first of all, let me apologise for my inexperience with electronics.

Now that's over with, I'm trying to follow a guide that specifies the use of a specific temperature sensor with an output of 5v DC. The problem is that the sensor I currently have provides different values and that a resistance of 14.2k ohms is required. The problem I have is that this step provides no further information. The actual step is as follows: -

The outlined box above, which currently is set to 14.2, represents the resistance in to be added in parallel

I'd like to know how to obtain this specific resistance to place between the sensor and the controller that reads this value. I assumed I could build the resistance in a series.

I am under the impression that resistors can be placed in series, i.e. I could add a 10k ohm resistor and a 4k ohm resistor and then a 0.2k ohm resistor. Forgive my ignorance here, please, if this is completely idiotic. So I went to http://www.crownaudio.com/ohms-law.htm and entered 14200 and 5 into the respective fields and it specified that the watts was '0.00176'

Does this mean that when shopping for resistors, I can buy: -


2.2k - 0.6w


10k - 0.6w


2k - 0.6w

I assume I need the same wattage for all but as above, adding all the resistances in series will create 14.2k ohms but they look so flimsy.

Thank you!

  • 1
    \$\begingroup\$ 14.3K is a standard 1% resistor value, it is likely close enough, though without knowing anything about the specifics of what you're trying to do it's impossible to say for sure. If you really need 14.3K for accuracy or whatever other reason there are lots of ways to series standard resistor values to get there. \$\endgroup\$
    – John D
    Jan 25, 2015 at 2:17
  • \$\begingroup\$ both good answers provided, it's a shame I can only mark one as the right answer \$\endgroup\$
    – ed_me
    Jan 25, 2015 at 10:24
  • \$\begingroup\$ and sorry, adding them in parallel at the end was a mistake at 1am! I did actually mean series :) \$\endgroup\$
    – ed_me
    Jan 25, 2015 at 10:32
  • \$\begingroup\$ @chrisw69 then update your question. \$\endgroup\$
    – tcrosley
    Jan 25, 2015 at 13:53

3 Answers 3


You're off to a good start. Yes, you can buy three resistors and wire them in series (not parallel!), although the first resistor you link to is a 22k, rather than a 2.2k, and if you connect the three resistors you've chosen, you'll get 34 k, rather than 14.2 k.

Despite the fact that they seem flimsy, the units you show will handle at least 100 times the power you expect to dissipate. After all, your calculator (correctly) showed that the desired resistor will need to handle .0018 watts, and each of your examples will handle 0.6 watts.

Finally, you really ought to be able to find a single 14k resistor to do the job, and you can forget building up combinations. Use this link http://www.daycounter.com/Calculators/Standard-Resistor-Value-Calculator2.phtml to determine that you can use a 14.3 k, 1% resistor and save yourself some effort.

And finally, finally, I'd recommend you stay away from Maplin. They have no search ability whatsoever, and this will make your attempts at finding appropriate parts a misery. As an alternative, try Digikey. I'm sure there are other suppliers in the UK as well, but that is one of the convenient go-to sources for parts in the US, and their website indicates they have facilities in the UK.

  • \$\begingroup\$ The website in your post is showing a few php exceptions and isn't working too well. Could you elaborate on why 14.3 would suffice? I'm basically showing that 14.2k ohms is the most optimal resistor to match the expected input of the controller. 14.3k would work but it changes the curve slightly. So, why 14.3k? thanks! \$\endgroup\$
    – ed_me
    Jan 25, 2015 at 11:23
  • \$\begingroup\$ It's a question of tolerances. If you want a given resistor to be accurate to 5%, for instance, the standard sequence (starting at one) is 1, 1.1,1.2,1.3,1.5, etc. Notice that 1.4 is missing. Try en.wikipedia.org/wiki/Preferred_number#E_series and look for the section on "E series". The short version is that the difference between 14.2k and 14.3 k is less than the tolerance of a 1% 14.3 k resistor. You can't be expected to know this at your experience level, but it if better than 1% accuracy were required, the data sheet would be sure to mention it. \$\endgroup\$ Jan 25, 2015 at 13:58
  • \$\begingroup\$ I've actually learned a bit more now and found a resistor with 0.1% tolerance. I'm going to buy a few and see if I can get the closest accuracy with a multimeter \$\endgroup\$
    – ed_me
    Jan 26, 2015 at 14:55
  • \$\begingroup\$ While you're learning stuff, you might want to learn about the accuracy limits of your multimeter. Before you spend too much time checking your 0.1% resistors. \$\endgroup\$ Jan 26, 2015 at 17:27

First of all, don't put the resistors in parallel, put them in series.

$$R_p = \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2} + \; ... + \dfrac{1}{R_n}}$$

Where \$R_p\$ is the equivalent resistance of the individual resistors in parallel, and \$R_1\$, \$R_2\$, ..., \$R_n\$ are individual resistors.

If you put them in series then they just add up:

$$R_s = R_1 + R_2 + \; ... + R_n$$

The latter is what you want.

Another thing, you can use two E12 resistors not three: 12K + 2K2 = 14.2K

You've already worked out that 5V across 14.2K results in trivial power consumption, so don't spend money on high-wattage parts. 1/4W resistors would be fine. Also, if you have to get resistors with different wattages then that would also be fine. There are circumstances where that wattage rating might matter, but this isn't one of them.


You can also put the resistors in parallel, but resistors in parallel have a lower total resistance than the individual resistors. The formula is \$1/R_t=1/R_1+1/R_2\$. A parallel combination of 15k and 270k will be 14.21k. The math is harder, but it works just as well. The power rating only needs to exceed the requirement, not match it, so 1/4 watt resistors will work fine (and be very cheap).

Those resistors you linked are all 1% tolerance, so you shouldn't try to get closer than that by just using values. You likely don't need to be that accurate, but if you need to you will have to buy precision resistors or measure the ones you get and adjust the series/parallel combination accordingly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.