ELOG - heater circuit calculations

40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop

40m Log

Not logged in

List | Find | Login | Help

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-