Why have an inductor between 2 parallel capacitors? (LED driver)

Question

This is the driver for an 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?

Answer 1

Driving low voltage high current LEDs off an AC line requires a smoothened current pump and 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 the Vac peak, so C1 is small.

• Yet it must have large enough energy capacity 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

  • current pump decay time to store the energy Ec=1/CV^2 must just exceed DC energy of load
  • yet peak power for 1/2 cycle AC depends on 50% duty cycle .

• 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

• The impedance for input and output are high input

• The Zout is low and as LPF ripple attenuation and Q are limited by Requiv. / Zc(f)

• Here C1 does not do much and the LCL filter resonates around 1.8 kHz with high Q when conducting

• When the LED’s are OFF, it is critically damped with an LED ~ < 15 ohms each) + Rs for a total of about 100 ohms for 5mm parts

• The 2nd L or T1 is now switching on and off so the PWM and frequency depend on the hysteresis, Vac input and Power output.

• If the freq is 18kHz, the ripple is reduced 40dB

• It is mostly a 2nd order filter in this range

• C1 also reduces some radiated EMI voltage noise levels but raises EMI conducted current levels so it is chosen 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.

I hope this reads a bit better.

I suggest you learn the Falastad SImulator and test this out. Beta for the FET can be changed calculated for RdsOn with FET properties. Start with Beta=1 (higher for lower Ron) e.g. http://www.falstad.com/circuit/circuitjs.html or my example

Answer 2

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.