Total resistance tolerance calculation

[mariuszoll]

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.

 

[sandipm14]

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