I am using the LT3845 buck regulator to step down a voltage from 9 to approx 1.2V.Now, I see that the formula for configuring the Vout is as stated below -


This means I cannot go below 1.231 as Vout. Due to this I change my Vout to 1.3V, this makes my Vout equation as R2 = 0.054*R1.Now, R2 is the upper resistor(conected to the inductor of the buck) and R1 is the lower resistor.

Normally, the upper resistor is higher than the lower one. How is this possible or am I missing something in here please ?


1 Answer 1


You want an output of 1.3 volts but you must have 1.231 volts across R1. This means R2 has a voltage of 1.3-1.231 volts = 69 mV.

Given that current thru R1 = current thru R2, R2 must be smaller than R1 or, put another way R2 = \$\dfrac{0.069}{1.231}\times R1\$ = \$\dfrac{R1}{17.84}\$

  • \$\begingroup\$ How can current throught R1 = current thru R2 ? There is a small current via the fb pin into the amplifier inside the buck no ? \$\endgroup\$
    – Board-Man
    Dec 8, 2015 at 8:30
  • \$\begingroup\$ But I understand your logic. Thank u. \$\endgroup\$
    – Board-Man
    Dec 8, 2015 at 8:31
  • 1
    \$\begingroup\$ VFB Pin Input Current is 25 nA - just make sure your feedback resistors take at least 100 x this for best accuracy. \$\endgroup\$
    – Andy aka
    Dec 8, 2015 at 8:32
  • \$\begingroup\$ But how can you say current through R1 = current through R2 ? This is quite confusing. \$\endgroup\$
    – Board-Man
    Dec 9, 2015 at 6:49
  • \$\begingroup\$ Ignoring the 25nA that might enter the VFB pin, all the current that flows through R1 also flows thru R2 so. if R1 is 1k, R2 will be 56 ohm and, the total voltage across the pair is 1.3 volts. This is a current of 1.230998 mA plus 25 nA = 1.231023 mA or a ratio of 1.00002. So me saying the currents are the same is inaccurate to 2 parts in 100,000 or 20ppm. \$\endgroup\$
    – Andy aka
    Dec 9, 2015 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.