I am trying to use a L4978 to create a 6 V regulator. I, however, am unable to find how they calculate the resistor divider in Figure 4 to get a given output voltage. Does anyone know how they are calculating this?

Figure 4. Test and evaluation board circuit.


1 Answer 1


The chip will be happy when the feedback voltage on pin 8 is 3.3 V. So for a 3.3 V output set the upper resistor, R3, to zero. Any other voltage can be obtained by simple voltage divider using R3 and R4.

The sample circuit uses a constant 4.7k for R4 and at 3.3 V there will be 0.7 mA flowing through it. For 6 V you will want the output 2.7 V higher than 3.3 V and since we know the current we can calculate R3.

R4's value will be mid-range. Too low and it will waste current. Too high and it may be loaded by pin 8 or susceptible to noise.

We can work out the formula. The relationship between \$ V_F \$, the feedback voltage on pin 8, and \$ V_O \$, the output voltage is given by

$$ \frac {V_F}{V_O} = \frac {R4}{R3+R4} $$

Rearranging we get

$$ V_O R4 = V_F(R3+R4) $$

$$ (V_O - V_F) R4 = V_F R3 $$

$$ R3 = \frac {(V_O - V_F) R4}{V_F} $$

Sample test for 12 V out:

$$ R3 = \frac {(12 - 3.3) 4k7}{3.3} = \frac {40k9}{3.3} = 12k4 $$

The datasheet recommends the nearest E12 value of 12 kΩ so our maths looks right.

We could work out the error with this rounding from

$$ V_O = \frac {R3 + R4}{R4}V_F = \frac {12k + 4k7}{4k7} 3.3 = 11.7 \, \mathrm V$$

You can keep on going with this by examining the conditions with max and min combinations of resistor tolerance and \$ V_F \$ for the chip but most of us would be happy enough with 11.7 V.

  • \$\begingroup\$ I don't believe this answers my question, or perhaps I worded it incorrectly. But a resistor divider requires 3 variables in order to calculate the resistors values, which three variables do I use to calculate R3 and R4? Or in this case just R3, as R4 seems constant. I'm looking to generate 6v. \$\endgroup\$
    – Alex
    Sep 25, 2016 at 22:25
  • \$\begingroup\$ Any better now? \$\endgroup\$
    – Transistor
    Sep 25, 2016 at 22:34
  • \$\begingroup\$ ah so if im understanding correctly, we just want to add voltage to the 3.3v to get to the desired output. such that the voltage source in a way is 3.3, R3 = n, R4 = 4.7, and output will be the difference between desired and 3.3? \$\endgroup\$
    – Alex
    Sep 25, 2016 at 22:42
  • 1
    \$\begingroup\$ Not sure exactly what you mean, but given those numbers, the output voltage will be ((3.3/4.7)*n)+3.3) Where n is in kOhms \$\endgroup\$
    – BeB00
    Sep 25, 2016 at 23:13
  • 1
    \$\begingroup\$ @Alex. Answer expanded further. Let me know if anything is unclear. Thanks for accepting the answer. \$\endgroup\$
    – Transistor
    Sep 26, 2016 at 6:44

Your Answer

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

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