins (which has exactly same size as Celsius). Dissipation comes from P=Vce*I as in BJT. \$\endgroup\$
    – user924
    Jul 25, 2011 at 18:26
  • \$\begingroup\$ @Rocket Surgeon: Yes, but none of the thermal resistance values are "***-to-ambient". Is it because a heatsink is always required? \$\endgroup\$ Jul 25, 2011 at 19:41
  • 1
    \$\begingroup\$ You can add arithmetically junction to package, and package to heatsink. The result will be junction to ambient. \$\endgroup\$
    – user924
    Jul 25, 2011 at 19:45
  • 5
    \$\begingroup\$ @Rocket Surgeon - [junction to package] + [package to heatsink] != [junction to ambient]. The first two thermal resistances are conductive and low (~1K/W), for the last one heat exchange is through convection and that thermal resistance is usually much higher than the others added together, often more than 10 times as high for small heat sinks. \$\endgroup\$
    – stevenvh
    Aug 8, 2011 at 9:05
  • 1
    \$\begingroup\$ @stevenvh: I guess it depends on your heat sink. Also, you beat me by 8 seconds. \$\endgroup\$ Aug 8, 2011 at 9:06

1 Answer 1


For a 48 V design with a BLDC motor, you want to use MOSFETs. The reason is that low voltage (< 200 V) MOSFETs are available with an extremely low on-resistance: RDS, on < 10 \$m\Omega\$ for VDS = 100 V is something you can get from at least three different manufacturers in a 5 x 6 mm2 SuperSO8 package. And you get the added benefit of the MOSFETs' ability to switch really fast.

IGBTs become the parts of choice when you want to switch high currents at high voltages. Their advantage is a fairly constant voltage drop (VCE, sat) vs. a MOSFET's on-resistance (RDS, on). Let's plug the respective devices' characteristic properties responsible for the static power losses into two equations to get a better look (static means we're talking about devices that are turned on all the time, we will consider switching losses later).

Ploss, IGBT = I * VCE, sat

Ploss, MOSFET = I2 * RDS, on

You can see that, with rising current, the losses in an IGBT rise in a linear way and those in a MOSFET rise with a power of two. At high voltages (>= 500 V) and for high currents (maybe > 4...6 A), the commonly available parameters for VCE, sat or RDS, on tell you that an IGBT will have lower static power losses compared to a MOSFET.

Then, you need to consider the switching speeds: During a switching event, i.e. during the transition from a device's off-state to its on-state and vice versa, there is a brief time where you have a fairly high voltage across the device (VCE or VDS) and there is current flowing through the device. Since power is voltage times current, this is not a good thing and you want this time to be as short as possible. By their nature, MOSFETs switch much faster compared to IGBTs and will have lower average switching losses. When calculating the average power dissipation caused by switching losses, it is important to look at your particular application's switching frequency - that is: how often you put your devices through the time-span where they will neither be fully on (VCE or VDS almost zero) or off (current almost zero).

All in all, typical numbers are that...

IGBTs will be better at

  • switching frequencies below some 10 kHz
  • voltages above 500...800 V
  • average currents above 5...10 A

These are merely some rules of thumb and it's definitely a good idea to use the above equations with some actual devices' real parameters to get a better feeling.

A note: Frequency converters for motors often have switching frequencies between 4...32 kHz while switching power supplies are designed with swithing frequencies > 100 kHz. Higher frequencies have many advantages in switching power supplies (smaller magnetics, smaller ripple currents) and the main reason why they're possible today is the availability of much improved power MOSFETs at > 500 V. The reason why motor drivers still use 4...8 kHz is because these circuits typically have to handle higher currents and you design the entire thing around rather slow-switching IGBTs.

And before I forget: Above approximately 1000 V, MOSFETs are simply not available (almost, or... at no reasonable cost; [edit:] SiC may become a somewhat reasonable option as of mid-2013). Therefore, in circuits that require the 1200 V class of devices, you just have to stick with IGBTs, mostly.


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.