Calculating effective temperature coefficient for two resistors in series - Page 1
Forum section: electronics / beginners

author: JoeN
author_rank: Frequent Contributor
timestamp: March 11, 2015, 01:35:09 am
content: Let's say you need a resistor for gain setting on an op-amp and since you want to be able to adjust or zero the output of the opamp you choose a resistor plus a pot (rheostat) in series. Since you want this arrangement to be stable, you choose low temperature coefficient parts. However, pots generally have much poorer temperature coefficients than precision resistors. On the other hand, let us say that the application allows the pot to be chosen and set at a resistance such that it has a small resistance compared to the resistor. How do you calculate the effective temperature coefficient of the resulting series resistor? Is the contribution of each discrete resistor linear as would be expected?Example: 1500 ohm 5ppm/C resistor plus a 100 ohm 100ppm/C pot that, let us say, is set to 50 ohm.My guess: ((1500*5)+(50*100))/1550 = 8.1ppm/C effective.Sane, insane, or what?

author: codeboy2k
author_rank: Super Contributor
timestamp: March 11, 2015, 01:16:44 pm
content: I think your calculation is reasonable, it should be linear like that.However, I never do just a resistor and a pot in a divider, I always put the pot in the middle between two precision resistors, that way it has no effect on the gain ratio. I do it like this:So if you find that you need to trim the pot to 40%/60% to set the gain you need, it will always contribute 40% to one side and 60% to the other side, no matter what the ambient temperature is. This is assuming that the resistive material of the pot changes equally across it's entire surface area with temperature, which is a reasonable assumption to make given today's process lines.So even if the pot doubled in value from 100 ohms to 200 ohms, the divider ratio would remain the same, and entirely dependant on the precision resistors being tempco matched (ratio matched), i.e. like an LT5400 (or two resistors on the same substrate, or in close proximity)Effectively, the pot tempco is meaningless.I like to set my gain as close as possible using the precision resistors, then choose a pot that is 2 orders of magnitude smaller (provided it gives enough gain swing to be meaningful; if not I go a little larger). With this configuration, even if the pot does have minor variations over temperature from the left to right sides of the wiper, it will contribute very little to the overall gain error.You can do something similar for offset nulling.

author: mzzj
author_rank: Super Contributor
timestamp: March 11, 2015, 01:38:09 pm
content: Quote from: codeboy2k on March 11, 2015, 01:16:44 pmYou can do that also but in my eyes its kind of ugly design. Intermitted contact or noise on the pot wiper causes the opamp to go open-loop.

author: JoeN
author_rank: Frequent Contributor
timestamp: March 11, 2015, 04:47:33 pm
content: Quote from: mzzj on March 11, 2015, 01:38:09 pmThis is Linear's example on how to use the trimpot and opamp to gain ~6.95V to 10.000V. In your opinion, better?

author: Phaedrus
author_rank: Frequent Contributor
timestamp: March 11, 2015, 04:49:56 pm
content: That's just a different way of drawing the same circuitEDIT: Well, it's slightly different, but it will probably have the same issues.

author: codeboy2k
author_rank: Super Contributor
timestamp: March 11, 2015, 05:01:15 pm
content: Linear's way is good, that won't go open loop.You can do it that way as long as your trimmer is really small valued compared to the other divider resistors. You want it to contribute an order of magnitude (or 2 orders!) less than the others do (to the divider), so that it's TC is not important anymore.Other factors always come into play, like how much error can you afford, how much tempco can you afford, etc.You just need to make sure that the trimmer doesn't cause you to exceed your allowed error budget.EDIT: I think for ppm and the elusive sub-ppm stability, I'd still do it with the trimmer in the middle, despite the possibility of going open-loop; that's just an error condition, caused by a dirty wiper, it's not an operating condition. Clean the wiper or replace the pot and it goes away for another 20 years

author: Rog520
author_rank: Regular Contributor
timestamp: March 11, 2015, 05:06:47 pm
content: Quote from: JoeN on March 11, 2015, 01:35:09 amAre you looking to zero the offset of the amp? If so, there are lots of precision op-amps (auto-zeroing, even) that will do the job without any external parts. You can then leave your gain fixed.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 11, 2015, 09:24:43 pm
content: I wrote a spreadsheet that calculates the required series-parallel combination to take a component of X tempco and Y value, and reduce it to a lower tempco and a desired total resistance.The exact example I was using was:I have a limited selection of temperature sensitive capacitors. Here's a 47pF "N330" (-330ppm/K) capacitor.But what I need is -100ppm/K at 60pF.What's the (C1 || C2) + C3 network required for this solution?(The solution, by the way, requires CALCULUS FOR ENGINEERING, so there)Since capacitances and resistances (or inductances) behave inversely: where it says "in series", do it in parallel. And vice versa.Note that the transformation (i.e., finding the tempco of a series-parallel network) depends upon the values of the components in the network -- as Calculus is wont to do, it's only true locally. If you have a massive tempco, like an NTC thermistor, you're probably better off doing a temperature sweep simulation, or something like that. (One example I've simulated before: an NTC to compensate a Vbe drop in a foldback current limit -- the functional range (of useful current output) was extended from +/-25C to the full military temperature range.)Tim

author: JoeN
author_rank: Frequent Contributor
timestamp: March 11, 2015, 10:55:55 pm
content: Quote from: Rog520 on March 11, 2015, 05:06:47 pmWhy does Linear suggest this then? I assume they have very little to gain by selling external parts, but that is exactly what they are doing here. What would be a good auto-zeroing opamp that would provide as much precision as Linear's solution with an LT 399?

author: Rog520
author_rank: Regular Contributor
timestamp: March 12, 2015, 04:33:30 am
content: Quote from: JoeN on March 11, 2015, 10:55:55 pmMy guess is that the value of the trimpot is small enough relative to the 19k resistor, so as to make the drift due to the tempco of the pot negligible. Might be a good question for an LT application engineer, though. I would probably try to make it even smaller (less than 2k), and increase the value of the 19k resistor a little. The LT1001 is actually a very capable op-amp for this application -- I didn't take a close enough look at the schematic you posted the first go-round to notice what they were using. Anyway, if you're really concerned about the trimpot, you can find low tempco versions (usually wirewound or foil) from the likes of Vishay (they make some down to 5ppm, but they're pricey). I probably wouldn't worry about it unless you can observe a problem after building the circuit.http://www.vishaypg.com/docs/63056/12801285.pdf