??? 08/22/07 07:17 Read: times |
#143500 - Getting closer Responding to: ???'s previous message |
For 100 µW/0.1°C
So maximum this much power (100 µW) dissipation is valid for thermistor to not degrade the performance. The thermistor resistor at 37 °C= 6.2 Kohm If I take 6.2 k( Precise resistor as suggest by Kai) then the current is (5/(6.2+6.2)Kohm)=4.0322E-4 Amp and so power dissipation in thermistor is (((4.0322E-4)^2)*6.2Kohm)=1.008E-3 watt You have done the maths, so you are nearly there. You have deduced that the dissipation is 1mW, but your error budget limits you to 0.1mW. The straight forward solution is to use a lower excitation voltage, so instead of using 5V, use 1.5V. I strongly suspect that from your comment Ans2:The thermistoer is in Contact of solid aluminum block If the tolerance on the device is much greater than 0.1 deg C, then you will need to calibrate your system. In that case a fixed amount of self heating is not significant. |
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