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:
I = V / R = 3.75 / 390,000 = 9.61538 uA
9.61538 - 0.15 = 9.46538 nA
4.0 - 3.75 = 0.25 Voltsfor the above current.
R2 = 0.25 / 9.46538e-6 = 26412 Ohms. Closest E12 value = 27 kOhms.
If stability is more desirable than saving quiescent current, try the above sequence with a starting value of R1 as 22 kOhms.
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.