Resistance Ladder Calculation - J E Patterson - 20220209
J E Patterson - jepspectro.com - 20231119
Description
This program calculates the resistance of a ladder-like array of resistances arranged so that R1 is distributed along one rail, R2 occupies the rungs and the remaining rail is conductive. Arrays like these are found in analog to digital converters. In linear A/D converters R2 usually has twice the resistance of R1. In the general case the following formula applies:
R = R1 x (1 + √(1 + 4 x R2 / R1)) / 2
This equation arises from the solution of a quadratic equation noting that the first cell is in parallel with the total resistance of the remaining cells.
R = R1 + (R2 x R)/(R2 + R) since the remaining resistance after the first cell is also R, for an infinite array.
Rearranging:
(R - R1) x (R2 + R) - (R2 x R) = 0
R2 - (R1 x R) - (R1 x R2) = 0
R2 / R1 - R - R2 = 0
Using the Quadratic Formula:
x = - b +/- √(b2 - 4ac) / 2a
a = 1 / R1
b = - 1
c = - R2
A positive solution is:
R = R1 x (1 + √(1 + 4 x R2 / R1)) / 2
An equivalent arrangement by Richard Feynman (The Feynman Lectures on Physics, Volume II, eqn 22.27) is:
R = R1 / 2 + √(R12 / 4 + R1 x R2)
To use the program:
R2 ENTER R1 GSB A
Example R1 = R2 = 1 ohm
1 ENTER 1 GSB A
The result is R = 1.6180 ohms which is the Golden Ratio. Consecutive Fibonacci numbers also converge to the Golden Ratio.
If R2 = 2 ohms then R = 2 ohms. If R2 = 3 ohms then R = 2.3028 ohms.
Program Resources
Labels
| Name |
Description |
|
| A |
Calculate R from R1 and R2 |
|
Storage Registers
Program
| Line |
Display |
Key Sequence |
|
| 000 |
|
|
|
| 001 |
42,21,11 |
f LBL A |
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| 002 |
44 1 |
STO 1 |
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| 003 |
20 |
× |
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| 004 |
45 1 |
RCL 1 |
|
| 005 |
43 11 |
g x² |
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| 006 |
4 |
4 |
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| 007 |
10 |
÷ |
|
| 008 |
40 |
+ |
|
| 009 |
11 |
√x̅ |
|
| 010 |
45 1 |
RCL 1 |
|
| 011 |
2 |
2 |
|
| 012 |
10 |
÷ |
|
| 013 |
40 |
+ |
|