View Full Version : jumper to put two resistors in Series or Parallel
02-18-2012, 07:16 PM
Is this the best way to get get the 3 possible R values?
One pin is never used.
Top two version have one of the jumpers just parked.
Using just one jumper is probably impossible?
I guess i could use the last pin for zero ohm.
02-18-2012, 08:19 PM
If you are trying to figure out the values of unknown resistors, then that would give you all the information you need to determine the value of each resistor.
And note GND and GND would be electrically connected (even though it is not shown).
And using an ohmmeter to take the readings between GND on the left and connection below that.
Warning: If this is for school, my answer may be wrong!
For example in the 1980's, the correct answer for "What is a mainframe computer?", was a computer with 1 megabyte of RAM memory or more. I had a PC on my desk with 1 megabyte of RAM memory, but the teacher wanted the answer which was in the old textbook!
Phil Pilgrim (PhiPi)
02-18-2012, 08:21 PM
You can do it with a 2x2 header array:
Four R values (incl 0) from 4 pins & one jumper. That's what keeps my hero PhiPi on top!
02-19-2012, 04:52 PM
With one jumper I see 0, 1R, 1R, and 2R. I can only see how to get R/2 with two jumpers. What am I missing?
02-19-2012, 04:58 PM
Yes, you need 2 jumpers for R/2
And that you can not park it when not in use, is bothering me :)
I guess you could glue a black and a blue jumper togheter and you will just let it hang outside when not used.
4 combinations: 0,R1,R2, and R1+R2.
Phil Pilgrim (PhiPi)
02-19-2012, 07:08 PM
I wondered if Tony could get more bang for his buck by using different values for the two resistors. That way he could get four distinct combinations (not including zero and infinity), instead of three. Ideally, the ratio of each successive combination to the one before it will be the same. These are, from lowest to highest:
A = r1r2/(r1 + r2), B = r1, C = r2, D = r1+r2
So I used an online equation solver (http://www.numberempire.com/equationsolver.php) to solve the following for r2:
B / C = A / B, i.e. r1 / r2 = (r1r2/(r1+r2)) / r1
And guess what result I got?
r2 = (1 + √5) / 2 * r1 = 1.61803399... * r1
That's the Golden Ratio! But it gets better. To test the series combination for consistency, I set r1 = 1 and r2 = 1.61803399..., then computed the following on my calculator:
(r1 + r2) / r2
And, again, I got the Golden Ratio, 1.61803399..., which shouldn't be surprising, since (x + y) / y = y/x is the equation from which the Golden Ratio is typically derived.
So, using 1% resistor values of r1 = 1.00K and r2 = 1.62K, you can get the following four resistance values:
A = 618, B = 1000, C = 1620, and D = 2620
whose pairs of successive values stand in nearly constant ratio to each other.
Addendum: I assumed that this "discovery" was nothing groundbreaking, so I Googled resistors "golden ratio". It turns out that the Golden Ratio also crops up in infinite networks of identical resistors: