Re: Formula for calculating temp in C
- From: Huub <hdotvdotniekerkathccnetdotnl>
- Date: Wed, 29 Jun 2005 10:41:08 +0200
Rich Webb wrote:
Still assuming that I can use Steinhart - Hart for all thermistors:On Tue, 28 Jun 2005 20:10:45 +0200, Huub <hdotvdotniekerkathccnetdotnl> wrote:
Rich Webb wrote:
On Mon, 27 Jun 2005 19:19:58 +0200, Huub <hdotvdotniekerkathccnetdotnl> wrote:
Hi,
Is there any formula for converting a given resistance of the (Siemens) NTC K164/1K5 into temperature in Celsius? I'm not familiar with components, but I need to use it in software.
You don't give enough information. How precise does your conversion need to be? A simple linear approximation works just fine for some ranges and for some applications. A second order curve handles many others.
For a more precise curve fit, search for the Steinhart-Hart thermistor equation.
This formula gives 1/T with T in Kelvin. When I rework this to Celsius by (1/(1/T))-273.15 I get a rediculous temperature in Celsius. Hence, I suppose I'm doing something wrong. Can you tell me how I should do it?
Don't assume that because S-H is the "best" approximating formula that it's appropriate for real world use. Do you really need (and can you afford the computational burden) of a 0.1 degree approximation? What's the spec'd accuracy of the thermistor and associated components?
For A = 1.139456E-3, B = 2.323002E-4, C = 9.579158E-8, and R = 10000 then 1/T = 298.164 and Tc = 25.014. It's just plugging in the numbers.
taken that A (diameter of NTC disc), B (width of NTC disc) and C (distance between the legs) are the physical measurements of the thermistor (Siemens NTC K164/1K5), I used metrics mm: A = 4mm, B = 1.5mm and C = 5.5mm. When I convert these to inches: A = 0.15748031", B = 0.059055118" and C = 0.21653543".
Tried in my workroom at 19C:
Tc = (1 / (0.15748031 + (0.059055118 * ln(1686.64)) + (0.21653543 * (ln(1686.64)^3)))) - 273.15 = -273.139 C
So, where do I go wrong? .
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