The h Bar

12:51 AM

@enumaris Answers have to be pretty far out there to be worthy of moderator action for being non-mainstream. We (mods) really don't want to be in the position of deciding what information is presented as right or wrong, at least not any more than an ordinary user of the site (i.e. by voting up/down).

We're generally reluctant to delete answers for being mainstream unless it really looks like it's clearly nonsense, and harmful to have the information associated with that question.

@manooooh thanks for asking here :)

As you've heard from other people, a question that doesn't ask anything more than whether your work is right isn't really on topic for us.

@Blue I took care of that question... I think a lot of questions just take a while to accumulate enough votes to go on hold.

user351417

3:32 AM

@DavidZ My flag on that post is still pending; I raised it 11 hours ago. I included links to other supersolid DM answers, all of which use the exact same wording, and all have been deleted. That answer is surely really really far out there, and I strongly believe that these are trolls, since they always show up with new accounts but the same words for the same supersolid-DM theory.

user351417

3:54 AM

@enumaris @DavidZ Never mind, it's been deleted now.

rob

@enumaris I want to echo @DavidZ that the moderator team is very cautious about unilaterally removing answers just because they may express some non-mainstream or personal theory. I've just removed that particular answer, but that was due to some issues unrelated to its physics content. Thanks for flagging; keep 'em coming.

user351417

Thanks (insert name of the mod)!

rob

Oh, good evening @Chair

user351417

Oh dear @rob I thought I was being all subtle and stuff :P

rob

@Chair You're fine. How's things?

user351417

3:57 AM

Kind of obvious who handled that flag though... you were on chat a few minutes ago so you'd have seen the message, and after that those stuff are there in my review queues :)

user351417

@rob I'm great, thanks! I have a week of vacations now because there's a Dussera festival here. Must apply to college though, and I have to prepare some stuff for some chem research I'm doing at school. How're you?

user351417

I'm curious, do you guys handle flags by scrolling through users' flag history and dealing with them, or is there a separate mod queue? because I suddenly had about 4-5 flags cleared over a couple of minutes.

rob

@Chair I'm a little tickled. The album I collaborated on was officially released today. I hadn't heard that piece since we performed it, and I'm very pleased with the outcome.

@Chair The mods have a separate flag queue. I just went through it and processed everything but some VLQ flags that I'd rather let the userbase handle.

David Z

The way the UI is designed, it would be very very inefficient to handle flags by going through a user's flag history. (Even assuming we happened to stumble upon a user who had pending flags at all.)

user351417

@rob Ooh that's cool stuff! Is that like orchestral stuff? My sister participates in classical chamber orchestras, so I've heard a bit of that.

rob

4:08 AM

@Chair Yes, it's a big chorus, a few soloists, and an orchestra. It's not an easy piece of music to listen to the first time, but that's perhaps a good quality for a requiem mass.

Akash. B

hi

rob

@Akash.B Greetings

2

Akash. B

4:28 AM

@rob same to you

user351417

4:39 AM

@rob That's cool! You play an instrument for it?

rob

@Chair No, I'm in the chorus.

Sir Cumference

5:25 AM

Howdy yall

user 2646

5:45 AM

hi

Akash. B

6:09 AM

@user2646 hi

user 2646

how are you?

Secret

8:10 AM

Paradigm shift string theory dark energy

Novalium Company

8:42 AM

Hi guys, I want to make an elektret mic amplifier and I found this: instructables.com/id/Pre-amp-to-electret-mic but I can't understand the circuit. I know what the components do individually, but everything connected seems so messy for me.

I thought you could just hook up the elektret mic into a transistor somehow and then the output to a speaker. What is the point of all these capacitors and resistors crossing over and connecting in weird ways?

Novalium Company

9:00 AM

I'm trying to follow the current throught the circuit but I don't know where to start from or even if I should do it that way.

I'm generally trying to understand how the signal gets amplified. I know the transistor can act as an amp but what do all the resistors and capacitors bring to the table and why are they hooked up the way they are?

John Rennie

@NovaliumCompany which resistors in particular are you puzzled about?

Novalium Company

Can I break the circuit into parts and analize them individually?

John Rennie

Not really ...

R3 is just there to form a voltage divider with the transistor. i.e. as the transistor resistance changes the voltage at the juntion between the transistor and R3 changes. OK so far?

Novalium Company

Yep

John Rennie

And likewise R1 and the microphone form a voltage divider. So as the microphone resitance changes (due to sound) the voltage at the mic/resistor changes. That changing voltage is then fed into the base of the transistor.

Novalium Company

9:08 AM

Got it

John Rennie

And finally R2 is the feedback resistor. It is there to control the overall gain of the amplifier.

The two capacitors are there to allow through oscillating signals, i.e. the signal due to the sound, but block constant voltages.

Novalium Company

What would happen if we remove the capacitors?

John Rennie

Without C1 there would be a constant DC voltage on the base, and that would turn the transistor permanently on i.e. zero resitance.

Novalium Company

Got it, how C1 allows only signals to pass and not constant voltage?

I mean, I can imaging turning on the circuit and C1 charging up to it's maximum but then?

John Rennie

@NovaliumCompany That's what capacitors do. The impedance of a capacitor is inversely proportional to frequency so the resistance to DC is (in principle) infinite.

Novalium Company

9:14 AM

Does speaking into the mic increase or decrease base voltage?

John Rennie

It makes the base voltage oscillate.

Novalium Company

So talking into the mic will increase the mic resistance, and therefor, decrease the R1/Mic junction voltage which will then sligthly discharge the capacitor... which will result in alternating base voltage. (What I'm saying is probably wrong)

John Rennie

To be honest I'm not entirely sure how to calculate the voltage at the base. My memory of circuit theory is a bit sketchy these days.

Novalium Company

Ok, so it all breaks down to pattern recognition, thanks! One last thing, I don't understand how exactly the feedback resistor works. What does it decrease or regulate?

John Rennie

But in broad terms you get a small oscillating voltage at the base, and that makes the transistor resistance oscillate. That makes the voltage at the collector/R3 junction oscillate, and that's the output.

Suppose the voltage at A increases. That turns on the current flow in the resistor, so the transistor resitance falls and the voltage at B decreases.

Novalium Company

9:22 AM

I suppose the feedback resistor plays an important part in the actual transistor amplifiaction and that's why I don't understand how exactly the signal amplifies since the alternating voltages between the R1/MIC and R3/Trans look the same to me.

John Rennie

But the point B is connected back to point A through R2, so the fall in the voltage at B reduces the voltage at A and this reduces the current flow through the transistor again.

Novalium Company

"Suppose the voltage at A increases" - You mean as we speak into the mic, that triggers the capacitor and it passes the alternated (increased) voltage?

John Rennie

@NovaliumCompany a small change in the base current creates a large change in the current flowing from the collector to the emitter. So a small input voltage change at the base causes a large voltage change at the collector/R3 junction. That's why the transistor amplifies the signal.

Novalium Company

@JohnRennie Yes but that only works if the voltage into the base is lower than into the collector, and from what I see they are the same if we image the feedback resistor doesn't exist?

John Rennie

@NovaliumCompany aha, that very circuit has been asked about on the Stack Exchange.

3

Q: transistor mic preamp

Hi I built this mic preamp circuit for use with an electret mic. The mic I have is this PUI AOM4544 electret mic I have plugged this mic preamp circuit to my soundcard line in. But I think the gain is so low that I cannot hear anything. I built this to use as a measurement mic. I tried measuri...

Novalium Company

9:29 AM

@JohnRennie As always, I forget to google first :D

John Rennie

> That's a crappy circuit. Don't believe everything you find lying around on the internet.

:-)

Novalium Company

So the circuit doesn't work?

John Rennie

@NovaliumCompany the answer says it works but has limitations. I can't make any useful comment since my memory of circuit theory is too tenuous.

Novalium Company

@JohnRennie Ok, thank you so much for helping, it means a lot.

John Rennie

@NovaliumCompany back when I was your age I used to spend days and days making electronic circuits like this. In those days the Internet didn't exist so we nerds played with electronics instead :-)

Novalium Company

9:33 AM

I haven't got the idea to look for voltage dividers in circuits. Now resistors don't look useless :D

@JohnRennie I can only imagine how hard it was...

John Rennie

@NovaliumCompany truly those were the dark ages :-)

Akash. B

10:59 AM

the new findings of Stephen Hawking was revealed ,right?

Avnish Kabaj

11:15 AM

Sotheby's, October 5th 2018

►

The banksy shred incident was quite something

Akash. B

3:30 PM

@JohnRennie is this transistor stable?

bolbteppa

4:30 PM

Assuming we know what an $SO(3)$ spinor of rank $1$ (and so one of rank $2j$) is and that it transforms under $SU(2)$, and using the fact that, for $j$ an integer, the $2j+1$ spherical harmonics $Y_{lm}$, the $2j+1$ components of an irreducible tensor of rank $j$, and the $2j+1$ components of a symmetric spinor of rank $2j$ all give the same irrep of $SO(3)$, we find for the case of $j = 1$ (vectors) that we can express a spinor in terms of a vector and vice versa, thus from the form of the three spherical harmonics (up to normalization)

(Should be $Y_{10} = i \cos \theta = i z$)