PFC boost transistor power loss calculation

[Miyuki]

Hi,
I want to ask about calculating power loss at PFC transistor
switching loss I can take from rms current, output voltage ad frequency (it is simple)

but is here any simple formula to calculate conductive loss (average duty)?

[Miyuki]

I found this equation, results looks reasonable, but can please someone explain me where it come form ?

[danadak]

I am thinking you cannot just use DC and load current to get at that,
as that does not account for switching losses in power switch. Usually
the vendor, like TI, have a simulator that will allow you to determine that
for various loads. Even includes thermal considerations.

Regards, Dana.

[Miyuki]

Infineon converter design guide says that this simple equations are accurate enough

𝑃𝑆.𝑜𝑛 = 0.5 ? 𝐼𝐿.𝑎𝑣𝑔 * 𝑉𝑜 * 𝑡𝑜𝑛 * 𝑓
𝑃𝑆.𝑜𝑓𝑓 = 0.5 ? 𝐼𝐿.𝑎𝑣𝑔 * 𝑉𝑜 * 𝑡𝑜𝑓𝑓 * 𝑓
and
𝑃𝑆.𝑐𝑜𝑛𝑑 = 𝐼𝑆.𝑟𝑚𝑠² * 𝑅𝑜𝑛

[danadak]

The equations seem to avoid the issue of gate drive Pdiss, but that would be OK if
Id x Id x Rdson x DC dominates the overall dissipation.

This might help -

https://www.power.com/sites/default/files/product_document/application_note/AN-1001_IGBT_and_MOSFET_Drivers_Correctly_Calculated.pdf

Regards, Dana.

[orolo]

I found this equation, results looks reasonable, but can please someone explain me where it come form ?

I'm no expert in this, in fact I'm a mathematician, not an EE, but the coefficients in your formula rang a bell, so perhaps I can help a bit.

If you have a sinusoidal input current, $$V_{peak}\cos\omega t$$ and pass it trough an ideal rectifier, you get $$V_{peak}  \vert \cos\omega t \vert$$. Taking the Fourier series of the rectified voltage, you get, to the second harmonic:

$$ \frac{2V_{peak}}{\pi} + \frac{4V_{peak}}{3\pi}\cos 2\omega t + \ldots $$

If your $V_{ac.min}$ is RMS, then, the input voltage is:

$$\frac{2\sqrt{2}V_{ac.min}}{\pi} + \frac{4\sqrt{2}V_{ac.min}}{3\pi}\cos 2\omega t + \ldots $$

The amplitude of the second harmonic is clearly recognizable in your formula, and in looks to be like a duty cycle format (disregarding efficiency, and taking the peak-to-peak voltage):

$$ D = 1 - \frac{V_{in, min}}{V_o} = 1 - \frac{8\sqrt{2}V_{ac.min}}{3\pi V_o} $$


So, somehow, your formula looks like:

$$I_S = \frac{P_o}{V_{ac.min}}\sqrt{D}$$

This is the part I don't really understand. I tried the computation for a normal booster, and found something along the lines of $$I_S = \frac{P_o}{V_{ac.min}}\cdot D^2$$ (continuous mode), so I don't know where the square root comes from. I also don't know why the zero-order harmonic doesn't come to play into the duty cycle, unless your output voltage equals the zero-order rectified voltage for some reason.

Edits: LaTeX cleanup.