ashrafsazid
Advanced Member level 4
I would like to discuss about resistor mismatch citing a very old post, even when I was not a student of Electronics. The post link is here: Resistor matching calculation
Let's say, I have two equal Resistor (R1, R2), 10k each, having a standard deviation (sigma ) is around 20. Therefore, the mismatch percentage we can say 0.2%. (sigma divided by mean, 20/10k).
What will be the sigma for the sum (R1+R2)?
What will be the sigma for the ratio (R1/R2)?
What will be the sigma for the resistive divider (R1/(R1+R2))?
as the variance are added, therefore the deviation of the sum is sqrt(20² +20²) = 28.28. in percentage 0.14%, (sigma divided by mean 28.28/20k)
The variances also add during the calculation of ratio. Therefore, I can expect same 28.28 for (R1/R2). But in this case I get from simulation that it is valid for the percentage. that means sqrt(0.2² + 0.2²) = 0.28
Therefore, it is unclear, when to calculate variance ( hence standard deviation) with percentage and when dirctly with the absolute number.
Could somebody please explain it to make me understand? Thanks in advance.
Let's say, I have two equal Resistor (R1, R2), 10k each, having a standard deviation (sigma ) is around 20. Therefore, the mismatch percentage we can say 0.2%. (sigma divided by mean, 20/10k).
What will be the sigma for the sum (R1+R2)?
What will be the sigma for the ratio (R1/R2)?
What will be the sigma for the resistive divider (R1/(R1+R2))?
as the variance are added, therefore the deviation of the sum is sqrt(20² +20²) = 28.28. in percentage 0.14%, (sigma divided by mean 28.28/20k)
The variances also add during the calculation of ratio. Therefore, I can expect same 28.28 for (R1/R2). But in this case I get from simulation that it is valid for the percentage. that means sqrt(0.2² + 0.2²) = 0.28
Therefore, it is unclear, when to calculate variance ( hence standard deviation) with percentage and when dirctly with the absolute number.
Could somebody please explain it to make me understand? Thanks in advance.