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This is the driver for a LED Bulb. I don't understand why there is an inductor between the 2 capacitors C1 and C2? I understand that C1 and C2 are used to handle low frequency ripple and the noise. But what does the 3mH inductor do?

schematic diagram

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    \$\begingroup\$ Google "pi filter". \$\endgroup\$ – John D Jan 30 at 20:49
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    \$\begingroup\$ @JohnD Thanks! I lost a lot of time because I didn't had the right term and couldn't find any information! \$\endgroup\$ – MagTun Jan 30 at 20:51
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    \$\begingroup\$ Note that, because the inductor and resistor are there, the capacitors are not in fact parallel. \$\endgroup\$ – Hearth Jan 30 at 22:42
  • \$\begingroup\$ @Hearth, so what are they if they aren't in serial or in parallel? \$\endgroup\$ – MagTun Feb 11 at 16:46
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    \$\begingroup\$ @MagTun Ohm's law doesn't care about the context. If you're talking about circuit transformations, for a situation like this you may need to use the wye-delta transform (or the general form, the star-polygon transform), which is a little bit more complicated than the basic series and parallel transformations people learn at the beginning of EE101. In addition to that, Kirchoff's laws are your friend. \$\endgroup\$ – Hearth Feb 11 at 18:10
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Driving low voltage high current LEDs off AC line requires a smoothened current pump then a precision constant current sink to regulate LED heat over a wide range of input Vac.

The bridge has a low resistance so surge currents into C1 must be low at turn on during Vac peak, so C1 is small. Yet large enough to store enough power to reduce the ripple voltage and thus current in L1. L1 must be large enough to feed the current to the LED with minimal DCR for current inductor rating to drive the LED string voltage, Vf, so I(Lmin) > I(led)

If L is too small it will rapidly decay with a load ESR of ~ 0.5 n/Pd for n LEDs in string total power, Pd and current pump decay time while C2 is large to store the energy =voltage * current * 1/2 AC cycle time and stay above Vf.

T1 further helps to reduce the slew rise in current but when If is sensed the drain cuts off with hysteresis, so you now have a hysteretic buck Pulse duration and frequency modulated step-down constant current sink with a bit of ripple determined by the %hysteresis.

You may consider it in the frequency domain as a Pi filter or CLC filter but the impedance for input and output are high input and low output Z as well as LPF ripple attenuation and limited Q. Here C1 does not do much and the LCL filter resonates around 1.8 kHz with high Q when conducting and LED’s off then critically damped with an LED Rs of about 100 ohms for 5mm parts and less with more power well damped current. But the 2nd L or T1 is now switching on and off so the pulse width and frequency depend on the hysteresis and Vac input and Power output. So if the freq is 18kHz the ripple is reduced 40dB.p as it is mostly a 2nd order filter in thus range and C1 also reduces some EMI voltage noise levels but raises EMI conducted current levels so it is kept smaller by that selected ratio.

The open drain regulates the current to be constant with residual current between on cycles at >10x the Pi filter with the shunt cap across the LEDs depending C value RsC=T being dt= dV*C/Ic for the hysteretic dV value.

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C2 handles the low frequency ripple, for power into the LEDs.

The inductor+C1 is to stop RF from the power conversion switching from getting back onto the mains, and making the module fail emission standards.

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