Help designing a thermistor circuit

[skeptic2]

Using that formula: 1/T = 1/To + (1/B)*ln(R/Ro) I get B = ln(R/Ro)/(1/T - 1/To).
Substituting values, and using 298 and 313 instead of 298.15 and 313.15, I get B = 5301.947. Using 298.15 and 313.15, B = 5307.158.

 

[Phrak]

That looks like the right equation, rusty.

But i should have said T_o = 298.15+25 in the definition of T_o. It was a typo.

I derived B using T_o = 298.15+25 \ \ \ \ R_o = 2815

The temperatures are in Kelvin. 25C = (298.15+25)K

323.15K should be the reference, not 0C nor 298.15K. This is just a property of an exponential fit to two data points. There's nothing magical about any particular temperature. You are free to choose any temperature/resistance reference pairs. When B=6211 the curve passes through the given data points.

If you plug T=298.15+40 into the equation I put up, you will get R=1200.0 ohms.

 

[rusty009]

hey guys, I think I'm sticking with my above circuit as my next task is to use the B parameter equation, so Phrak, I will use your equation for that question. But for the linear approximation, how do i calculate the error with just two points?

 

[skeptic2]

Once you calculate B correctly
* you can find the value of the thermistor at various temperatures
* you can calculate the voltage that will be seen by the A/D.
* Take the A/D reading and find a way to calculate the temperature from it.
* Calculate the temperature from the A/D value for each temperature.
* % error = Abs(T - Tcalc)/T * 100
* You may be interested in both average error and maximum error.

 

[Phrak]

If I were you rusty, I would do it the same way you came up with that straight line plot.

You really want to know the error in V, the voltage feeding the ADC.

You have V as a function of R, and R as a function of T. Plot V as a function of T. Then compare it to a straight line with 25C and 40C at the endpoints.

Do you have MATLAB or something? If not, do it in excel (so I've heard).

 

[Phrak]

Using that formula: 1/T = 1/To + (1/B)*ln(R/Ro) I get B = ln(R/Ro)/(1/T - 1/To).
Substituting values, and using 298 and 313 instead of 298.15 and 313.15, I get B = 5301.947. Using 298.15 and 313.15, B = 5307.158.

skeptic, if you calculated your B correctly, and plug 298 into T, you get 2815 ohms. But it should be 2815 ohms at 25C. See how it goes? It's just curve fitting though two data points.

 

[Redbelly98]

Hey Phrak, when I calculate B i get 4043.92, I use this equation,
http://upload.wikimedia.org/math/8/d/6/8d649a9ef06d12e8bf4f65e7d32cad0d.png

I think its because we use different K values, I used 298.15K for 25 degrees and 318.15 for 45 degrees, have I made a mistake? I am also not sure how you got the 30.1k value?

Isn't the 2nd temperature value supposed to be 40 C = 313.15 K?

EDIT:

I get B = 5302 K (agrees with Phrak) using 298 & 313 K
and I get B = 5307 K using 298.15 & 313.15 K:
http://www.google.com/#hl=en&q=ln(1200/2815)+/+(+(1/313.15)-(1/298.15)+)&fp=0k_STq_SEOg

 

[skeptic2]

"skeptic, if you calculated your B correctly, and plug 298 into T, you get 2815 ohms. But it should be 2815 ohms at 25C. See how it goes? It's just curve fitting though two data points."

Good point Phrak, does the scale used make a difference in the values one gets for the curve? If so, which scale is best? Let's find out.

B = ln(R/Ro)/(1/T - 1/To)
For R = 1200 & T = 40 and
Ro = 2815 & To = 25
B = 56.84271

For R = 1200 & T = 313.15 and
Ro = 2815 & To = 298.15
B = 5307.158

For Celsius we get:
Temp...25...30...35...40
Res...2815...1927...1470...1200

For Kelvin we get:
Temp...298...303...308...313
Res...2815...2099...1580...1200

Apparently the scale does make a difference but which is correct? The curves intersect at 2815 and 1200. If we extend the range a little...

For Celsius we get:
Temp...-5....0...5...10...15...20
Res...0.0033...undefined...25079259...85245...12817...4970

For Kelvin we get:
Temp...268...273...278...283...288...293
Res...20623...14355...10124...7228...5221...3814

...we can see which is probably the correct curve.

 

[Redbelly98]

skeptic, T is the absolute temperature in these equations, so use Kelvins.

(Note the problem with division-by-zero that would occur at 0 C, if Celsius were used.)

 

[skeptic2]

skeptic, T is the absolute temperature in these equations, so use Kelvins.

(Note the problem with division-by-zero that would occur at 0 C, if Celsius were used.)

The temp values under Kelvin are Kelvin degrees. I was trying to demonstrate that if Celsius were used as Phrak suggested in post #36, you would get a division by zero error at 0 C. Is this not clear?

 

[Phrak]

I think its because we use different K values, I used 298.15K for 25 degrees and 318.15 for 45 degrees, have I made a mistake? I am also not sure how you got the 30.1k value?

I missed this question earlier.

That was a mistake. I should have written 3.01K not 30.1K.

You want 3.01K because you don't want to overflow your ADC, but you want it as small as possible to get the best engineering linear output. Assume your thermistor is good to 5% of the manufacturer's given reististance of 2815 ohms. It could be as large as 2955.8. Given V_out = 2.55 volts, solve for the value of the series resistor.

You should get 2840 ohms. You need a 2868 ohm 1% resistor to get at least 2840 ohms. 3.01K is the most common 1% resistor greater than 2.84K ohms.

If the thermistor is really a 10% part, 3.01K wasn't conservative enough.

 

[Phrak]

Isn't the 2nd temperature value supposed to be 40 C = 313.15 K?

EDIT:

I get B = 5302 K (agrees with Phrak) using 298 & 313 K
and I get B = 5307 K using 298.15 & 313.15 K:
http://www.google.com/#hl=en&q=ln(1200/2815)+/+(+(1/313.15)-(1/298.15)+)&fp=0k_STq_SEOg

40C = 238.15K. I use a B of 6211 at a reference temperature of 25C.

If you use a different reference temperature you get a different B. But whatever reference you use, the curve it still has to pass through the two data points.

 

[rusty009]

If I were you rusty, I would do it the same way you came up with that straight line plot.

You really want to know the error in V, the voltage feeding the ADC.

You have V as a function of R, and R as a function of T. Plot V as a function of T. Then compare it to a straight line with 25C and 40C at the endpoints.

Do you have MATLAB or something? If not, do it in excel (so I've heard).

I understand what your saying here, but then I will be using another linear approximation which again carries an error ?

 

[Phrak]

does the scale used make a difference in the values one gets for the curve? If so, which scale is best? Let's find out.

Nicely done. This business of rescaling and offets comes up all the time. I wonder if there's some general study of it?

You're holding my feet to the fire--which is a good thing. I made a mild 'engineering assumption' based upon the magnitude of the error values that were inherent in rusty's design. I assumed I could get the same curve (or very close) by using a reference temperature other than 0C--namely 25C.

To resolve this, we need to use a reference at 0C. But we don't know the resistance at zero C, so now there are two simultaneous equations to solve.

Using R_T = R_o \cdot exp[ B_o(1/T - 1/T_o) ] where R_o and B_o are unknowns,

R_{T_1} = R_o \cdot exp[ B_o(1/T_1 - 1/298.15) ]

R_{T_2} = R_o \cdot exp[ B_o(1/T_2 - 1/298.15) ]

T_1 = 323.15 , R_{T_1}=2815

T_2 = 338.15 , R_{T_2}=1200

Solve for B_o and R_o

How does this curve compare my curve,

R_T = 2815 \cdot exp[ 6211(1/T - 1/323.15) ] , where B_{25} = 6211 and R_{25} = 2815 ?

 

[Redbelly98]

skeptic, T is the absolute temperature in these equations, so use Kelvins.

(Note the problem with division-by-zero that would occur at 0 C, if Celsius were used.)

The temp values under Kelvin are Kelvin degrees. I was trying to demonstrate that if Celsius were used as Phrak suggested in post #36, you would get a division by zero error at 0 C. Is this not clear?

Sorry skeptic, I was skim-reading much of your post and missed your point. My apologies.

40C = 238.15K. I use a B of 6211 at a reference temperature of 25C.

Umm, but:

40 C = (40 + 273.15) K = 313.15 K

238.15 K would correspond to (238.15-273.15)C = -35C

If you use a different reference temperature you get a different B. But whatever reference you use, the curve it still has to pass through the two data points.

Agreed that the curve does need to pass through the two data points. But it seems that B does not change, even when the reference values Ro and To change:

For example, the equations

R = 1200Ω * exp[5302K * (1/T - 1/313K)] <-- reference at 313K or 40C
and
R = 2815Ω * exp[5302K * (1/T - 1/298K)] <-- reference at 298K or 25C

both agree with the 2 known data points

2815Ω at 25C = 298K
1200Ω at 40C = 313K

In either case, B is the same.

 

[Phrak]

40 C = (40 + 273.15) K = 313.15 K

: oh cr*p

 

[Redbelly98]

: oh cr*p

LOL, happens to everybody.

 

[Phrak]

Redbelly, using your more better 273.15K = 0C, I calculated the error incurred for various bias resistors. But PF doesn't accept MathCad files, so I can't attach it.

The results are pretty impressive. A 3.01K will deviate ~ -3.75% from a straight line passing through the end points at 25C and 40C. The optimal bias resistor is 1.45K with ~ +/-0.3% error, but the voltage across the thermistor is above the 0 to 2.55 volt ADC range.
Windows won't display a MathCad 7 plot correctly. Oh well.

 

[rusty009]

Redbelly, using your more better 273.15K = 0C, I calculated the error incurred for various bias resistors. But PF doesn't accept MathCad files, so I can't attach it.

The results are pretty impressive. A 3.01K will deviate ~ -3.75% from a straight line passing through the end points at 25C and 40C. The optimal bias resistor is 1.45K with ~ +/-0.3% error, but the voltage across the thermistor is above the 0 to 2.55 volt ADC range.
Windows won't display a MathCad 7 plot correctly. Oh well.

How did you calculate the error ?

 

[Phrak]

How did you calculate the error ?

I graphed it. But that's not the total error. Everything that can deviate from ideal can be a source or error. It's the end of the day for me. See the post I made with a square root in it.

 

[Phrak]

I've been very curious about how the nonlinearity of resistor biasing could go toward canceling the thermistor nonlinearity. Check-out these curves.

(hope this works this time)

 

[Redbelly98]

I've been very curious about how the nonlinearity of resistor biasing could go toward canceling the thermistor nonlinearity. Check-out these curves.

Nice!

 

[skeptic2]

If instead of using one 1% percent resistor you used two 5% resistors the error could be reduced as shown. The circuit with two resistors has a 2.2k resistor connected to +5V. The other side of that resistor is connected to the parallel combination of the thermistor and a 8.2k resistor. Even at the extremes of the 5% rating the error is still less than with one 1% resistor. The disadvantage of this method is that the temperature range is covered with fewer levels of the ADC, 82 steps instead of 99.

 

[Phrak]

If instead of using one 1% percent resistor you used two 5% resistors the error could be reduced as shown. The circuit with two resistors has a 2.2k resistor connected to +5V. The other side of that resistor is connected to the parallel combination of the thermistor and a 8.2k resistor. Even at the extremes of the 5% rating the error is still less than with one 1% resistor. The disadvantage of this method is that the temperature range is covered with fewer levels of the ADC, 82 steps instead of 99.

You're error curve looks very nice. I'm suprised that adding the parallel resistor helped rather than hindered. I'll paste it into MathCad and see if I get the same. I'm still skeptical, as you might have guessed.

The step count reduction is a miniscual increase in step error comparatively.

Using 5% parts adds error of their own. So does a 1%, of course. But I'll use your valued and see what happens. Do you think you could calculate the resistor divider error? Independent errors are like independent probabilities.

 

[Phrak]

By the way, Skeptic. Sorry if I caused you any confusion after I mistook the value of zero Kelvin.

Here's what I get for your input circuit. And good job! I would have never guessed. Did you do all your calculations in Excel??

I think our error values differ slightly from using different values of B.

 

[skeptic2]

Yes I used Excel. I have MathCad but I find Excel easier to use although MathCad certainly has more capabilities and the graphs look better.

The parallel resistor works because it limits higher resistances more than lower resistances. Thus at low temperatures as the thermistor resistance rises exponentially, the parallel resistor tends to straighten the curve a little.

The precision of the measurements can be improved by taking multiple measurements and averaging. If the voltage is between two steps of the ADC, averaging will interpolate between the two steps. For this reason a little noise is a good thing.

Another thing that improves the performance slightly is that when multiple components are combined and contribute individually to the tolerance of the output, because of the Central Limit Theorem, the tolerance at the output generally is less than the tolerance of the individual components. For instance if two 100 ohm resistors with 5% tolerance are paralleled, the 50 ohm resultant value has a tolerance of 3.5%. (5% / sqrt(2)) This is because the error of one resistor tends to compensate for the error of the other. When the components have different values or different tolerances or contribute different amounts to the output tolerance, the output tolerance is more difficult to calculate.

 

[Phrak]

I've shyed away from Excel because of the formula presentation in the earlier versions. I don't know if it's changed any. Unfamiliarity has something to do with it too.

I've played around with resistor values, and your values seem to be the near optimum trade-off between ADC resolution error and input nonlinearity error.

I looked over Wikipedia's Central Limit Theorem. It's hard to see if any of the contents pertains to combination of errors. Would you know how to calculated the combined error of two resistors in parallel of variable values?

 

[skeptic2]

I was working at Motorola at the time they introduced six sigma around 1984 and discovered this effect trying to characterize circuits in terms of sigma. This is really the same thing as doing sensitivity analysis and finding that some components contribute a lot more to the output tolerance than others. When I mentioned it to my brother-in-law who has a PhD in statistics, his reaction was to the effect of, "isn't that obvious?" I never did pursue it further.

 

[Corneo]

I was working at Motorola at the time they introduced six sigma around 1984 and discovered this effect trying to characterize circuits in terms of sigma. This is really the same thing as doing sensitivity analysis and finding that some components contribute a lot more to the output tolerance than others. When I mentioned it to my brother-in-law who has a PhD in statistics, his reaction was to the effect of, "isn't that obvious?" I never did pursue it further.

I work for a high tech company manufacturing certain test instruments. We set our specs based on 3 sigma criteria. Just about once a month I have to put on my statistian's hat to crunch all the variances and means across 3 different variables. Who knew statistics would be so helpful, I only wished I stayed awake in class.

 

[Phrak]

...For instance if two 100 ohm resistors with 5% tolerance are paralleled, the 50 ohm resultant value has a tolerance of 3.5%. (5% / sqrt(2)) This is because the error of one resistor tends to compensate for the error of the other. When the components have different values or different tolerances or contribute different amounts to the output tolerance, the output tolerance is more difficult to calculate.

I did a little preliminary work in MS Word combining sources of independent error to obtain total error for sum, product, reciprocals, and parallel resistors too. It's not as turn-key as I was hoping--at least not yet.

For the two equal resistors in parallel, I obtained your value-- 3.5% for 5% parts.

In general, where epsilon is relative error (percent error/100), and

R_{parallel}= \frac{R_1 R_2}{R_1 + R_2}

\epsilon_{parallel} = R_{parallel}\sqrt{ \frac{\epsilon_{1}{}^2}{R_{1}{}^2} + \frac{\epsilon_{2}{}^2}{R_{2}{}^2}


Taking exponents of temperature could be more difficult.

Maybe I should start a thread on this topic, what to you think?