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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?

schematic diagram

JRE
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MagTun
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    Google "pi filter". – John D Jan 30 '19 at 20:49
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    @JohnD Thanks! I lost a lot of time because I didn't had the right term and couldn't find any information! – MagTun Jan 30 '19 at 20:51
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    Note that, because the inductor and resistor are there, the capacitors are not in fact parallel. – Hearth Jan 30 '19 at 22:42
  • @Hearth, so what are they if they aren't in serial or in parallel? – MagTun Feb 11 '19 at 16:46
  • @MagTun What do you mean? They're not in series and they're not in parallel, that's certainly allowed. It would be a boring world if everything had to be one of the two. – Hearth Feb 11 '19 at 16:59
  • @Hearth, ok but then how to you know which context (series vs parallel) to use to apply Ohm's law to find out the current / resistance / voltage in each "branch"/composant? – MagTun Feb 11 '19 at 18:07
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    @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. – Hearth Feb 11 '19 at 18:10
  • Thanks for your reply! By context, I was talking about this https://www.allaboutcircuits.com/textbook/direct-current/chpt-5/correct-use-of-ohms-law . Thanks for the wye-delta transform and the Kirchoff's laws, I think that was what I was missing! – MagTun Feb 11 '19 at 18:13

2 Answers2

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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.

Neil_UK
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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

Tony Stewart EE75
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