HTSOP8 Thermal Pad - Page 1
Forum section: electronics / beginners

author: JonPyro
author_rank: Regular Contributor
timestamp: March 10, 2020, 11:04:46 pm
content: Hi All,I am struggling so would really appreciate some help.For a hobby project I am designing it really has three different power modes:1) All circuitry asleep : <100uA2) Main circuitry awake : <150mA3) Main circuitry awake with xBee module transmitting short bursts: <300mAIt runs off 12V and I have chosen this 3.3V regulator:https://docs.rs-online.com/6991/0900766b81676e5e.pdfIt is an HTSOP mount so it does not fit onto a normal heat sink.Using the formula in the datasheet : Power consumption  = Input voltage (VCC) - Output voltage (VO) ×IO (Ave)The max situation of (12-3.3)*0.3 gives me 2.61W.The datasheet also says that the junction temp must never be above 150 degrees and ambient no higher than 85 degrees (that's not too difficult in England!).The datasheet also says:Calculation based on ambient temperature (Ta)Tj=Ta+θj-a×W＜Reference values＞θj-a: HTSOP-J8153.2℃/W 1-layer substrate (copper foil density 0mm×0mm)113.6℃/W 2-layer substrate (copper foil density 15mm×15mm)59.2℃/W 2-layer substrate (copper foil density 70mm×70mm)33.3℃/W 4-layer substrate (copper foil density 70mm×70mm)Substrate size: 70mm×70mm×1.6mmI am right in understanding that foil density is the area of copper the chipset is attached to?so for 2 layer with 70x70mm copper fill Tj = (30+59.2) * 2.61= 232.8 degrees - way too high!my PCB is not going to be bigger than 70x70 and is filled with over components so how would I get over this issue? I am not interpreting this correctly? I have never considered thermal design before so I suspect I am way off the mark.Thank you

author: thm_w
author_rank: Super Contributor
timestamp: March 10, 2020, 11:25:04 pm
content: The absolute maximum rating PD for that package is 2.1W, if you actually expect a continuous burn of 2.6W, that is not going to work.Can find somewhere else to dissipate the power (eg a series resistor), choose a larger package, change to switch mode converter.It is possible the continuous dissipation will not reach 2.6W, if you can guarantee a transmit delay time or something.Is the 12V supply fixed as well, no way to get that lower?

author: ImABeginner
author_rank: Contributor
timestamp: March 11, 2020, 03:46:55 am
content: If you're looking for simple solutions, you could use multiple regulators in parallel. This will usually cause one to work harder than the other, but it's not cause for concern. At 1.3 watts of heat, you could dissipate that with a heatsink with a surface area no bigger than a dime passively.You seem to be needing more precision, however. If that is the case, you can use a zener diode to help soak some power. Most common through hole zener can handle 1/4 watt, which is 0.8V at your specified current. Incidentally, this isn't far from the drop off a sillicon diode, so they may be an appropriate substitute.

author: JonPyro
author_rank: Regular Contributor
timestamp: March 11, 2020, 11:28:19 am
content: Thank you both for your suggestions. I hadn't even considered pairing regulators.The 12V supply is not something I can change as the circuit actually drives high current devices run from the same source.All things considered, you are both spot on and my solution is very sub-optimal. I think the best solution would be to change to a switch mode regulator.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 11, 2020, 12:09:04 pm
content: No, paralleling regulators usually won't work, they end up not sharing the current at all (so one does most of the job, or thermally runs away until destruction). Exception is, some rare regulator ICs support parallel operation, but this will be explicitly advertised. Workaround is using extra resistance after each regulator to make them share current more equally, but this comes with the cost of losing load regulation accuracy. I would never do it unless I have to MacGyver something from the parts I have.Assuming you run at high duty cycles, i.e., won't sleep all the time, I would definitely consider using a switch-mode converter for such a load (12->3.3V @ 150mA = 0.5W average). Modern super-integrated 12V->3.3V switch mode regulators cost tens of cents and come in tiny packages, and only require an inductor and a few ceramic capacitors; datasheets come with layout suggestions. This way, you easily have 90% efficiency. Today, with cheap and advanced power silicon, it's often easier to not generate heat, instead of figuring out how to sink it away! Getting to say the "saving the planet" argument is an added bonus.If you really want to do it with linear regulator of your choice, I would start by considering if you canguaranteeto yourself that the 300mA peaks areA) short compared to the thermal mass of the regulator, say, < 100 ms,B) low in duty cycle, say, less than 10% of the time,this allows you to do the math using the average consumption of 150mA, dissipation down to 1.3W, and resulting Tj of 122 degC at 45 degC ambient with the 2-layer board. Note, while doable, this is still very marginal; you need to consider thelocalambient. The chip doesn't know the temperature of the weather; assuming the board sits in a box of any kind, the local ambient (referenced against the board temperature) starts slowly rising.For this nearly-there design, a few cheap diodes in series, each taking approx. 0.7V off the voltage, would be a simple way to take some heat off the regulator. The total heat is still the same, but now part of it is dissipated in the diodes, which you can place as far from the regulator as possible to spread the heat. An added bonus is a reverse voltage protection which comes with the diodes.But I'd definitely go for the switching regulator. They used to be "difficult for beginners" when you needed 20 external components, calculation of loop compensation parts, etc. Nowadays it's all integrated, unless you need some exotic voltages or configurations. But going from 12V input to 3.3V digital logic is one of the most widely needed configurations, hence it's all solved for you.