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Here's a sample datasheet from a FET driver:

http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=ucc27211a&fileType=pdf

Section 9.2.2.5 describes calculating power dissipation. Obviously the dynamic power dissipation goes up with switching frequency.

However, what I am unable to determine is should I use a full-period or a half-period for this calculation.

I've always assumed that "switching frequency" means "what is the frequency of switching events"

For example if the minimum time between changing states (either from low to high, or high to low) is 2us. This means a switching event happens every 2us, so the frequency of switching events is 500kHz. However I've seen several places where the term "switching frequency" seems to imply a full cycle, so the frequency for this would be 250kHz as it takes 4us for it to go from one "low" to the next "low" state.

Which definition is the one that TI is using in this data sheet?

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  • \$\begingroup\$ Yes, this is the frequency of "events". But "events" are usually some waveforms, pulses, returning back to some initial baseline. This is the necessary part of definition of an "event", some steady state, and then an "event" occurs at certain "frequency". Therefore the simplest "event" (pulse) consists of minimum TWO edges. There could be more switching edges in one "event" that is repeating "frequently". \$\endgroup\$ Jun 28 '18 at 15:44
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The linked datasheet tells us this:

PG = CLOAD × VDD^2 × fSW = QG × VDD × fSW (4)

This power PG is dissipated in the resistive elements of the circuit when the MOSFET/IGBT is being turned on and off. Half of the total power is dissipated when the load capacitor is charged during turnon, and the other half is dissipated when the load capacitor is discharged during turnoff. When no external gate resistor is employed between the driver and MOSFET/IGBT, this power is completely dissipated inside the driver package. With the use of external gate-drive resistors, the power dissipation is shared between the internal resistance of driver and external gate resistor.

(emphasis mine)

So they consider the whole cycle. I've not yet come across the interpretation of switching frequency like you had in mind.

Note that they already calculated that you have to take double the frequency in calculation of the power:

The energy to charge the gate would just be: \$\frac{1}{2} C U^2\$ and that energy is wasted at every edge (like you said). So the power dissipation is: \$\frac{1}{2} C U^2 \times 2 f_{SW}\$

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  • \$\begingroup\$ +1 because it may help the OP read the datasheet, a generally useful skill. \$\endgroup\$ Mar 23 '18 at 12:20
  • \$\begingroup\$ I think the confusion may stem from devices which are fed from mains power using full-wave rectifiers. There, even though the mains power cycles sinusoidally 50 or 60 times per second, the downstream circuitry will see an irregular 120Hz waveform. \$\endgroup\$
    – supercat
    Mar 23 '18 at 13:02
  • \$\begingroup\$ @supercat which is usually called a rectified AC waveform. All encounters I had with switching frequency were exactly like it is described in all the answers here. \$\endgroup\$
    – Arsenal
    Mar 23 '18 at 22:27
  • \$\begingroup\$ Yeah, the gate energy formula does have a 1/2, I missed that, so this is definitely correct. I don't agree that your emphasized sentence implies one or the other definition of switching frequency though, because regardless of how you define your terms, it is still true that half of the power will be dissipated during turn-on and half during turn-off. \$\endgroup\$
    – Jason
    Mar 25 '18 at 5:55
  • \$\begingroup\$ @supercat I wouldn't call that a switching frequency since it's sunusoidal, and "switching" to me implies an on/off square wave. The confusion for me is that there are two switching events in a single period. This is good news for me because I had less thermal margin than I wanted using my wrong definition, and I just halved the power I need to dissipate from this part which means I may not even need a gate resistor. \$\endgroup\$
    – Jason
    Mar 25 '18 at 6:06
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You need to apply the concept of period: the time interval from the beginning of the first pulse to the start of the second.

In the picture below I tried to depict this more clearly. enter image description here

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  • \$\begingroup\$ This is a great visual for illustrating it. The only part I wasn't sure on before asking the question was whether T or T/2 was the correct period to use for calculating the switching frequency. \$\endgroup\$
    – Jason
    Mar 25 '18 at 5:57
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I've always assumed that "switching frequency" means "what is the frequency of switching events"

"Switching frequency" is a real frequency and not an implied value based around "on" or "off" periods. Your definition (or belief) is wrong.

I've seen several places where the term "switching frequency" seems to imply a full cycle

That is the correct definition.

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  • \$\begingroup\$ periods and frequencies are just inverses of each other. If you have a period, you have a frequency and the reverse. \$\endgroup\$
    – Jason
    Mar 25 '18 at 5:53
  • \$\begingroup\$ @Jason yes I know. \$\endgroup\$
    – Andy aka
    Mar 25 '18 at 9:38

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