I was expecting Hal

On the other hand, an experienced driver might forget it’s there since they never use it. Add in a high-stress situation, and you get a problem.

Now imagine you’ve been driving the Tesla for a long time and don’t ever use the manual release because you’re not supposed to so you don’t mess up the window. And then imagine you’re in a high-stress situation. That’s how having an unmarked backup can fail.

Plus, that handle doesn’t even look like a normal handle - I have never see a car where you pull up to exit instead of sideways away from the door.

On the other hand, if you never use the mechanical release and have spent a long time only driving your Tesla, wouldn’t it be possible to forget it’s there while in a high-stress situation?

Do I just connect thermal pads to the ground plane and call it a day?

Yup.

Wouldn’t that make the components hard to solder with hot air?

Sorry, I’ve never tried hot air assembly.

Do I make an isolated polygon that only acts as a thermal pad?

Ideally the copper area is big to spread out the heat. If you have an isolated polygon it can’t spread very far and buys you less cooling.

The 7333A is a linear regulator, which means it drops voltage by converting power to heat. Typically those make sense when the input voltage is close to the output voltage or the load is very small. If it’s getting too hot, the load is high enough that the efficiency will be very bad…whether or not this is a problem depends on your application.

Some random site claims 170mA and another claims up to 400mA. 170mA * 8.7V (12V in minus 3.3V out) = about 1.5 watts, which is too much for a TO-92 package.

Can you use a tiny buck converter instead? Or a larger package for the linear regulator that can add a small heat sink?

As for your actual circuit, the second transistor is an interesting idea (you’re using it to invert the state so you can have the GPIO pulled in the non-problematic direction?) and I don’t have enough experience to give further suggestions.

I’m not entirely clear on the problem, but yes - the circuit as drawn makes the microcontroller pin start high, then fall after some time. Do you need the microcontroller pin to have a different voltage than the transistor base (I assume when you said gate you mean base…gates are for FETs), or is this good enough?

Would a circuit like this power-on reset circuit work for your application?

It’s because of the “lift to drag ratio”. Airplanes in level flight at ordinary speeds generate about 15x as much lift as drag meaning if the engine spends 1 unit of work moving the plan forward, the wings give 15 units of work* upwards. So flying level needs about 1/15th the engine power of going straight up. (I’m using “work” very sloppily here, not in a precise physics sense.)

You can see this in sailboats too, which can travel faster than the wind when they’re sailing at an angle to the wind. Efficient boats travel faster when they’re going almost perpendicular to the wind, not straight downwind! This is because the “lift” of the sail pulling the boat forward even more strongly than the push of the wind in the downwind direction.

While I can’t give an intuitive explanation for why this is, there’s a very easy demonstration that it’s true: kites. If a kite had a lift-to-drag ratio of 1, then it would fly at 45° up. It would fly 50 meters downwind of you when it’s 50 meters up. But any decent kite can fly at a much steeper angle than that; sometimes they look like they’re right over your head! That’s because with a lift to drag ratio of e.g. 10, the 1 unit of drag gives 10 units of lift; if it’s 10 meters downwind it will be 100 meters high.

What is this list sorted by?