Electronic – LED diodes as both diode and load for boost converter

boostcobled

I'm doing an LED project with a Cree CXB3590 LED that I intend to underpower to experiment with efficiency and learn about boost converter design. I've made a 5uH inductor with 14 gauge magnet wire wound onto a T50 toroidal yellow/white(iron powder) core. I plan to use a 555 clock circuit and a 555 pulse generator with a horrible oscilloscope to determine the maximum pulse length for the inductor core and then experiment with frequency to compare total light output from the CXB3590 at different power levels to the output from single LEDs such as an XP-G or XR-E at the same power level. I'll be using PWM with a frequency limit ~1Mhz due to the iron powder core inductor. The large wire on the inductor and a decent MOSFET are intended to keep resistive losses as low as possible, hopefully negligible.

Because the CXB3590 is capable of using much more power than the boost converter circuit is likely to provide, I'm hoping to use it as both output diode and load, making the assumption that with each pulse the inductor will drive voltage up to the on voltage of the LED (~33-34V) and when that voltage is reached because the LED's effective resistance will become extremely low, it will not rise significantly higher as the inductor dumps its stored energy.

I'm hoping that this will result in the LED operating in a favorable efficiency band, operating in discontinuous mode and at close to the lowest voltage that will allow it to dissipate the power being pumped into it.

schematic

simulate this circuit – Schematic created using CircuitLab

So my question is this: For what reason(s) I would benefit from using the topology shown on the right, such as the inclusion of a capacitor helping clamp down the voltage the LED sees? the CXB3590 is rated at 86.4W, so at the ~3-10W I intend to operate it at, as I mention above it will likely operate discontinuously.

Best Answer

ADD: The left cct cannot run in CCM (continuous conduction mode) because the switch also shorts out the LED to 0V. But besides that now you have a low ESR path with a discharged cap that draws as much current as the inductor can swing V+/DCR and causes a high Q current resonant cct and adding pulses adds fuel to the fire with burnout on the series resonant circuit.

That's a serious flaw on the left. 1st it starts a low frequency resonance, the as the current builds up in the inductor the cap voltage is discharged at the same time. So the series diode switch is essential by lowering the Q switching the Cap off and ON. and allowing the voltage to accumulate with current switched pulses **


Examine the impedance of every part and try to choose parts towards 0.1% of LED ON resistance. (incremental V/I) to minimize losses. ( i.e. 2 mOhm ballpark )

Assuming 36V LEDs: (although you would be wiser to use 72V LEDs)

Rled = 1.7Ω= 1V/0.6A = 35.5V/2A - 36.5V/2.6A (slope)
Fast Power Schottky Diode: ~ 2 mOhm @ 20A STPS5045SG-TR 50A rated
FET Switch: <5m Ohm = RdsOn @ 10Vgs

Operate gate drive frequency at 10k to 50kHz max ( lower is more efficient)
Operate duty cycle < 75% pref. <50% ON time.

enter image description here Choose large cores e.g. T50-26 = 0.5"D , AL= 33 nH/N² with 12 turns AWG16 magnet wire 3.2mΩ to get L = 33nH*240 = ~ 7.9 uH

Then with a very tight layout, attempt to achieve these current pulses in Falstad simulation.

LED Current = Cap Current (-ve ramp discharge) + Diode pulse current (+ve ramp charge)

Compute the efficiency from the floating power graphs with Average power.
+12V =-65.827 W
LED+ = 64.256 W
**loss = 1.571 W
/ 66W 100% = 2.5% @ 33kHz mostly in FET.
using low ESR , DCR, Rs (diode) components.

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