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I am trying to use an LM2596-adjustable to make an inverting Buck-Boost power supply as per below:

Inverting Buck-Boost from ON semiconductor datasheet

The circuit requires a voltage divider that is used to adjust the output voltage. In the diagram, the resistors R4 and R3 somehow adjust the output.

In my application I have not used C1 or R2; the ON/OFF pin is just connected to the output.

If I do not connect any resisters, just connect the feedback pin to ground, I get -1.23v output, as expected.

My question is what values to use for R3 and R4 to get -5v output.

Thanks for any ideas. Bob

gratefulfrog
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2 Answers2

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You may be confused because 'ground' is now connected to the top end of the voltage divider rather than the bottom. However the LM2596 always regulates voltage relative to its own GND pin. The regulator still 'thinks' that it is regulating a positive voltage, and it considers the -12V output to be ground.

the LM2596-ADJ maintains 1.23V between its feedback and GND pins, producing a fixed current through R3. That same current also flows through R4, so its value is adjusted to drop the remaining output voltage. The resistor values are calculated using the same formula as in positive output configuration, but substituting R3 for R1 and R4 for R2 like so:-

R4 = R3 ((Vout/Vref)-1.0)

R3 should be between 1k and 5k. If you want to use standard resistor values then you may have to try various values for R3 until you get a standard value (or close enough to it) for R4.

Bruce Abbott
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  • Are you saying that to get -5v output, with R3 = 1K, then R4= 1((5/1.23)-1) = 3k? I have been using 100k and 33k for my +5v psu, do you think that these values will work? Thanks for your help, sorry to be so ignorant, still...sigh... – gratefulfrog Nov 03 '14 at 08:34
  • When I tried the above circuit with LM2956 feedback pin connected to gnd, I get -1.23v. But if I connect the feedback pin to the output of the voltage divider, I get +0.58v output and the lm2596 heats up. The circuit is powered from a 9v battery which also heats up if I connect the voltage divider, but not if the feedback is connected to battery gnd. Any thoughts would be very welcome :-) – gratefulfrog Nov 03 '14 at 10:58
  • 33k is too high. R3 should be between 1k and 5k (R3=1k and R4=3K is OK). In this configuration the LM2956 can draw a lot of current when starting up, perhaps too much for your battery. You should use the recommended delayed startup circuit (C1, R1, R2) on pin 5. Read the datasheet! – Bruce Abbott Nov 03 '14 at 17:47
  • Thanks again! I have read and re-read the datasheet, in fact many versions of the datasheet. That's why I'm asking questions here ;-) I tried the circuit as you suggested, with C1,R1,R2 setting delayed startup on pin5, R3=1k, R4=3k, and two 9v batteries in parallel to supply more current. Then, with feedback pin4 to ground, I get -1.23v as before. With feedback pin4 connected to the R3/R4 junction, I get 0.2v output and the LM2596 gets burning hot! I am at a loss as to what is wrong, still - but THANKS for your HELP! I really appreciate it! – gratefulfrog Nov 03 '14 at 22:21
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Using small batteries as the DC input will be problematic. I've built a circuit using the same LM2596 part, but without the soft-start circuit and protection diodes. The datasheet really means what's written and can confirm that it is important to have an input source that can supply the high initial current (1A+) and not drop-out. Otherwise the LM2596 will keep drawing an excessive current, the DC input supply drops out, and the whole circuit never works.

Even though that the inverting buck-boost topology is very similar to the buck circuit, keep in mind the design equations and max voltages are slightly different.

python
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