The LM26420 has a Vref accuracy of 1.5% (I assume this is +/-1.5%, correct?) - so if you are using a +/- 1% resistor to set the feedback voltage, what is the output voltage tolerance?
Our formula would look something like (equation 1 of the datasheet plus known errors):
R1 +/-1% = (Vout +/-X / Vref +/-1.5% - 1 ) * R2 +/-1%
So assuming we are trying to find Vout error it would be:
(R1 error / R2 error) * Vref error = Vout error
According to the principles of error prop:
sqrt((del_R1_error/R1)^2 + (del_R2_error/R2)^2 + (del_Vref_error/Vref)^2) * Vout = Vout_error
Is this correct - can someone double check my thought process? I'm relatively certain this is an error propagation question, but please correct me if I'm wrong in that assumption.