Electronic – Need help interpreting a basic circuit question

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: R_DS, on < 10 \$m\Omega\$ for V_DS = 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 (V_CE, sat) vs. a MOSFET's on-resistance (R_DS, 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).

P_loss, IGBT = I * V_CE, sat

P_{loss, MOSFET} = I² * R_DS, 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 V_CE, sat or R_DS, 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 (V_CE or V_DS) 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 (V_CE or V_DS 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.

A Coulomb is a unit of electrical charge, and electrical charge is a property of electrons. So moving electrons from A to C via B means there's an electrical charge passing point B. How much charge? Well, the charge of a single electron is quite small: \$-1.6 \times 10^{-19}\$ Coulomb. That explains why there have to pass so many electrons before a full Coulomb is transferred. The speed at which this happens determines current, in Ampere.
So, by definition, if you see \$ 6.24 \times 10^{18}\$ electrons pass in 1 second you have 1A. If you see twice as many pass in 1 second, or \$ 12.48 \times 10^{18} \$ that's 2A. If only 1 electron passes your checkpoint every second you have a current of \$1.6 \times 10^{-19}\$A, or 0.16 attoAmpere (aA).

(By the way, the number you've read as expressed here is wrong. Expressed the way you write it, with all the zeros, it suggests 19 significant digits. The real value is more like \$ 6.241 509\times 10^{18} \$. Writing it this way means that these are the first 7 digits of the number, but it expressly says the following digits probably aren't zeros. It's the scientific way to express rounded numbers. Writing out all digits is only done if all digits are correct. Here you can see that there's already a discrepancy at the third digit. )