Most of the LED drivers and flyback converters I have come across operate in the 60-100 kHz range. Is there any particular reason behind this - example - ferrite cores not being able to handle higher frequencies, etc.?

Regular buck converters go up to 1-2 MHz switching frequencies and boast about small magnetics that can be used which means smaller size and/or lower costs. Why aren't LED drivers and other flyback ICs coming in higher frequencies?

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    \$\begingroup\$ The chip makers boast about smaller cheaper magnetics because the MHz chips themselves are often outrageously expensive for consumer applications. \$\endgroup\$ Nov 2, 2019 at 11:57
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    \$\begingroup\$ slower switching allows cheaper transistors, and reduced EMI. \$\endgroup\$ Nov 2, 2019 at 12:16
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    \$\begingroup\$ Switching losses.... \$\endgroup\$
    – MadHatter
    Nov 3, 2019 at 2:54
  • \$\begingroup\$ @spehro - Haha funny and true. I have a follow-up question. A lot of people pointed out that EMI and switching losses are the reason behind 100 kHz frequency mark. Will the same not apply to buck converters? \$\endgroup\$ Nov 3, 2019 at 4:33
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    \$\begingroup\$ @Whiskeyjack In part, yes, but the design considerations for those are often different. \$\endgroup\$
    – Mast
    Nov 3, 2019 at 14:55

2 Answers 2


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.

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    \$\begingroup\$ Thanks a lot for this well-written answer. 0.05 USD for LED driver IC seems crazy low. I have received similar feedback from other people in the industry but no one mentioned this low of a price. As the biggest company in Turkey, you guys must be needing to maintain a decent quality. Are these Chinese chips good enough to match your quality standards? If yes, why are the big guys like TI, Power integrations, Diodes Inc, etc so over-priced? Do they have anything extra-ordinary to offer for the price they charge? \$\endgroup\$ Nov 2, 2019 at 12:49
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    \$\begingroup\$ @Whiskeyjack Very good question. We maintain the quality right at the supplier-side. For the LED driver market, we work with China's biggest wafer design and packaging companies directly (e.g. this one is a Chinese state-funded company and this one is founded by former MAXIM employees but located in China). We tell them our requirements then they offer us a decent chip. We purchase millions per year so they can chop off the price more aggressively without lowering the quality. \$\endgroup\$ Nov 2, 2019 at 13:04
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    \$\begingroup\$ ... @Whiskeyjack The big guys are not over-priced. Actually, the workmanship should be expensive. For example, TI's factories are in the US, Japan, etc. and also they hold a lot of patents. These could make their products expensive. Just an example: TI's Turkey distributor offered us the same chip 10x expensive :) \$\endgroup\$ Nov 2, 2019 at 13:08
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    \$\begingroup\$ Also, RF interference. When you go up in switching frequency you start causing radio interference. FCC won't like that. FAA won't either. \$\endgroup\$
    – Mike
    Nov 3, 2019 at 2:52

Mainly because they don't generally need to run any higher than 100 kHz. At the lower frequencies, it's easier to keep the efficiency high with lower-cost components, and the applications they're generally used in don't require the extreme tiny form factors that the higher frequencies allow.

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    \$\begingroup\$ Also switching losses scale linearly with switching frequency. \$\endgroup\$ Nov 2, 2019 at 12:14
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    \$\begingroup\$ @HarrySvensson: That's what I meant by "high efficiency". \$\endgroup\$
    – Dave Tweed
    Nov 2, 2019 at 12:16

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