# Resistor values for adjustable voltage regulator

I'm using a LT1529 voltage regulator and I'm having a hard time understanding the process for determining the resistors to use in the voltage divider. The datasheet provides the following schematic and calculation:

ADJ (Pin 2): Adjust Pin. For the LT1529 (adjustable version) the ADJ pin is the input to the error amplifier. This pin is internally clamped to 6V and – 0.6V (one VBE). This pin has a bias current of 150nA which flows into the pin. See Bias Current curve in the Typical Performance Characteristics. The ADJ pin reference voltage is equal to 3.75V referenced to ground.

I want an output voltage of 4.0V. The typical ADJ pin bias current (which I think is the desired current for the adj pin) is 150nA. I'm not sure if I should use these values to determine the R2 resistor. It gives a rather large resistance value (like 26.6M). Once I know R2, R1 should be easy to solve for but I would appreciate confirmation on that value too.

Turn the problem around: Once you know R1, then R2 is easy to solve for. R1 is pretty easy to decide upon.

It is a bit of a balancing act, really, but here's how it goes:

• The datasheet suggests R1 be kept below 400 kOhms for stability. Lower the value of R1, higher the quiescent current required by the voltage splitter R1+R2. Higher the value of R1, higher the instability of the output.
• We know that the upper leg of R1 in the diagram is biased at 3.75 Volts for steady state.
• Hence, let us start with the maximum standard E12 series resistor value for R1 within the datasheet constraints, i.e. 390 KOhms
• IR1 can be calculated thus: I = V / R = 3.75 / 390,000 = 9.61538 uA
• Current through R2 is given as the sum of current through R1, and bias current 150 nA. IR2 is thus 9.61538 - 0.15 = 9.46538 nA
• For a desired output voltage of 4.0 Volts, R2 must thus develop 4.0 - 3.75 = 0.25 Volts for the above current.
• Therefore R2 = 0.25 / 9.46538e-6 = 26412 Ohms. Closest E12 value = 27 kOhms.
• Vo with R1 = 390 k and R2 = 27 k is 4.01367 Volts, less than 0.5% deviation from target voltage (assuming perfect resistor values, of course).

If stability is more desirable than saving quiescent current, try the above sequence with a starting value of R1 as 22 kOhms.

• IR1 = 170.455 uA
• IR2 = 170.305 uA
• R2 = 1468 Ohms, nearest E12 value 1.5 kOhms
• Vo = 4.00591 Volts.

Using the above calculation steps, choose any value for R1 as long as it is less than 400 kOhms, to obtain the value of R2.

• This looks like exactly what I need. Thank you. One other question while I have your attention. Do you have any suggestions for the value of the capacitor? I suppose it depends on my load? Jul 26, 2013 at 16:37
• The minimum recommended value is 22µF with an ESR of 0.2Ω or less, says the datasheet. Start with that value, increase it if ripple is unacceptable. Jul 26, 2013 at 16:55

The datasheet seems to be clear. See Fig 2 (pg 8) and the equation under this figure shows that it's a standard non-inverting amplifier gain equation. If you pick a value for R1, say 100k, the equation can be re-written as 100K(Vout/3.75 - 1) = R2. They suggest making R1<400k to minimize errors due to the bias current, that's why I picked 100k.