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Does anyone know if it's possible to connect the feedback pin of a DC-DC boost converter chip like Texas Instruments' LM2733 in a different manner than the one described in the datasheet, namely on the collector of the output side of a standard BJT current mirror, like so :

schematic

simulate this circuit – Schematic created using CircuitLab

In which case
formula
With Vfb=1.23V for this specific chip.
This would allow minimizing losses in Q2 by setting Vce a little bit above 0.7V, without having to know the exact voltage of the LED string at a given current.
However I'm way out of my depth here, so I'm wondering :

  1. Is this worth trying?
  2. Is there a risk that the chip will go berserk due to the nonlinearity of Vce as a function of the voltage across the load?
  3. Where should the forward compensation capacitor be connected? What should its value be?

Thank you very much.

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    \$\begingroup\$ as long as the loop gain remains about the same, as well as the bandwidth, why would this not work? \$\endgroup\$ – analogsystemsrf Mar 31 '19 at 2:39
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    \$\begingroup\$ If you are trying to regulate current through the LEDs using a switching regulator, why not use a resistor in place of the transistor and save a lot of hassle? \$\endgroup\$ – Andy aka Mar 31 '19 at 11:02
  • \$\begingroup\$ @Andyaka Because there are several strings I want to match. \$\endgroup\$ – Oliver J. Whale Mar 31 '19 at 13:54
  • \$\begingroup\$ @analogsystemsrf How do I make sure that they do ? \$\endgroup\$ – Oliver J. Whale Mar 31 '19 at 13:56
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I'd suggest what you are trying to do would work with the changes shown below.

The boost convertor would threshold the Vce of Q2 at about 1.5V.
D2 ensures you can't swing FB beyond supply.
C2 is for ripple reduction on Q2 collector.
C1 is for storage for the boosted supply.

I'm not sure about operating this at over 1MHz, but it should work at 600kHz.

I also assume you want to run multiple strings of LEDs from the same boosted supply (it's actually quite clever).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you very much for your answer. How bad would it be if Vfb ever exceeded Vin ? Wouldn't the IC adjust the duty cycle accordingly (to 0 if the transistor fails open in which case Vfb=Vout absent D2) ? \$\endgroup\$ – Oliver J. Whale May 8 '19 at 1:03
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    \$\begingroup\$ From the datasheet it seemed inadvisable to allow V(fb) to go above the supply. For example if there was a wiring error that shorted out a LED string. \$\endgroup\$ – Jack Creasey May 8 '19 at 2:19

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