Electronic – How to interpret this thermal curve graph for the heatsink

You don't use the original Tja anymore because it's not valid - there are new thermal resistances to place between the junction and the ambient air.

Have a look at this Analog Devices application note MT-093.

You can see that: $$\Theta_{JA} = \Theta_{JC} + \Theta_{CA}$$

Now when you add a heatsink, you are inserting another thermal resistance between the case and ambient air, so your equation becomes:

$$\Theta_{JA} = \Theta_{JC} + \Theta_{CS} + \Theta_{SA}$$

Where Tcs is thermal resistance of the case to your sink. This depends on your interface material, sometimes known as "thermal grease". Various factors (surface area, applied pressure) change how well the thermal grease will conduct heat. An estimate of .25°C/W seems reasonable.

Tsa is the thermal resistance of the sink to ambient. (3°C/W as stated, with some airflow)

So to calculate your maximum allowable ambient temp:

$$ T_A = T_J - P * (\Theta_{JC} + \Theta_{CS} + \Theta_{SA}) $$

Maybe, maybe not, but I'd ask why you are not correlating hot chips with power supply currents, and why you're not putting a temperature sensor on the heatsink. If the thermal path from the die to the heatsink is impaired you'll get a different temperature differential between the die and the heatsink. Likewise, if the chip is drawing more current you should be able to predict the final temperature of the die based on normal thermal behavior. And measuring the heatsink temp doesn't require a dedicated contact sensor: a temporary one will do, or a non-contact IR unit should work, since the emissivity of the heat sinks should be pretty uniform.

As to why the maybes, consider the following model:

^{simulate this circuit – Schematic created using CircuitLab}

If the thermal resistance from the die to the heatsink is much larger than the thermal resistance of the heatsink to ambient, and the thermal capacity of the die is much less than the capacity of the heat sink (and I would guess both to be true), the latter is the dominant factor in determining the thermal time constant of the heatsink, and thus of the die. In this case, increases in the die/HS thermal resistance will have only small effects on the time constant of the die, but will cause the die to get hotter. You'll have to figure the values for your board to see if this is the case.