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Help designing a thermistor circuit 
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#37
Feb309, 08:08 AM

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P: 12,071

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(12...fp=0k_STq_SEOg 


#38
Feb309, 09:29 AM

P: 1,815

"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. 


#39
Feb309, 09:37 AM

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P: 12,071

skeptic, T is the absolute temperature in these equations, so use Kelvins.
(Note the problem with divisionbyzero that would occur at 0 C, if Celsius were used.) 


#40
Feb309, 10:41 AM

P: 1,815




#41
Feb309, 01:23 PM

P: 4,512

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. 


#42
Feb309, 01:36 PM

P: 4,512

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. 


#43
Feb309, 02:32 PM

P: 70




#44
Feb309, 02:34 PM

P: 4,512

You're holding my feet to the firewhich 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 0Cnamely 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 [tex]R_T = R_o \cdot exp[ B_o(1/T  1/T_o) ] [/tex] where [tex]R_o[/tex] and [tex]B_o[/tex] are unknowns, [tex]R_{T_1} = R_o \cdot exp[ B_o(1/T_1  1/298.15) ] [/tex] [tex]R_{T_2} = R_o \cdot exp[ B_o(1/T_2  1/298.15) ] [/tex] [tex]T_1 = 323.15 [/tex] , [tex]R_{T_1}=2815[/tex] [tex]T_2 = 338.15 [/tex] , [tex]R_{T_2}=1200[/tex] Solve for [tex]B_o[/tex] and [tex]R_o[/tex] How does this curve compare my curve, [tex]R_T = 2815 \cdot exp[ 6211(1/T  1/323.15) ] [/tex] , where [tex]B_{25} = 6211[/tex] and [tex]R_{25} = 2815[/tex] ? 


#45
Feb309, 03:37 PM

Mentor
P: 12,071

40 C = (40 + 273.15) K = 313.15 K 238.15 K would correspond to (238.15273.15)C = 35C 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. 


#46
Feb309, 03:42 PM

P: 4,512




#48
Feb409, 04:49 AM

P: 4,512

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. 


#49
Feb409, 05:29 AM

P: 70




#50
Feb409, 05:42 AM

P: 4,512




#51
Feb409, 03:32 PM

P: 4,512

I've been very curious about how the nonlinearity of resistor biasing could go toward canceling the thermistor nonlinearity. Checkout these curves.
(hope this works this time) 


#53
Feb409, 10:19 PM

P: 1,815

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.



#54
Feb509, 04:00 AM

P: 4,512

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. 


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