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I'm trying to make my first voltage regulator circuit using the LM5143A-Q1 Buck IC. The datasheet does include examples and all the values required to get the 3.3/5V fixed output, but it barely has any information on how to get the 0.6 - 55V output. This is the datasheet, and I attached the schematic for the regulation below:

Datasheet schematic for output voltage regulation

I'm still new to this so can you explain me why they used a voltage divider? Isn't that going to always give a FIXED voltage output? Shouldn't the source of feedback be something like a potentiometer like we use in the LM2596? How does this one work?

Output Voltage Settings chart Also, the output voltage chart mentiones FB1/FB2 must be connected to the Rdivider, but, again, that's fixed so how can we for example change the voltage from 12 to 24V with a 50V Input?

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I'm still new to this so can you explain me why they used a voltage divider? Isn't that going to always give a FIXED voltage output? Shouldn't the source of feedback be something like a potentiometer like we use in the LM2596? How does this one work?

  • Most circuit applications (greater than 99% I estimate) for all buck converters will not use a potentiometer but, will use a fixed ratio potential divider. That's the first thing.

  • The second thing is that a small PCB-mounted trimmer potentiometer can usually be used with most buck converters without causing unwanted output instability.

  • The third thing is that using a potentiometer like this....

enter image description here

.... is liable to cause significant instability in the output voltage and, it's unlikely that this is easily predictable beforehand.

Most buck converters operate close to performance limits when it comes to the feedback signal and, any potentiometer wiring (with its associated stray capacitance and ability to pick up unwanted interference) will likely cause significant problems.

It's pot-luck really but you might get lucky. So, if you feel lucky the two resistors in the feedback potential divider might be replaced with a potentiometer.

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  • \$\begingroup\$ Thank you! Only now I get it! I didn't fully understand Equation (5) and that's why I asked, since for my application I need 7.4V Vout. Then this brings another question: If Vref is 0.6, and Rfb1 = (7.4/0.6-1) * 15000 (they said anything between 10K - 20K ohms) = 170kOhms. Is this calculus correct? With the above values (Rfb1 = 170kO, Rfb2 = 15kO) the output voltage should be 7.4V? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 12:50
  • \$\begingroup\$ @Mito please be aware of this: electronics.stackexchange.com/help/someone-answers. 170 kohm is the correct value. \$\endgroup\$
    – Andy aka
    Commented Jul 22, 2023 at 12:54
  • \$\begingroup\$ ok, but just any 170kohm resistor will work? If I go to Mouser I also get Power Rating, Tolerance and Temperature Coefficient as properties. Could you briefly explain these so that I can try and pick the right SMD resistor? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 13:38
  • \$\begingroup\$ The 170 kohm resistor will have 7.4 minus 0.6 volts across it therefore, the power it will dissipate is really rather small at 270 microwatts. Virtually any resistor will suffice but, choose one with a decent tolerance, a low-ish temperature coefficient (100 ppm/degC or lower) and small size to avoid the problems I alluded to in my answer @Mito \$\endgroup\$
    – Andy aka
    Commented Jul 22, 2023 at 13:44
  • \$\begingroup\$ what do you mean by decent tolerance? How much is good enough? I see options from 0.05% to 5%, also the Temperature coefficient is from 5 to 200 ppm/deg C. What exactly does the Temperature coefficient mean? \$\endgroup\$
    – Mito
    Commented Jul 22, 2023 at 15:40

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