# Evaluating junction temperature from thermal imaging data

I have a SOD-123F diode through which a current of ~1A flows, creating a forward voltage drop of around 300mV. The datasheet does not specify Rjc but gives Rja = 200 C/W and Rjl = 70 C/W for a 10cm x 10cm copper pad area (!). I only have 12 mil traces going to and from this diode, so I am far from this ideal heatsink scenario.

From thermal imaging data I have a top-side casing temperature of 65C (~40C elevation).

What can I do to evaluate the junction temperature ?

Since I do not have access to Rjc, I thought I could evaluate a worst case scenario with Rjc = Rja = 200 C/W (since in reality Rja > Rjc) and assuming ambient to be 65C instead of 25C:

Tj = Rja * P + 25C = 200C/W * 1A * 300mV + 65C = 125C


Which is just on the 125C limit allowed by the datasheet. There should be some margin in there since I used Rja > Rjc.

Any other thoughts ?

• That is a question that always troubles me too and one I unfortunately don't the answer! I think in reality you can't be sure of how much is the junction temperature. Even if you had the $R_{jc}$ number, this refers to a very specific characterization test board that is mostly far away from the reality. The best would be to have the $\psi_{jc}$ parameter or a maximum value of the case temperature that is allowed. Jul 19, 2017 at 15:38
• If it is a silicon diode, you can for sure estimate junction temperature using Vf and If. While I am sure it is possible and accurate, I do not know the exact way to do it off the top of my head. If I remember right, the temperature vs Vf relationship also depends on the current. Maybe the same thing is possible with Schottky's. Jul 19, 2017 at 19:50
• I just looked at datasheets for several parts in that package. I found two which listed theta junction to lead. One was 23 K/W and the other was 70 K/W. Either way, the part definitely dissipates much more power through the lead than through the case. So another option for you is to put a thermocouple directly on the case, then put foam insulation on top of the thermocouple. The insulation will prevent heat transfer, so the case will be very close to die temperature. The insulation will only have a slight effect on temperature because most of the heat goes out through the leads. Jul 20, 2017 at 5:17
• This may also explain why they don't give you theta to case. Because most of the heat is transferred to the board. Jul 20, 2017 at 5:19
• Oh yes, I did misread your comment. I am not sure if tlead = tcase. It would be great to measure the lead temperature directly, if possible. Maybe you can glue a thermocouple directly to the lead, or even solder it there. If you use an imaging technique, be sure to accurately enter the emissivity. It is not possible to obtain an accurate result for temperature without using the correct emissivity. Jul 20, 2017 at 18:32

@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.

The thermal timeconstant of 1micron cube is 11.4 nanoseconds.

For 10 micron cube...... 11.4nS * 100

For 100 micron cube...... 11.4nS * 100 * 100

For 1,000 micron cube..... 11.4nS * 100 * 100 * 100

Your diode junction is likely 10microns (based on breakdown voltage and max current).

Your pulses need to be << 1uS.

• Could you share a reference that would relate die size to diode specs ? Jul 20, 2017 at 17:22