Electronic – Evaluating junction temperature from thermal imaging data

One or more of your assumptions is not correct.

You should have \$\theta_{ja} = \theta_{jc} + \theta_{ca}\$, giving \$\theta_{ca}\$ of 30.76 C/W for IC1 and 21.76 C/W for IC2.

With those values, IC1 should have a case temperature of

23 C + (1.2 W) (30.76 C/W) = 59.9 C,

which is pretty close to what you measured; but IC2 should have

23 C + (0.7 W) (38.23 C/W) = 38.23 C,

which is way off the measurement.

A likely cause are if the datasheet's \$\theta_{ja}\$ value makes some assumption about board layout that is not met in your application. If your board doesn't provide sufficient copper area attached to the chip to spread heat away, \$\theta_{ja}\$ will increase substantially. It's also possible that the datasheet value assumes some forced air cooling over the chip.

No. You have to measure power dissipation in some way.

You can use some hacks: one method is to install temperature sensor (small NTC resistor) to an opposite (to heat sink) side of semiconductor device. Then you have to use some thermal simulation SW (Ansis or similar) to get a temperature difference between heat sink side and the point where your sensor is placed.

For example, Ansis tells that 100 W dissipation causes 20 C temperature difference between the sensor and heat sink. You measure 10 C temperature difference. Then your power dissipation is 50 W.

The junction temperature is then a sum of case temperature and a product of Junction-to-case thermal resistance by 50 W.

This method requires to have an adequate model for Ansis, to install the sensor with good thermal contact, to know case temperature with reasonable precision etc.