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Help designing a thermistor circuit 
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#55
Feb509, 05:30 AM

P: 4,513

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. 


#56
Feb509, 09:19 AM

P: 1,803

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. 


#57
Feb509, 03:34 PM

P: 4,513

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 tradeoff 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? 


#58
Feb509, 05:13 PM

P: 1,803

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 brotherinlaw who has a PhD in statistics, his reaction was to the effect of, "isn't that obvious?" I never did pursue it further.



#59
Feb509, 11:50 PM

P: 322




#60
Feb709, 04:12 AM

P: 4,513

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 [tex]R_{parallel}= \frac{R_1 R_2}{R_1 + R_2}[/tex] [tex]\epsilon_{parallel} = R_{parallel}\sqrt{ \frac{\epsilon_{1}{}^2}{R_{1}{}^2} + \frac{\epsilon_{2}{}^2}{R_{2}{}^2} [/tex] Taking exponents of temperature could be more difficult. Maybe I should start a thread on this topic, what to you think? 


#61
Feb1109, 03:53 PM

P: 4,513

The tolerance of the bias resistor, or bias circuit The error of resistance with respect to temperature of the thermistor itself The curve fitting error you are asking about 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 (leastsquares 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 [tex]V = F(R_b,T) \ ,[/tex] then [tex]\delta V = \frac {\partial F(R_b,T)}{\partial T}\delta T +\frac {\partial F(R_b,T)}{\partial R_b} \delta R_b\ .[/tex] [itex]\delta V, \ \delta R_b, [/itex] and [itex]\delta T [/itex] represent absolute error. We need relative error (And we need to represent relative error as vectors.) [tex]\hat{ \epsilon}_{V} = \delta V / V[/tex] [tex]\hat{ \epsilon}_{R} = \delta R_b / R_b[/tex] [tex]\hat{ \epsilon}_{T} = \delta T / T[/tex] I could have written [itex]\delta T[/itex] and [itex]\delta R_b[/itex] 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, [itex]\hat{ \epsilon}_{R} [/itex] and [itex]\hat{ \epsilon}_{T} [/itex] would add like perpendicular vectors. [itex]\hat{\epsilon_{V}}[/itex] , on the other hand is some combination of these vectors, and pointsoff in a direction within the plane spanned by these two other vectors. The norm (magnitude) of the vector [itex] \hat{\epsilon_{V}}[/itex] is the relative error in voltage to the ADC input. In general, the form of the result will involve a square root. [tex]\epsilon_{V} = C \sqrt{ A \epsilon_{R_b}\ ^{2} + B\epsilon_{T}\ ^{2} }[/tex] where A, B and C are constants to be determined by applying the above procedure. 


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