40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
  40m Log  Not logged in ELOG logo
Message ID: 13292     Entry time: Tue Sep 5 09:47:34 2017
Author: Kira 
Type: Summary 
Category: PEM 
Subject: heater circuit calculations 

I decided to calculate the fluctuation in power that we will have in the heater circuit. The resistors we ordered have 50 ppm/C and it would be useful to know what kind of fluctuation we would expect. For this, I assumed that the heater itself is an ideal resistor that has no temperature variation. The circuit diagram is found in Kevin's elog here. At saturation, the total resistance (we will have a 1\Omega resistor instead of 6\Omega for our new design) will be R_{tot}=R+R_{h}=1\Omega +24\Omega =25\Omega. Therefore, with a 24V input, the saturation current should be I=\frac{V_{in}}{R_{tot}}=\frac{24V}{25\Omega}=0.96A.  Therefore, the power in the heater should be (in the ideal case) P=I^2R{_{h}}=22.1184W

Now, in the case where the resistor is not ideal, let's assume the temperature of the resistor changes by 10C (which is about how much we would like to heat the whole thing). Therefore, the resistor will have a new value of R_{new}=R+50ppm/C\times 10C\times 10^{-6}=1.0005\Omega. The new current will then be I_{new}=\frac{V_{in}}{R_{new}}=0.95998A and the new power will be P_{new}=I_{new}^{2}R_{h}=22.1175W. So the difference in power going through the heater is about 0.00088W.

We can use this power difference to calculate how much the temperature of the metal can we wish to heat up will change. \Delta T=\Delta P\times (1/\kappa) /x where \kappa is the thermal conductivity and x is the thickness of the material. For our seismometer, I calculated it to be 0.012K.

ELOG V3.1.3-