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June 27[edit]

N-channel MOSFETs vs P-channel MOSFETs[edit]

Modified for readability 2606:A000:4C0C:E200:C915:F679:15DB:E494 (talk) 18:05, 27 June 2016 (UTC)[reply]

I remember learning that for general switching applications, always use N-channel MOSFETs, since they’re cheaper, more common and have less R_DS(on) than their P-channel counter parts. Basically P-channel MOSFETs are for complementary pairs and niche applications only; for everything else, there's N-channel.

And yet today I came across this document[1]:

   When using an N-channel MOSFET in a load switch circuit, the drain is connected directly to the input voltage rail and the source is connected to the load.

BIG diagram
I've always connected the drain to the load and the source to the ground, like this: .

   The P-channel MOSFET has a distinct advantage over the N-channel MOSFET, and that is in the simplicity of the on/off control block. The N-channel load switch requires an additional voltage rail for the gate; the P-channel load switch does not.

I have no idea what this is about. Why does the N-channel load switch requires an additional voltage rail? Using TTL logic is plenty enough to drive the gate high with respect to the ground. Johnson&Johnson&Son (talk) 12:36, 27 June 2016 (UTC)[reply]

The data sheet considers only Common drain configuration of a load switch transistor so your Common source circuit (which inverts the gate control signal) is not considered. It also does not limit applications to those where standard TTL 5-volt supplies are available. The switch-off time of your circuit drawn is prolonged by the Miller effect, see Common source#Bandwidth and the supply voltage must not exceed the transistor's V_GSmax. AllBestFaith (talk) 13:56, 27 June 2016 (UTC)[reply]

The quoted paragraph basically says to use the transistor as a high-side switch (transistor on top of load on the diagram.) Naturally, as there's now something between source and ground, and it drops voltage (ideally most of it), source is positive wrt ground and you'll need a second voltage to make the gate more positive still, by at least Vth. I don't know what the advantage of using an N-channel type as a high-side switch is, one obvious one (in addition to what AllBestFaith said about bandwidth) might be that the load's ground is the same as the circuit's (no Rds in between.) Asmrulz (talk) 15:47, 27 June 2016 (UTC)[reply]

Ok, I think I get it now. That datasheet is dealing with a very specific case then. I guess I was mislead by its very board title of "Using MOSFETs in Load Switch Applications" and thought it was talking about using MOSFETs as load switches in general.

So let's a take a simple example, using a CMOS logic level output of a microcontroller to drive a LED. The bandwidth doesn't matter. Free choice between Common drain and Common source since the gate drive is under our control. Free choice between high side drive and low side drive since the LED doesn't care about being on the ground or not.

In this case, would the right choice be a N-channel MOSFET or a P-channel MOSFET? Johnson&Johnson&Son (talk) 18:39, 27 June 2016 (UTC)[reply]

I'd personally say that in this case it doesn't even matter if it's a FET or BJT, nevermind its type of conductivity. But N-channel types tend to be "better." For example, the IRF540 has a smaller Rds(on), smaller rise and fall times, a higher ampacity and is also cheaper than its P-channel complement the IRF9540. Sometimes the N-channel will even have a smaller Vth (absolute value) Asmrulz (talk) 20:34, 27 June 2016 (UTC)[reply]

Visibility of a nebula from nearby[edit]

As I understand it, the Orion Nebula takes up basically a degree of the sky (24/1344 * 360/2pi) or 60+ arc minutes, and therefore is twice the size of the full Moon. But only people with very good eyesight even notice it. I assume it would be no brighter from nearby, and this backs me up. However, maybe the middle is more visible, or maybe other nebulae are more visible.

The Earth has been wandering around the galaxy for (almost) five billion years and had a lot of chances to wander close to such events. What I'm wondering is -- what is the brightest that this has ever made the night sky? Has it ever been equal to a permanent moonlight or more? I imagine someone somewhere must have run simulations, so I thought I might as well ask.

If people can back up the source above as true, I should also probably put something about it high up in Nebula so people understand that you don't have planets where you just look out and ooh and aah at the beautiful colors. Wnt (talk) 15:54, 27 June 2016 (UTC)[reply]

Your source says

Yes, if you get closer, your eye will receive more total power from the nebula. But the nebula will also look bigger, so that energy will be spread out over a larger visual area (technically: “solid angle”). The physics tells you that the power per solid angle in fact stays exactly the same, and this quantity is precisely the “brightness” of an object. So if nebula are too faint for to see from Earth with the naked eye — and they are — getting up close and personal doesn’t help any.

While the source says this is a paraphrasal of an astronomer's email, I fing it perplexing. Suppose you apply the same reasoning to a star that is invisible from Earth to the naked eye, instead of a nebula. If we got closer (say 1AU from the star) would it really still be invisible? What am I missing here? Loraof (talk) 16:18, 27 June 2016 (UTC)[reply]

It is true so long as the image is on a significant part of your retina. However, once it gets down to a single cell being lit up (if you have perfect vision) or to whatever point that a blurry image gets dimmer rather than smaller, then it gets dimmer instead.

It is also not true for very bright light - the Sun on a normal day is less bright than a sliver of sunlight seen during an eclipse, because the iris clamps down on sunlight but might not react properly to just a small searing point on the retina. Wnt (talk) 16:26, 27 June 2016 (UTC)[reply]

The source is correct. If you have very good eyesight and you're at a very dark place, you may be able to see such a nebula with your naked eye. As light pollution has increased dramatically in the past 25 years, I doubt anyone can still see those nebulae with the naked eye from my small Western-European country.

Nebulae (not including the galaxies, some of which are still called nebula because they were discovered and named before the difference was known) work by scattering starlight. So, if you're inside one, it will at best be as bright as a piece of paper you hold in your hand (maybe a little more, taking fluorescence into account). If you're some distance away from a star, that's quite dim. Moreover, nebulae consist of a mixture of dark dust and gas that only scatters very specific wavelengths. So that would be a dark, coloured piece of paper. Now you could of course imagine you're very close to a bright star sitting in that nebula, but stars have the habit of using their wind to blow a hole in the nebula. If the Sun were in a nebula, the nebula would only begin at the heliopause, which is so far away that it's quite dark there. PiusImpavidus (talk) 19:50, 27 June 2016 (UTC)[reply]

I agree with what's been said, above - but take a more practical example: The Milky Way galaxy shines with the light of maybe 200 billion suns - an unimaginable amount of light - and yet, here we are, out on the very edge of it (you really couldn't get any closer!) - and yet we can only just barely see it at all with dark, clear skies as a softly glowing smudge. The exact same thing would happen with a nebula if you got up close to it. SteveBaker (talk) 20:23, 27 June 2016 (UTC)[reply]

Nitpick: we're not at the edge of the Milky Way. We're about halfway out from the galactic center (I've heard it put that we're "in the suburbs"). --71.110.8.102 (talk) 22:33, 27 June 2016 (UTC)[reply]

While I agree with the answers given so far, one has to note that M44 is a very bright object. Only if there is extremely severe light pollution where you live will it be hard to spot. You probably have something like Bortle class 8 skies: "City sky. The sky glows whitish gray or orangish, and you can read newspaper headlines without difficulty. M31 and M44 may be barely glimpsed by an experienced observer on good nights, and only the bright Messier objects are detectable with a modest-size telescope. Some of the stars making up the familiar constellation patterns are difficult to see or are absent entirely. The naked eye can pick out stars down to magnitude 4.5 at best, if you know just where to look, and the stellar limit for a 32-cm reflector is little better than magnitude 13."

Without light pollution you're at Bortle class 1: "Excellent dark-sky site. The zodiacal light, gegenschein, and zodiacal band (S&T: October 2000, page 116) are all visible — the zodiacal light to a striking degree, and the zodiacal band spanning the entire sky. Even with direct vision, the galaxy M33 is an obvious naked-eye object. The Scorpius and Sagittarius region of the Milky Way casts obvious diffuse shadows on the ground. To the unaided eye the limiting magnitude is 7.6 to 8.0 (with effort); the presence of Jupiter or Venus in the sky seems to degrade dark adaptation. Airglow (a very faint, naturally occurring glow most evident within about 15° of the horizon) is readily apparent."

What you need good eyesight for is not M44 but M81 see here for Brian Skiff's account of how he spotted this with the naked eye. Even Neptune is in theory visible to the naked eye, it's just that only a few people have made a serious attempt to spot it, Brian Skiff tried but failed, he was observing from too far North, Neptune is then difficult to spot due to being too low in the sky.

If you compare these facts to what reliable sources have to say about what is visible or not to the naked eye, so see just how much severe light pollution is assumed as the normal viewing conditions. Count Iblis (talk) 21:23, 27 June 2016 (UTC)[reply]

Someone correct me if I'm wrong, but photos you see of nebulae are long exposures that gather in much more light than your eye can, so they would not look like the photo if you were up close. Bubba73 ^{You talkin' to me?} 03:53, 1 July 2016 (UTC)[reply]

Are SpaceX's rocket and/or engine designs certified by licensed P.E.s?[edit]

75.75.42.89 (talk) 22:39, 27 June 2016 (UTC)[reply]

I doubt it. Cars aren't signed off by PEs, aircraft aren't, why would rockets be? The whole PE restriction of trade scam is sternly resisted in the manufacturing industries. Greglocock (talk) 04:20, 28 June 2016 (UTC)[reply]

It'd be ideal if someone could explain what P.E. stands for in the context of this question. --Tagishsimon (talk) 17:05, 28 June 2016 (UTC)[reply]

Professional engineer. shoy (reactions) 17:39, 28 June 2016 (UTC)[reply]

The work a double A battery can do[edit]

If you made a device with gears that slowly raised a weight, how heavy a weight could be raised up one metre, by a single double A battery and end up exhausting that battery? I have a buck on this, so lucky lucky lucky. And if this is hard to figure out, or is too trivial, I am sorry. We are curious. Many thanks if anyone can answer this. Anna Frodesiak (talk) 23:22, 27 June 2016 (UTC)[reply]

It depends on the AA battery. Heavy duty or alkaline or lithium? 175.45.116.105 (talk) 23:28, 27 June 2016 (UTC)[reply]

Oooh, I don't know. We were talking about one of those copper-black duracell ones that everyone buys. Anna Frodesiak (talk) 23:31, 27 June 2016 (UTC)[reply]

shameless copied from this url

https://www.baldengineer.com/9v-battery-energy-density.html

Calculating Energy in an AA Battery

A single Energizer AA, or E91, is rated for just under 3000 mAh at 25 mA drain. As a conservative value, let’s use 2750 mAh.

we can calculate the amount of energy in a single 1.5 V AA battery.

E = 1.5V * 2750mA * 3600s = 14850J

Thus it can do the work of a force of 14850 newtons moving a distrance of 1 metre. Or move 1485 kilogram directly up for one metre.

175.45.116.105 (talk) 23:38, 27 June 2016 (UTC)[reply]

Thank you!!! I win!!!!. Ha! Many thanks!!! :) Anna Frodesiak (talk) 23:46, 27 June 2016 (UTC)[reply]

And by the way, 175.45.116.105, that was pretty snazzy maths. I couldn't have figured that out in a million years. Anna Frodesiak (talk) 03:58, 28 June 2016 (UTC)[reply]

The math produces a great theoretical maximum...but of course a lot depends on the efficiency of the motor - and you'd need some pretty impressive gearing to lift that much weight. Almost certainly, a motor that could be powered by a single AA battery would have so little torque that it wouldn't be able to turn a gearbox strong enough to lift a car. So the issue comes down to how efficiently you could gear it down. SteveBaker (talk) 17:01, 28 June 2016 (UTC)[reply]

The 1485 kg raised to 1 meter result is surprisingly large. Friction would seem to be a big problem in achieving it. One way to lift a large mass with an AA battery might be by using it to pump hydraulic fluid into a hydraulic cylinder like a car jack. The case described above would involve the battery delivering a constant 25 milliamperes to the motor for 110 hours. This will extract more energy from he battery than if it had o deliver a more typical motor load such as 500 milliamperes. The data sheet specified that the battery was exhausted when the voltage was down to 0.8 volts, but there is still some energy in the battery even at that point, if a circuit were able to operate on the low voltage. Possibly an electronic circuit could drive the motor and use up he battery more completely. Additionally, more energy can be extracted from an alkaline battery when it is allowed to "rest" and cycled on and off than when it is operated continuously to exhaustion. Edison (talk) 17:29, 28 June 2016 (UTC)[reply]