Modify the divider of Example 4.8 so that self-heating
is reduced to 0.1 degrees Celsius. What...
A thermistor is to monitor room temperature. It has a resistance of 3.5 kΩ at 20°C with a slope of-10%/C. The dissipation constant is Po-5 mw/oC. It is proposed to use the thermistor in the divider of Figure 4.6 to provide a voltage of 5.0 V at 20°C. Evaluate the effects of self-heating. Solution It is easy to see that the design seems to work. At 20°C, the thermistor resistance will be 3.5 k2, and the divider voltage will be EXAMPLE 4.8 3.5 kΩ 3.5k3.5kn emisor i be given by RTH 3.5kn The temperature rise of the thermistor can be found from Equation (4.13): FIGURE 4.6 Divider circuit for Example 4.8. 10 V 3.5 kn