# Power dissipation in Fets synchronous switcher

I'm working on a design that includes a synchronous buck switching regulator and I've been trying to understand how much power each FET will have to dissipate. I've been looking at different formulas and several seem to be in the form below. The answers I keep getting don't make sense to me. I feel like they're off by more than an order of magnitude and that I end up with the wrong units in the end:

Here's the formula taken from powerelectronics.com

Edit: Not enough rep to post image so the formula is on that page half way down as:

$J = 10^{-9} \cdot (\dfrac{V_{in} \cdot I_{out}}{I_{drive}} + \dfrac{Q_g}{Q_{sw}}\cdot V_{drive})*f_{sw}$

(in W/nC)

So just for the high-side I know the following from my design: Vin = 12V Iout = 30A Vdrive = 5V Idrive = 300mA Fsw = 1Mhz

Then from the FET datasheet I get: Qg = 11nC Qsw = 5.6nC

If I plug all that in to the equation I get 1.215 V/sec. Even if my units came out as W/nC that would still be 11*1.215 or 13.365W. Maybe this is right and I need more Idrive but the units come out wrong so I'm not confident in my math so far. What am I missing here?

Power electronics formulae copied from here

They define J & K as follows (their text)

• Since switching losses are proportional to MOSFET size and conduction losses have an inverse relationship to MOSFET size, there exists a point where their sum, total loss, is minimized. This drives the MOSFET optimization process, selecting the MOSFET that has minimum total losses. To that end, we will define two parameters. J will be the switching losses per unit switch charge (W/nC) and K will be the conduction losses per unit drain-source on resistance (W/mΩ). Losses that are independent of MOSFET size have little impact on MOSFET optimization, so are not included in J or K equations.

• Actually, W/nC and V/sec are exactly equivalent. You can think of W as IV and you can think of C as It. The I terms cancel out. – Dave Tweed Jun 11 '13 at 19:05
• Thanks Dave I should have thought of that. The other thing I wasn't sure of was if this number is in W/nC. Do I then multiply that by the Qg in nC to get total number of Watts? – confused Jun 12 '13 at 14:23
• Yes, that seems to be the intent. – Dave Tweed Jun 12 '13 at 15:14
• Yes, for a 30A transistor, you very likely need more than 300 mA drive. 1A drive is common, and there exists driver chips that can dump 6A into the gate to switch it very quickly. – Jon Watte Jun 12 '13 at 19:52

## 1 Answer

The "J" equation includes two terms that are related to switching losses for the upper transistor (Q1).

The first term represents energy lost in the MOSFET channel during the two switching transitions, while VDS is making the transition from Vin to zero and back again. This term is proportional to the transition time, which is determined by QG/Idrive.

The second term represents the energy consumed by the gate driver itself. This is basically proportional to Vdrive×QG. It represents the charge you dump into the gate to turn the MOSFET on, and then dump to ground to turn it off again.

Increasing your driver's current capability will improve the first term by reducing the transition time, but will have essentially no effect on the second term.

I'm not sure if this addresses your actual question; if not, let me know.