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.

## Best Answer

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:

below 400 kOhmsfor 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.390 KOhms_{R1}can be calculated thus:`I = V / R = 3.75 / 390,000 = 9.61538 uA`

_{R2}is thus`9.61538 - 0.15 = 9.46538 nA`

`4.0 - 3.75 = 0.25 Volts`

for the above current.`R2 = 0.25 / 9.46538e-6 = 26412 Ohms`

. Closest E12 value =27 kOhms.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._{R1}= 170.455 uA_{R2}= 170.305 uA1.5 kOhms4.00591 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.