Why do most flyback converters / LED drivers operate under 100 kHz?
There are many reasons:

- Cost: This is probably the biggest reason. The lighting market is quite competitive. There are lots of Chinese players offering aggressively cheap prices. This forces the other brands (even the bigger European ones like Osram, Philips, etc.) to use "cheaper" components to keep themselves in the game. As an employee for one of those companies (honestly, the largest one in Turkey), let me give an example: Our target electronic BoM cost for a 36W constant-current LED driver (36V/1A for 60x60 panel lights) is under 2 USD. This is unbelievable. The ICs that we are currently using in our designs have a maximum switching frequency of 67kHz and they are as cheap as 0.05 USD (As a side note: None of the Chinese IC manufacturers that we work with has higher-speed controller ICs).  I'm asking you: Do we need to use expensive ICs to decrease the size of the transformer or the driver? If the market requires smaller sizes for the same power levels then we can discuss to use higher-speed controller ICs.



- EMC: Electromagnetic Compatibility (especially the radiated emission) can be problematic at higher frequencies. And the PCB should be designed more carefully.



I also want to talk about the skin effect. As you might know, skin depth for copper can be approximated as $$d_s [mm] = \frac{72}{ \sqrt{f_{SW[Hz]}}}$$

For fSW = 60kHz, ds=0.29mm.

For fSW = 100kHz, ds=0.23mm.

For fSW = 200kHz, ds=0.16mm.

For fSW = 500kHz, ds=0.1mm.

This means that the current will flow through the outer ring having a thickness of ds. So the inner circle of the conductor will be empty.

Let's talk on a practical example: To determine the wire diameter for a flyback converter's transformer (actually, it's a coupled-inductor), current density (J) can be selected as 420A/cm² = 4.2A/mm². For example, for a secondary current of Isec=1 Arms, the minimum cross-sectional area of conductor should be \$S_{[mm^2]} = 1_{[A]} / J_{[A/mm^2]} = 0.24mm^2\$ and the required wire diameter (from S = pi w²/4) is \$0.55mm\$.

Of course, we can use a single 0.6mm-dia wire, but the inner half of the conductor will not be filled thus we couldn't use the wire effectively. This will lead to conduction losses. 

To keep the conduction losses minimum, you need to use wires having a diameter of less than or equal to ds so that the conductor is completely filled. And since the cross-sectional area requirement should be met as well, you have to use paralleled wires.

Let's assume the switching frequency is  60kHz. ds=0.29mm at that frequency. So if we use 0.25mm-dia wires, since the cross-sectional area of one wire is 0.05mm², we need to use 0.24/0.05=5 wires parallelled for secondary.

If we increase the frequency to 200kHz then the secondary should be wound using 14 parallelled 0.15mm-dia wires.

Increasing the frequency may cost you more expensive transformer due to the construction difficulties.