Electronic – Is it possible to calculate junction temperature from case temperature without knowing the power dissipation

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.

@mkeith is correct. If you measure the Vf with very short 1A pulses, to avoid self heating, whilst controlling the ambient temperature with an oven, you will be able to get the Vf @ 1A over the temperature range. If short pulses can not be acheived, steps between 0A and 1A are acceptable if you take the measurement very quickly and obviously keep the duty cycle very low; this can be done with a scope and preferrably also a current probe to see when the current has settled.

I have done a similar experiment in the past to characterise the temperature dependance of the Vf of various LEDs over a range of currents in order to calculate the junction temperatures in normal operation.