{"id":333,"date":"2012-08-04T21:07:25","date_gmt":"2012-08-05T02:07:25","guid":{"rendered":"http:\/\/tonykordyban.com\/?page_id=333"},"modified":"2014-04-13T20:25:33","modified_gmt":"2014-04-14T01:25:33","slug":"everything-you-know-is-wrong-december-2001","status":"publish","type":"page","link":"http:\/\/tonykordyban.com\/?page_id=333","title":{"rendered":"Everything You Know Is Wrong December 2001"},"content":{"rendered":"

Answers to those Doggone Thermal Design Questions<\/strong><\/p>\n

By Tony Kordyban<\/strong><\/p>\n

Copyright by Tony Kordyban 2001<\/p>\n

 <\/p>\n

Dear Thermal Guru,<\/em><\/p>\n

I recently had to attach a thermocouple to the lead of a component.\u00a0 I recalled a story about a guy named Herbie who had done something similar and ended up with erroneous temperature measurements (because the thermocouple wires picked up voltage from the component lead).\u00a0 To avoid repeating Herbie’s mistake I decided to apply a bit of Kapton tape to the lead before attaching my wire, with the understanding that the Kapton would also affect the measurements, although (I hoped) in a minor way.<\/em><\/p>\n

My tests showed that the component was very close to its operating limit, so I needed to know how big an error the Kapton tape might have introduced.\u00a0 So I computed the thermal resistance of the tape using the trusty formula R = t \/ (k*A).\u00a0 My result was about 100 deg C\/ watt (see details in next paragraph), and the component was dissipating 1 watt.\u00a0 That seemed to imply a 100 deg C rise across the Kapton, and that just didn’t seem right.<\/em><\/p>\n

Calculation details: the component was in a SOT-223 package; I assumed the Kapton tape\u00a0 was the same size as the lead (see dimension in the figure below).\u00a0 The Kapton was .0025 inch thick; I used k = 0.12 W\/mK (source: the DuPont web site);\u00a0 I neglected the adhesive and assumed that there was no air trapped under the tape.<\/em><\/p>\n

I know that I’m using a 1-D conduction approximation, which may not be great, but then I don’t need a computer to solve it.\u00a0\u00a0 But is it wrong to use the full package power dissipation of 1 watt for the heat passing through the Kapton tape?\u00a0 What is the right thing to do?\u00a0<\/em><\/p>\n

\u00a0Baffled in Boston<\/em><\/p>\n

 <\/p>\n

Dear Baffled,<\/p>\n

At first I thought, this guy must be nuts, thinking that all the heat generated in his component would pass through the single lead that he happened to be measuring the temperature of.\u00a0 There must be at least 23 other leads on the package, so the power going through any one of them must be pretty small.\u00a0 Then I looked at Figure 1, and now I understand what you are talking about.\u00a0 So I’ll reserve judgment on your nuttiness for now.<\/p>\n

\"\"<\/a>

Figure 1. SOT 223 geometry<\/p><\/div>\n

This package looks suspicious to me.\u00a0 Why is one lead so much bigger and fatter than the other three?\u00a0 My guess is that the fat one is meant to be the thermal path from the die to the printed circuit board (PCB), and might be pretty well connected to the die.\u00a0 The three smaller leads are probably only connected to the die by bond wires.\u00a0 So your assumption that most of the 1 watt generated inside the package comes out through that lead might be a pretty good one.<\/p>\n

I checked your math, and I get a thermal resistance of about 100 deg C\/watt for Kapton tape on top of the big fat lead.\u00a0 Now the big question is, does all of that 1 watt in the lead come out through the top surface where the tape is?<\/p>\n

Obviously, the end of the lead is soldered to a copper pad on the PCB, and some portion of that 1 watt will conduct through the lead and into the board.\u00a0 So that gives at least two major paths for heat to get out of the lead::<\/p>\n