Total resistance tolerance calculation

Hi,

I use 1% tolerance resistances in my electronic circuit that should work from -40C to 85C. First of all, I would like to calculate the total tolerance of my resistors ( including the tolerance given by the supplier, temperature, and the aging.
Could you give me please some pieces of advice how to get this total tolerance.

Thank you in advance.

In series, you sum all the impedances and determine the weighed average of the impedance tolerances. For your example:

R = R1+R2
R = 10 kΩ+20 kΩ = 30 kΩ

T = (T1*R1+T2*R2)/(R1+R2)
T = (10%*10 kΩ+20%*20 kΩ)/(10 kΩ+20 kΩ)
T = (5 kΩ)/(30 kΩ)
T = 16.7% tolerance

In parallel, you sum all the admittances and determine the weighed average of the admittance tolerances. For your example:

R = R1║R2 = (R1*R2)/(R1+R2)
R = (10 kΩ*20 kΩ)/(10 kΩ+20 kΩ) = 6.67 kΩ

T = (T1/R1+T2/R2)*(R1║R2)
T = (T1/R1+T2/R2)*(R1*R2)/(R1+R2)
T = (10%/10 kΩ+20%/20 kΩ)*(10 kΩ*20 kΩ)/(10 kΩ+20 kΩ)
T = (1%+1%)*(200)/(30)
T = 13.3% tolerance

- - - Updated - - -

In series, you sum all the impedances and determine the weighed average of the impedance tolerances. For your example:

R = R1+R2
R = 10 kΩ+20 kΩ = 30 kΩ

T = (T1*R1+T2*R2)/(R1+R2)
T = (10%*10 kΩ+20%*20 kΩ)/(10 kΩ+20 kΩ)
T = (5 kΩ)/(30 kΩ)
T = 16.7% tolerance

In parallel, you sum all the admittances and determine the weighed average of the admittance tolerances. For your example:

R = R1║R2 = (R1*R2)/(R1+R2)
R = (10 kΩ*20 kΩ)/(10 kΩ+20 kΩ) = 6.67 kΩ

T = (T1/R1+T2/R2)*(R1║R2)
T = (T1/R1+T2/R2)*(R1*R2)/(R1+R2)
T = (10%/10 kΩ+20%/20 kΩ)*(10 kΩ*20 kΩ)/(10 kΩ+20 kΩ)
T = (1%+1%)*(200)/(30)
T = 13.3% tolerance