Mentor
Blog Entries: 10

## Help designing a thermistor circuit

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

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.
Attached Files
 Error Comparison.xls (16.0 KB, 21 views)

 Quote by 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.
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.

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.
Attached Files
 Skeptics Thermistor Circuit.xls (55.5 KB, 22 views)
 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.
 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?
 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.

 Quote by 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.
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.

 Quote by skeptic2 ...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?

 Quote by rusty009 How did you calculate the error ?
The error in the circuit has about four contributing factors:

The tolerance of the bias resistor, or bias circuit
The error of resistance with respect to temperature of the thermistor itself
The step resolution error of the ADC

The curve fitting error may be reduced by chosing a different curve other than a straight line approximation to fit it to, and/or preforming a regressive analysis (least-squares fit).

The ADC error is usually quoted by the manufacture as 1/2 step over the span. I'd use an error value of one step to be practical.

The tolerances of the input circuity will tell us the error in the voltage actually applied to the ADC. This involves doing a little math--

We need to find the variation of the voltage as a function of temperature and bias resistor.

Given
$$V = F(R_b,T) \ ,$$
then
$$\delta V = \frac {\partial F(R_b,T)}{\partial T}\delta T +\frac {\partial F(R_b,T)}{\partial R_b} \delta R_b\ .$$

$\delta V, \ \delta R_b,$ and $\delta T$ represent absolute error. We need relative error (And we need to represent relative error as vectors.)

$$\hat{ \epsilon}_{V} = \delta V / V$$

$$\hat{ \epsilon}_{R} = \delta R_b / R_b$$

$$\hat{ \epsilon}_{T} = \delta T / T$$

I could have written $\delta T$ and $\delta R_b$ as vectors, but the form of the equation may already be somewhat unfamilar to begin with.

We are assuming that the errors of the independent variables are independent. The error in the resistor is independent of the error in the thermistor. Abit, $\hat{ \epsilon}_{R}$ and $\hat{ \epsilon}_{T}$ would add like perpendicular vectors. $\hat{\epsilon_{V}}$ , on the other hand is some combination of these vectors, and points-off in a direction within the plane spanned by these two other vectors.

The norm (magnitude) of the vector $\hat{\epsilon_{V}}$ is the relative error in voltage to the ADC input. In general, the form of the result will involve a square root.

$$\epsilon_{V} = C \sqrt{ A \epsilon_{R_b}\ ^{2} + B\epsilon_{T}\ ^{2} }$$

where A, B and C are constants to be determined by applying the above procedure.