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? – Matt Ruwe Jul 26 '13 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. – Anindo Ghosh Jul 26 '13 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.