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How do I pick resistors for a 12v-30v boost circuit using an SDB628 IC; Vout= 0.6(1+R1/R2). I am using a 50k pot for R1. How do I get that 12v-30v output voltage range? To get 30V, my calculation of R2 is 1K; to get 12v, R1 has to be 19K, but since this pot can go down to 18K ohm, even 0 ohm, so many precision range of the pot is wasted. Then let me assume that the 50K pot is at its 0 ohm position, then I need to add another resistor to get 12V. My calculation is 19K for R0 to get 12V. (R0+R1)-(R2); but the problem is when the position of the pot went up to 50K, Vout will be much higher than 30V. doesn't work either way.

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  • \$\begingroup\$ You can put a resistor in parallel with the pot, the pot will dominate at lower settings and the second resistor will dominate at higher settings. E.g. a 33K in parallel with the 50K pot would give a max effective resistance of just under 20K at one end and whatever the pot's minimum is at the other end of the pot's range. Then add that series resistor to limit the lowest resistance. Two resistors and one pot overall now make up R1. (Rseries + Rpot//Rparallel) \$\endgroup\$
    – Sam
    Jun 9 '16 at 6:12
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As your post is not very clear, I will assume what you want to do is to control the output voltage of your IC with a Potentiometer. So, I would suggest you to use a circuit like this one:

schematic

simulate this circuit – Schematic created using CircuitLab

To calculate the values for the resistor, just replace R1 in your formula with R11 + Pot. Consider the voltage to be maximum (30V) when Pot value is 50k, whereas the voltage is minimum (12V) when Pot is 0. This will give you the following equations to solve:

  • 0.6 x (1 + R11 / R2) = 12V
  • 0.6 x (1 + (R11 + Pot) / R2) = 30V

However, the values you will obtain are not standard. Here are some ideas to get more practical values, so the actual resistors will be easier to find:

  • Use several resistors in series or in parallel to do the job. Ex: for R2, 3 5k resistors in parallel will give the value
  • Change the potentiometer: a 150k potentiometer will give you 5k for R2 (easy to find) and 95k for R11 (more difficult, but can probably be done using 2 resistors in series)
  • Replace R2 by a resistor (R21) and the Potentiometer. I didn't do the calculation, so I don't know if it would be easier or not.
  • Put the resistors in parallel instead of in series, as suggested by Tom in the comments
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  • \$\begingroup\$ That is exactly what my question is. I checked the values, and they are all correct, but how do you get 31k and 1.7k? what math or equation did you use to solve it>? \$\endgroup\$
    – Atmega 328
    Jun 9 '16 at 18:29
  • \$\begingroup\$ The maths used are the 2 equations detailed at the beginning of the answer. it comes from the equation you provided - "Vout= 0.6(1+R1/R2)", where R1 is replaced by a resistor and a potentiometer in series. Then, you just consider the potentiometer is minimum (0 Ohm) for the lowest voltage (12V), and maximum (50 kOhm) for the highest voltage (30V). Then, it is just a system of 2 equations with 2 unknown to solve. \$\endgroup\$
    – Edesign
    Jun 10 '16 at 7:24
  • \$\begingroup\$ If you want to know where the equation comes from: a regulator will adjust its output voltage (Vout) in order to have a fixed voltage in its feedback pin (Vf). In your case, the feedback should be 0.6V. This is where this value in the equation comes from. Then, you calculate Vf from Vout using the formula of the voltage divider, and you invert it, in order to have Vout from Vf: Vf = (R2 / (R1 + R2)) x Vout ==> Vout = Vf x (1 + R1/R2) \$\endgroup\$
    – Edesign
    Jun 10 '16 at 7:33
  • \$\begingroup\$ For the resistors, maybe you will find these particular values from your "preferred distributor", but they are not standard. I would definitely suggest you to use a 150kOhm potentiometer (standard), R2=5kOhm (standard) and R11=92kOhm (not standard, but you can easily do it with 2 standard resistors in series, such as 82k and 10k for example) \$\endgroup\$
    – Edesign
    Jun 10 '16 at 7:38

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