Ask a physicist anything. (5)

[Upisoft]

It's a MOSFET transistor. N-channel, to be precise. In a simple integrated circuit.

Well, I've never seen such complex diagram of the thing. I've seen some simplified diagrams though. I don't know what most of the elements are. I suppose that some of them have to be protecting Z-diodes. But that out of my competence anyway.

 

[TheReasoner]

Well, I've never seen such complex diagram of the thing. I've seen some simplified diagrams though. I don't know what most of the elements are. I suppose that some of them have to be protecting Z-diodes. But that out of my competence anyway.

Not really. It's very simple. Colors denote function. Green: p-doped Si, red n-doped. the 'fuller' the color, the higher the dopant concentration. The bottom is your substrate, the lighter green is the epitaxial layer. 14 and 13 shows the shallow trench isolation of the transistor with an oxide film to protect against diffusion into the Si. The golden 'cones' are metal interconnects, usually wolfram. The thin silver/gray films around them is a Ti or TiN film - again, diffusion barrier and an adhesive which aids the W plugs. The gray bars connecting the plugs are metal interconnects - or "wires" running around to different parts of the circuit if you will, and the thicker gray areas are a low k dielectric material to reduce capacitance where you don't want it. At the top there's a metal pad, usually aluminum (for connecting the die to whatever packaging you want) and a passivation layer, for die protection, mostly.

The image is just something I threw together while reviewing, as a flash card.

This would be a part of a simple IC. In more complex circuits such as a CPU you get MANY more layers. A 3D view - without dielectric - can be seen here:

Wikipedia has an article which explains it rather well: Integrated circuit - Wikipedia, the free encyclopedia

 

[Upisoft]

This would be a part of a simple IC. In more complex circuits such as a CPU you get MANY more layers.

I never ever imagined these things are so complex. Good we have computers to help us designing bigger and faster computers.

 

[Brak]

It's a MOSFET transistor. N-channel, to be precise. In a simple integrated circuit.

That's a lot more than just a MOSFET. 6, 7, 8, 17, and 18 make up the MOSFET. Maybe 5 as well. The rest is just connectivity going to/from the transistor. 9 is some sort of packaging material and not part of the MOSFET. For that matter, there's nothing there that makes it N-channel, either, because the only difference is the type of material you use.

 

[Brak]

Oh! I've got a real question for a physicist. And this is not just a "try to stump the physicist" question...this is a real question:

The sensor arrays at CERN (or Fermi, or SSC, whatever). How do you induce an electrical current and/or electric potential from sub-electronic particles? You know, you're trying to track down the signature of a Higgs Boson (or strange quark...whatever). Somehow you capture these signatures in the form of an electric current or something. You amplify it, run that through ADC, digitize it, display it on a computer. What I can't figure out is, the particles are so much smaller than even a single electron. How do you produce a flow of multiple electrons out of that? Does a boson produce a magnetic field or something?

For that matter, the whole incident happens in the span of attoseconds. How do you capture that??? The smallest capacitance we can get in the circuitry is on the order of femtofarads. It seems like the whole reaction would get lost in the noise. Do you just massively over-sample and stagger the phase over attosecond-long increments or something?

 

[Upisoft]

Oh! I've got a real question for a physicist. And this is not just a "try to stump the physicist" question...this is a real question:

I'm not a physicist, so you can stump me if you like to.

The sensor arrays at CERN (or Fermi, or SSC, whatever). How do you induce an electrical current and/or electric potential from sub-electronic particles?

It's a guess, but I think it has to do with the fact a charged particle in magnetic field will emit photons. Another possibility is that charged particle with high energy can ionize the matter it goes through. The particles from a high energy collision has enough energy to ionize a lot of atoms.

What I can't figure out is, the particles are so much smaller than even a single electron. How do you produce a flow of multiple electrons out of that? Does a boson produce a magnetic field or something?

I don't think that the size has anything to do with it. The energy matters. Singe energetic (fast) electron can knock other electrons from a lot of atoms.

For that matter, the whole incident happens in the span of attoseconds. How do you capture that??? The smallest capacitance we can get in the circuitry is on the order of femtofarads. It seems like the whole reaction would get lost in the noise. Do you just massively over-sample and stagger the phase over attosecond-long increments or something?

I don't think the current produced by a single particle is what they measure. So, the capacitance does not matter.

 

[chris4243]

The original detectors were cloud chambers and photographic plates. When a very energetic particle hits an atom, it strips away electrons. A very energetic particle will cause a chain of reactions because the particles it hits will themeselves be energetic enough to leave trails themselves. In a cloud chamber, the ions act as nucleation sites that allow supersaturated air a place to condense forming trails of clouds, of which pictures can be taken. In a Geiger counter, the ions act as a conductor to transmit charge. In the photographic plates, the plate gets exposed (and these can be in stacks so you get a permanent 3D shape). I'm not sure quite how the modern detectors work, but as you noted the event itself is too quick to do in real-time, and I'm sure it reads as some sort of trail as well.

Furthermore, an electric field at the detector causes the particle to be attracted to one side or the other, and a magnetic field causes the particle to curve. Measuring the curvature of the trails and the trails of the cascade of secondary particles allows measuring the charge, mass, energy of the particle in question.

 

[Brak]

Measuring the curvature of the trails and the trails of the cascade of secondary particles allows measuring the charge, mass, energy of the particle in question.

Well that explains how you can interpret the signatures. You see one trail curve twice as much as another, then one's a top quark and the other's a bottom. You see a trail that does not mathematically match anything you've seen before, then you've discovered a new particle. Cool....

But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?

 

[Upisoft]

Well that explains how you can interpret the signatures. You see one trail curve twice as much as another, then one's a top quark and the other's a bottom. You see a trail that does not mathematically match anything you've seen before, then you've discovered a new particle. Cool....

But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?

I really doubt they detect quarks directly, as when you have the energy to make a quark escape a hadron then you have the energy to make new hadron. Thus you detect most of the particles by the way they decay.

When there is collision most of the particles don't continue around the ring and cross the detector that surrounds the ring of the accelerator.

Now, maybe all of them have to cross the detector, I just don't have enough knowledge to calculate where all the product particles must go. Considering the conservation of the momentum law they might trick them all to cross the detector.

 

[chris4243]

But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?

Yes, the collisions occur in a vacuum, and then you get particles flying off in all directions from the collision. (you can also smash a stationary target, but then due to conservation of momentum the particles need to have more kinetic energy and so there is less energy available for particle production). During the collision, there is a lot of energy available, and some particles to carry away energy and momentum, and also particle/anti-particle pairs are generated. I'm not sure how pair production works, but I heard that the rule is "If it is not forbidden it is mandatory".

Oh, and I forgot earlier that also the decay products of the particles give a lot of information about it (for the particles that decay, which will be all the non-boring ones).

 

[Seamus Riley]

Totally legit.

i think you've given me both my avatar and custom user title, thanks.

either that or that bed intruder song guy..."we gone find you."

 

[Wiccan_Child]

Well that explains how you can interpret the signatures. You see one trail curve twice as much as another, then one's a top quark and the other's a bottom. You see a trail that does not mathematically match anything you've seen before, then you've discovered a new particle. Cool....

They're not quarks, they sound more like electrons and positrons (opposite charge, same mass, so equal and opposite spirals). We can't isolate quarks yet

But I thought they were smashing H+ ions (a.k.a. protons) in a vacuum? How do you sense a curve if you don't sense anything at all until you collide with something?

Because we know what we should see if the Higgs boson (say) exists. A major tool is not in what it does, but in what its decay products do - if the Higgs decays into a shower of pions and kaons (I have no idea if it can or not), we can trace back the identified paths of paions and kaons to a single point - the Higgs.

 

[Wiccan_Child]

Oh! I've got a real question for a physicist. And this is not just a "try to stump the physicist" question...this is a real question:

The sensor arrays at CERN (or Fermi, or SSC, whatever). How do you induce an electrical current and/or electric potential from sub-electronic particles? You know, you're trying to track down the signature of a Higgs Boson (or strange quark...whatever). Somehow you capture these signatures in the form of an electric current or something. You amplify it, run that through ADC, digitize it, display it on a computer. What I can't figure out is, the particles are so much smaller than even a single electron. How do you produce a flow of multiple electrons out of that? Does a boson produce a magnetic field or something?

For that matter, the whole incident happens in the span of attoseconds. How do you capture that??? The smallest capacitance we can get in the circuitry is on the order of femtofarads. It seems like the whole reaction would get lost in the noise. Do you just massively over-sample and stagger the phase over attosecond-long increments or something?

We use multipliers to amplify what each particle does. Take a look at this video - you can see macroscopic effects of single atoms decaying:

YouTube - How to Build a Cloud Chamber!

 

[Lostalongtheway]

Thanks for that 28.1.11 video Wiccan Child. A bit daunting due to implants but I got through it .

"A scientific man ought to have no wishes, no affections, -- a mere heart of stone."
​

- Charles Darwin

You DO know he was joking?

 

[TerranceL]

Totally legit.

His real hair too!

 

[MattRose]

How high must one jump (in a vacuum obviously) so that you stay in the air for 1 second. Please give your answer in feet or meters please. Ok, Europeans can answer in metres, but furlongs are right out.

 

[mzungu]

How high must one jump (in a vacuum obviously) so that you stay in the air for 1 second. Please give your answer in feet or meters please. Ok, Europeans can answer in metres, but furlongs are right out.

If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.

If it takes you 0.5 seconds to climb then the total distance (start to finish) should be 5 metres thus the height should equal 2.5 metres if you exclude inertia. otherwise you have to calculate the deceleration time before reaching a speed of zero before gravity causes you to reverse direction.

Someone do the maths here because I am too full of Easter cholesterol and wine to think straight

 

[TheReasoner]

If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.

If it takes you 0.5 seconds to climb then the total distance (start to finish) should be 5 metres thus the height should equal 2.5 metres if you exclude inertia. otherwise you have to calculate the deceleration time before reaching a speed of zero before gravity causes you to reverse direction.

Someone do the maths here because I am too full of Easter cholesterol and wine to think straight

In vacuum you're approx. right. If you use the somewhat more accurate 9.81m/s^2 you get 2.4525 m (0.0121913028 furlongs) so it's right as rain I'd say. 2.5 meters does it. Gives you room to wriggle your toes a bit ;-)
It's a good approx even in atmosphere. For such short jumps the air resistance contribution can't be significant I'd imagine. Not unless you weigh next to nothing, are bloated like a blimp to boot and stand atop an air vent

 

[MattRose]

If you mean starting from the ground going up and then coming down then you have to specify the climb rate. At one g the drop rate is 10 metres per second squared.

If it takes you 0.5 seconds to climb then the total distance (start to finish) should be 5 metres thus the height should equal 2.5 metres if you exclude inertia. otherwise you have to calculate the deceleration time before reaching a speed of zero before gravity causes you to reverse direction.

Someone do the maths here because I am too full of Easter cholesterol and wine to think straight

Correct answer. Unfortunately, I can't give you full credit. You took it upon yourself to set gravity to 10m/s*s rather than the universal 9.8. Fittingly, I'm awarding you 98/100.

 

[TheReasoner]

Correct answer. Unfortunately, I can't give you full credit. You took it upon yourself to set gravity to 10m/s*s rather than the universal 9.8. Fittingly, I'm awarding you 98/100.

(pssst: It's not universal. It depends on the position within the gravitational field)