Comments on: EIA Resistor Values Explained

By: Jeff

Jeff — Thu, 15 Jan 2009 18:09:21 +0000

Eric – Thank you for providing the formulas to solve for R and the resistor ratio – I have to admit that I am guilty of using trial and error to find the same results – this is much more straightforward. Important to note that the R value in the first equation may not exactly match a given value in the series but will be within a few ohms.

Jason – You are correct – any geometric series will be a straight line on a log scale since R1/R2 is a constant.

By: Jason Gallicchio

Jason Gallicchio — Thu, 15 Jan 2009 06:08:16 +0000

The resistor values are equally spaced on a log scale — the resistor values make a straight line on a log plot. Using Eric’s formula,

log(R) = n/b

Where b is the series number (24 for 5%, 96 for 1%) and n is the number in the series.

By: Eric

Eric — Thu, 15 Jan 2009 03:36:27 +0000

One thing you might also want to discuss is the formula for obtaining the resistor value:

R = 10 ^ (n / b)

Where ^ is exponentiation, and b is the series number (24 for 5%, 96 for 1%). What is especially interesting about this formula is that it really helps calculate standard resistor ratios. These are used all the time for voltage dividers, gain setting equations in op-amp circuits, and so forth.

R1/R2 = [10^(n1/b)] / [10^(n2/b)] = 10^[(n1-n2)/b]

b * log (R1/R2) = n1-n2 = delta n

If we know the ratio we want, we can solve for delta n. This is a very useful result. For our resistor ratio, we can pick any resistor. Now the other resistor is (delta n) steps higher in the series. Using this trick you can change the impedance of the resistor divider to be whatever you want without changing the divider ratio.

By: Windell Oskay

Windell Oskay — Wed, 14 Jan 2009 20:28:36 +0000

Excellent.

By: Tony

Tony — Mon, 12 Jan 2009 17:46:24 +0000

Very interesting and makes sense, thanks for posting this.