Ive been reading some of the LT application notes on Jim William's Analog Circuit Design book, and there's something I don't quite understand. I'm attaching a scan of the circuit in question.

Negative to Positive Buck Boost Converter

The LT1070 has its ground pin at -12V, and Q1 is forcing a current of around 1mA through the 1.24k resistor, this produces a 1.24V drop through that resistor, which corresponds to the 1.24V reference inside the FB pin (the error amplifier).

Ok I understand that, what I don't understand is why use the transistor in the first place? Why not just use a simple resistive voltage divider? whats the advantage of using Q1 rather than a simple resistive voltage divider?


1 Answer 1


The transistor's base is tied to ground, which means that the current through R1 depends only on the output voltage (along with the base-emitter drop of the transistor). The transistor sets the current through the 1.24K resistor (R2?), but keep in mind that there's also a significant voltage drop (roughly Vin - 1.24V) across the transistor itself, which varies with Vin.

If you just set up a resistive divider without the transistor, then the current through it would depend on both Vin and Vout, which would eliminate the ability of the circuit to ignore changes on Vin (line regulation). Sometimes you do this on purpose, if your intent is to create a "tracking" supply, but that isn't the intent of this circuit.

  • \$\begingroup\$ Thats what I thought, however the -12V rail is the LT1070 ground so, if the -12V rail varies, isnt the ground of LT1070 making the LT1070 change the output? \$\endgroup\$
    – S.s.
    Aug 12, 2014 at 21:38
  • 1
    \$\begingroup\$ No, not really. Remember, the "Vin" pin of the LT1070 is tied to ground, so the output voltage really only depends on the energy stored in L1, and doesn't depend directly on the "ground" of the chip. Basically, the chip varies the duty cycle of its output in order to maintain 1.24V between its FB pin and its GND pin. The duty cycle controls how much energy is stored in L1 on each switching cycle, which determines Vout, which determines the current through R1, completing the feedback loop. Line and load regulation is the same as it would be in a non-inverting configuration. \$\endgroup\$
    – Dave Tweed
    Aug 12, 2014 at 22:08

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