Transistor Heatsink - where did I go wrong? - Page 1
Forum section: electronics / beginners

author: skinsman
author_rank: Newbie
timestamp: September 14, 2025, 04:55:44 am
content: Hi, I've built an LED dimmer for my car's dashboard which needs to switch a low-side 2A load. I used a BD682 PNP Darlington transistor in a TO-126 / SOT-32 package that can dissipate about 1W without a heatsink and up to 40W in ideal (unrealistic) conditions. So I thought it would be easy to add a heatsink that would dissipate the expected 2W (2A x 1V Vce), and I put together something (see attached photo) that I thought was overkill. That photo shows the heatsink bolted through the BD682 (with a high quality thermal pad) onto the outside of my enclosure. That's the side of the enclosure, with the downwards-facing side of the heatsink facing the camera, so I expected the fins were oriented correctly for airflow.When I tested this in a cool room my infrared thermometer showed the heatsink temperature rising pretty quickly, getting up to around 100C after 15 minutes and then the transistor failed. Early in the test I measured Vce and it was just under 1V as expected. My power supply showed a draw of 2A.Now I've made the sensible decision to change over to a P-channel MOSFET with low Ron (0.07 Ohm) so the problem will be solved. But for future reference I'd like to understand where I went wrong with the heatsink. It just feels like it shouldn't be so hard to dissipate 2W?Thanks,Steve

author: Andy Chee
author_rank: Super Contributor
timestamp: September 14, 2025, 06:15:17 am
content: Quote from: skinsman on September 14, 2025, 04:55:44 amYou're possibly asking the wrong question.The correct question might be; why does my circuit seem to dissipate more than 2W?To answer that, we need to see your circuit.

author: Benta
author_rank: Super Contributor
timestamp: September 14, 2025, 09:57:46 am
content: When you say dimming, I assume it's PWM control. Swirching losses can add a lot of power dissipation.Again, we need to see your circuit.

author: Sorama
author_rank: Frequent Contributor
timestamp: September 14, 2025, 10:21:12 am
content: Assuming 12V would mean 24W leds, which is a lot of power for just some (interior?) leds.Are these CC leds or voltage driven leds ?

author: mikerj
author_rank: Super Contributor
timestamp: September 14, 2025, 09:28:57 pm
content: Quote from: Sorama on September 14, 2025, 10:21:12 amI was also very surprised to see 2A for dashboard illumination, the OP's dashboard must look like a Heathrow runway at night.OP what is the thermal impedance of your heatsink (Celsius/Watt)?

author: kite31
author_rank: Regular Contributor
timestamp: September 14, 2025, 11:34:55 pm
content: Quote from: skinsman on September 14, 2025, 04:55:44 amIf I have understood your statement correctly, you would be better off with the heatsink in the orientation you see in the picture, fins vertical on the flat base rather than side on; not that it is likely to make a significant difference in your case.

author: Benta
author_rank: Super Contributor
timestamp: September 14, 2025, 11:45:40 pm
content: Seems to be a "shoot 'n forget" question.

author: kite31
author_rank: Regular Contributor
timestamp: September 15, 2025, 12:23:20 am
content: Quote from: Benta on September 14, 2025, 11:45:40 pmI suspect the clue to absence is in the "2A x 1V" original vs the more likely 24W as suggested by others. Without forced ventilation, that heatsink would gethot.

author: skinsman
author_rank: Newbie
timestamp: September 15, 2025, 01:35:22 am
content: Thanks for the replies everyone, plenty of food for thought. Wasn't so much a "shoot and forget" question as "shoot and sleep" - different timezone to most of you probably.The dashboard in question is in a classic car ('72 Volvo 1800ES) with about a dozen illumination bulbs that were very sooty incandescents (mostly T10 form factor) and are now LED replacements that have improved things a lot; no it doesn't look over the top like a runway! With the light spread across a lot of separate instruments it looks good.Though each bulb draws only about 130mA, that adds up. In the car it will be more like 1.5A total but I'm designing for more, and my test rig draws more.Yes the dimming is being done with PWM however at full brightness (which is what I was running when it overheated) the duty cycle is 100% as confirmed with my oscilloscope. So I don't think switching losses were the issue.The relevant part of the circuit is attached. As said before I've now modified this to use a P-channel MOSFET but still am curious what went wrong here.I don't know the thermal impedance of the heatsink sorry, it's just a cheap aluminium one that doesn't have any specs. The thermal pad I used however is a 0.5mm-thick Maxtor one rated at 14.8W/mk thermal conductivity and 0.625K/W thermal resistance.Steve

author: golden_labels
author_rank: Super Contributor
timestamp: September 15, 2025, 02:36:19 am
content: Quote from: skinsman on September 15, 2025, 01:35:22 amPeople, who come with problems, generally do very poor with telling what is relevant and what is not. Obscuring information is only impeding attempts to help.I can only guess that the long horizontal line is GND and that this GND is also shared with all other elements of the circuit. To what the emitter of BD682 is connected ­— I can’t infer. Yet this will determine, if it even opens fully. In particular if current through T1 causes a voltage drop above.If you connected BD682 as the pin numbering in the diagram suggests, base and emitter are swapped. Compare with the datasheet:(after: STMicro BD677–BD682 datasheet (Rev. 5; Jan 2008), p. 1)Independent of that, 2 W is power dissipated if BD682 is in saturation. Is it in your situation? In particular with Q2 having an emitter resistor, and R10 being 2 kΩ?Quote from: skinsman on September 15, 2025, 01:35:22 amUsing Wakefield’s 624-45AB (21×21×8.9 mm, anodized aluminum) as a reference, 20 °C/W with 200 ft/m convection.Quote from: Benta on September 14, 2025, 11:45:40 pmWhen you posted that comment, it was less than 20 h since the first response OP got. That’s hardly “and forget” for people, who have life beyond the internet.

author: Andy Chee
author_rank: Super Contributor
timestamp: September 15, 2025, 06:17:04 am
content: I think R9 should be deleted. It prevents the full turn-on of Q2, and by extension, Q1.Q1 should also have a pull up resistor from base to emitter. With the circuit as you have drawn it, when Q2 is off, Q1's base is indeterminate/floating.

author: skinsman
author_rank: Newbie
timestamp: September 16, 2025, 04:41:47 am
content: Quote from: golden_labels on September 15, 2025, 02:36:19 amFair enough, some things seem obvious when you're too close to a problem.Yes the long horizontal line is GND."T1 Dashlights" has "IN" of +13.8V (from the lights switch) and "OUT" going to the load - all the dash illumination bulbs in parallel and then to GND.Quote from: golden_labels on September 15, 2025, 02:36:19 amThe pin numbering came from the KiCad BD682 library component, and yes I did have some initial confusion due to the mismatch with the datasheet. However it did get connected correctly ... 2nd try.Quote from: Andy Chee on September 15, 2025, 06:17:04 amR9 is a feedback resistor for stabilisation. I understand that to be best practice but I may well have misunderstood.Quote from: Andy Chee on September 15, 2025, 06:17:04 amSure, I can see that would be safer and I should have done it. I've already included one along with the replacement MOSFET where it's essential. BTW in my testing with 100% duty cycle that led to overheating this wouldn't have been an issue anyway.Thanks for the help everyone! In summary what I'm hearing is that it's likely that somehow more than 2W was going into that heatsink due to an error somewhere, and that if it genuinely was only 2W it should have coped OK. That's the main thing I needed to know.BTW another potential source of error is the provenance of my BD682s. I checked and found that I had bought them some time ago from AliExpress which I now understand is a bad idea for semiconductors. Maybe that's the root of my problems

author: Andy Chee
author_rank: Super Contributor
timestamp: September 16, 2025, 05:01:22 am
content: Quote from: skinsman on September 16, 2025, 04:41:47 amAs Q2 is being used as a digital hard switch in your application (i.e. fully on or fully off, nothing in between), stability is not a concern.In linear analog applications however, such a feedback resistor is often necessary. Not required in hard switching applications.Additionally, after omitting R9, you could probably increase R3 to 1k in order to further reduce the strain on your PA8 output driver.

author: skinsman
author_rank: Newbie
timestamp: September 17, 2025, 01:47:05 am
content: Quote from: Andy Chee on September 16, 2025, 05:01:22 amOK understood. BTW I checked my notes & the actual build, and see now that I neglected to update in KiCad when I realised R3=220R was the wrong value and changed it some time ago - it is actually 1K in the build.Thanks again.

author: mikerj
author_rank: Super Contributor
timestamp: September 22, 2025, 04:13:47 pm
content: Quote from: skinsman on September 15, 2025, 01:35:22 amThat still seems like an exorbitant amount of current for an LED, the lamps used in older cars for instruments were typically only 2-3watts and a modern LED consuming similar power would have at least an order of magnitude more light output. Do you perhaps have the "CANBUS safe" style lamp that includes resistors to make the bulb draw similar current to an incandescent to prevent bulb failure warnings?

author: MrAl
author_rank: Super Contributor
timestamp: September 22, 2025, 05:24:33 pm
content: Quote from: skinsman on September 14, 2025, 04:55:44 amHi,You are right these things can be interesting even if we found another solution.This problem has a lot of variables that would have to be defined at least approximately. I'll list a few...1. We need to know the model of the heatsink, or at least the surface area. The effectiveness of these things varies wildly even if they look good.2. We also need to see what is switching the transistor. If the transistor could turn on and off immediately, then you would probably see 2 watts. If the transistor has to ramp up to 'on' and ramp down to 'off', that can eat up a lot more energy which means it could be dissipating a lot more, like 5 watts. You would need a scope to see this. The ramp up and ramp down should probably be less than 10us each, but we also need to know the PWM frequency so we can gauge that better.This ramp up and ramp down would be referred to as the mechanism that causes switching losses. We like to keep the switching losses low, and that means making sure the transistor turns on fast and turns off fast. If we know the ramp times we can estimate the switching losses, but as mentioned you'd need to use an oscilloscope. If the frequency is low you can get a cheap one as that will work ok, but we need to know that frequency so we know what to get.3. Where the heatsink is mounted and what kind of air flow does it get. If part of the heatsink is covered by some kind of material that does not conduct heat well, then the heatsink will get much hotter than it should.So in short, we need to know more about the heatsink or the at least surface area, and that means the size and maybe some more pics of that one.We also need to see the circuit so we can estimate the switching losses, or see snapshots of an oscilloscope picture of the waveforms, and we need to know the frequency. If you are testing with a particular duty cycle, we also need to know that.You might also not be aware of this, but using straight PWM without any energy storage element will cause a power dissipation equal to a resistor that causes the same current to flow as the duty cycle would. It's only when we combine PWM with an energy storage element like an inductor that we see lower power dissipation. The power distribution changes that's all. You may or may not need to worry about this right now though because after all PWM does cause a dimming effect without having to actually vary a resistor.