At the feedback voltage divider of a voltage regulator, is it ok to choose a large value for R2 (the bottom res in divider)?
I have read it makes FB pin noise sensitive ( This is from a customer provided datasheet under NDA). All I know is a large R2 can alter the divider ratio since op-amp internal impedance also starts playing a role in determining the effective ratio.
How does R2 particularly play a role in making the FB pin noise sensitive?
How does R2 particularly play a role in making the FB pin noise sensitive?
You would probably want to keep I1 much lower than Ib (and definitely never Ib>I1) and not on the same order of magnitude. Why? because the output of the power circuitry will be low impedance (usually lower than say 1Ω or around that range), this means that if R2 and R1 are in the kΩ range or above it won't be so much of a load for the 'power circuitry' that it would cause error to change vout.
The problem is if you make R1+R2 in the MegΩ range then it could start to cause error for VA (the voltage between R1 and R2) as Ib could start to pull enough current (through Ib if it were in the uA or nA range) to cause error for VA.
You can think of the V- terminal as a load on the voltage divider which could be approximated as a resistive load (you could include the capacitance of V- for simplicities sake I'll ignore it) :
VA/Ib = Rv-
You want Rv- to be much higher than R1+R2 (because it essentially parallels R1, so you want Rv- to be negligible)
simulate this circuit – Schematic created using CircuitLab
