How to calculate needed parallel resistor value to bring resistor in tolerance

Discussion in 'Math' started by Mrdouble, Nov 11, 2016.

  1. Mrdouble

    Thread Starter Member

    Aug 13, 2012
    86
    0
    I keep running into the issue of paralleling a resistor to bring existing resistor back in tolerance. I can (and have) use a parallel calculator and just jam numbers in until I get right value but it's really bothering me I forgot how to do simple algebra. (Rearranging formulas)

    Ok, I have a resistor of 25,000 ohms with 2% tolerance
    The actual measured value is 26,200 which is 5% out
    Brute force and online calculator told me I need about 500,000 ohms in parellel to bring that to 24,900 ohms.

    That's great but how do I calculate that my self. I should be able to setup a simple formula to solve for unknown value. The two knowns are 25,000 and 26000

    FYI : I'm not looking for an answer as I already have that using caveman logic

    Thank you in advance
     
  2. WBahn

    Moderator

    Mar 31, 2012
    18,087
    4,917
    If you have a 2%, 25 kΩ resistor that is measuring 26.2 kΩ then you probably should be looking at why that resistor is so far out of tolerance. It has likely been stressed and is therefore damaged in some way. Trimming it back into tolerance is quite possibly only masking an unresolved problem.

    Leaving that aside, you are correct that your problem is atrophied algebra skills.

    Start with the formula for resistors in parallel (we could start further back and derive this formula, but will start here).

    <br />
\frac{1}{R_t} \; = \; \frac{1}{R_1} \; + \; \frac{1}{R_2}<br />

    You know Rt and R1 (25 kΩ and 26.2 kΩ, respectively) and want to solve for R2. So subtract the first term on the right side from both sides.

    <br />
\frac{1}{R_t} - \; \frac{1}{R_1} \; = \; \frac{1}{R_2}<br />

    or

    <br />
\frac{1}{R_2} \; = \; \frac{1}{R_t} - \; \frac{1}{R_1}<br />

    At this point you really are done -- subtract the reciprocal of R1 from the reciprocal of Rt and then take the reciprocal of that to get R2.

    Or finish it off completely by taking the reciprocal of both sides algebraically. There are a few ways to go about doing this. Perhaps the easiest to follow is to first put the fractions on the right over a common denominator and then take the reciprocals.

    <br />
\frac{1}{R_2} \; = \; \frac{R_1}{R_t \cdot R_1} - \; \frac{R_t}{R_t \cdot R_1}<br />
\frac{1}{R_2} \; = \; \frac{R_1 \; - \; R_t}{R_t \cdot R_1}<br />
R_2 \; = \; \frac{R_t \cdot R_1}{R_1 \; - \; R_t}<br />
     
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  3. Mrdouble

    Thread Starter Member

    Aug 13, 2012
    86
    0
    It's an old capacitor tester I got on eBay. Somebody did a number on this thing. I think I've replaceed 6 resistors so I'm going through and all values and checking and replacing components one by one. Let's put it this way. Two wires were grounded to chassis that were not supposed to be.

    Thank you very much. Gonna break that all down and hopefully my schooling will come back to me lol
     
  4. WBahn

    Moderator

    Mar 31, 2012
    18,087
    4,917
    Good luck with it. I'd say that if someone did a number on it, then you have all the more reason to replace out-of-spec components rather than trying to trim them. You don't know why they are out-of-spec so it is best to assume that they are simply damaged sufficiently to require replacement with known good components.
     
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  5. Mrdouble

    Thread Starter Member

    Aug 13, 2012
    86
    0
    I agree and thank you very much
     
  6. #12

    Expert

    Nov 30, 2010
    16,685
    7,326
    Right now I'm rehabilitating an antique guitar amplifier, about 10 watts output and I replaced 8 resistors. The carbon composition resistors always drift to higher values. Just replace them with modern metal film or carbon film resistors at about 13 to 19 cents each and be done with the inevitable drift of the antique resistors.
    http://www.mouser.com/
     
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  7. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,273
    224
    A reminder, that when paralleling a resistor, you can only make the resistance smaller. The parallel combination will always be less than the smallest resistor.
     
  8. hp1729

    Well-Known Member

    Nov 23, 2015
    2,097
    232
    I have a solution for that in Excel, but I can't post an Excel file here. The same answer posted. I added a second step that lets me plug in the actual value of the resistor to see how close the result is.

    Example:
    25,000 target
    26,000 first resistor, actual measured value

    650,000 is suggested, but not having that I find a high 620K or low 680K, measure actual value.
    It tells me what that result would be..

    My email (e-mail address removed) if you are interested.

    MOD NOTE: Do not post e-mail addresses. Even if you don't mind it getting vacuumed up by spambots, we don't want to become a favorite target for spambots by being a good hunting ground.
     
    Last edited by a moderator: Nov 11, 2016
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  9. Mrdouble

    Thread Starter Member

    Aug 13, 2012
    86
    0
    Yes, there are plenty carbon composition resistors in this circuit and you are 100% right, and after 70 years they have all drifted far off target. The problem is some of these are 1 watt and larger resistors
     
  10. #12

    Expert

    Nov 30, 2010
    16,685
    7,326
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  11. S Haque

    New Member

    Jan 2, 2017
    12
    0
    I've often had to scrounge parts from other projects, and for me, mouser parts often came to my university lab later than expected (when I was at the university some years ago). I often ended up with ordering wrong parts anyway (or parts for a different project, with the same values). So -- if you are short on space, and money, this tip might work for you: Order the SMT version of the resistor parts, not throughhole. Stock up a lot of these taped parts (they come on a tape system) in various values, and then use as needed, by soldering wires to the ends if you need them to be used in regular PCBs for repairs. Then if you get PCBs where you need those same parts, with the pads for SMDs still available, use them as needed. Well, it won't be pretty, but it would work. I'm only talking about resistors here .. The worst is trying to find N number of resistors, and ending up with N-1 number of resistors because you used up the rest on an earlier project, and then forgot to order them. Order 2X the requirement is now my motto .... and keep ordering .. (my suggestion may not work for higher current valued resistors, but at least look at the wonderful mouser parts selection tool.
     
  12. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,273
    224
    Sometimes, in a pinch if you really need a value if you don't have for a test. Take a smaller carbon composition resistor and file it to make it a larger value. Clean and cover with nail polish. The wattage of your new resistor could be lower too.
     
  13. S Haque

    New Member

    Jan 2, 2017
    12
    0
    Capture.PNG
    I didn't realize higher power SMD type resistors are readily available. Mea Culpa. I'm not sure if the original poster is interested, but take a look at some of the power ratings of these devices (using the link, click and browse the parametric table of parts, then match the spec sheet (See attached screenshot example) with your desired values you need to keep in stock - ask if you have any issues).
     
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