scott wrote:
>> How many resistors can you see? Because I count NONE!
> 
> Sure, but that circuit does not show how you generate the inputs A,B,Cin 
> or how to connect the output to an LED.
> 
> Both of which require some basic knowledge of analogue electronics.

NNOOOOOOOOOO!!!! X_X



Well anyway, FWIW, the IC managed to drive the LED just fine. (I bought 
special 5V LEDs; apparently normal ones take roughly 3V.) It's just that 
I couldn't get the right truth table out of it. :-(

You think using a 2-pole switch with the poles connected to each of the 
power rails will do it?

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scott wrote:
>> Just for the record: I have never taken any A-levels. I only took 3 
>> GCSEs. I got a B and two Cs. This are unprecidentedly high grades for 
>> the school I went to. (Remember, I went to a school for mentally 
>> retarded people.)
> 
> OOC why did you go to that school, you don't seem mentally retarded!

Have you heard my voice? I certainly *sound* retarded! (I talk waaaaay 
tooo sloooowllllyyyy...)

Well anyway, the first school I went to was a rough school. Kids could 
beat up other kids just for entertainment. Whoever the weakest kid was, 
he or she got beat the most. Wanna guess who the weakest kid in my class 
was?

Anyway, when I reached the age of 9, I still couldn't read or write. At 
all. I mean, I spent every day at school in a state of absolute terror, 
I wasn't even paying attention to *lessons*! And the school repeatedly 
refused to lift a finger to stop the kids bullying me.

I can vividly remember being chased across the playground screaming blue 
murder begging for somebody to save me. But the staff just stood there 
smoking, not really paying attention. Afterwards I asked one of them 
"why didn't you save me?!" The disinterested individual casually 
responded "uw, I thought they woz just playin widja". Yeah, thanks for that.

So then I got moved to this school for weird kids. And my god, were they 
weird! Some of them couldn't even speak properly, or at all. Almost all 
of them were very obviously strange.

On the plus side, I learned to read and write. (Although it seems some 
people here have nothing better to do than point out my spelling 
mistakes.) On the minus side... I spent 7 years without being able to 
interact with normal humans, and without being able to interact with 
*girls*. As absurd as it seems, I was 21 years old the first time I had 
a real conversation with a girl my age who wasn't my sister.

(Actually, this turned out to be the first time I *touched* a real girl 
too. After weeks of pestering, she did eventually let me kiss her. Most. 
Terrifying. Thing. EVER! But DAMN she was ****ing good at it!! *sigh* 
She's married now. I can't even contact her any more...)

Originally I assumed that I went to this school because I couldn't read 
or write. But now, looking back, it has gradually become clear to me 
that *I* am not normal either. I do things that normal people don't do. 
There clearly *is* something wrong with me. And I know it's incurable. I 
find this extremely upsetting.

>> FWIW, my statements above were about college too. The class I was in 
>> contained a bunch of "lads" who were more interested in getting drunk 
>> and arguing about Cast vs Blur than actually learning stuff.
> 
> That mostly stops when you get to A-levels, because it's optional you 
> only get people there who actually want to be there.

Well, going to college is optional too. People told me it would be 
"different" at college. But from what I can tell, the guys who were 
there were too hopeless to get actual jobs, so they went to college 
instead. Wuh?! o_O

At least at uni the people weren't so *completely* stupid...

> Aww, would have been better if you had some hot girls in your group, 
> then I'm sure you wouldn't have minded completely rewriting their code 
> from scratch :-)

Girls? In a computing course...?

Which part of the world do *you* live in?? o_O

(Actually, in fairness, the full class was 80 people, and IIRC about 6 
of them were female. And 2 of them were my age. The other 4 were married 
with children. The two my age held a deep fascination for me... although 
obviously I was terrified of them. And, apparently, they were both 
spoken for anyway...)

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> Well anyway, FWIW, the IC managed to drive the LED just fine. (I bought 
> special 5V LEDs; apparently normal ones take roughly 3V.)

"normal" ones work at all sorts of voltages, it even varies from piece to 
piece quite significantly.  But you never drive an LED directly by applying 
a fixed voltage, you always drive it by regulating the current to a fixed 
amount (like 20 mA or whatever).  Adding a series resistor to a raw LED is a 
quick and crude method of fixing the operating current.

> You think using a 2-pole switch with the poles connected to each of the 
> power rails will do it?

Yup, or just connect a high value resistor (eg 10 or 100K) between the input 
and the other power rail - so you have your switch arranged like this:

http://www.seattlerobotics.org/encoder/mar97/basics4.gif

That way, when the switch is open, the input is connected to 0V, and when 
it's closed it is connected to V+.

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scott wrote:
>> Well anyway, FWIW, the IC managed to drive the LED just fine. (I 
>> bought special 5V LEDs; apparently normal ones take roughly 3V.)
> 
> "normal" ones work at all sorts of voltages, it even varies from piece 
> to piece quite significantly.  But you never drive an LED directly by 
> applying a fixed voltage, you always drive it by regulating the current 
> to a fixed amount (like 20 mA or whatever).  Adding a series resistor to 
> a raw LED is a quick and crude method of fixing the operating current.

This doesn't make sense to me.

Presumably the resistence of the LED is finite and fixed. How does 
adding another resistor help? There are several schematics in my 
electronics kit that involve LEDs and no resistors at all.

>> You think using a 2-pole switch with the poles connected to each of 
>> the power rails will do it?
> 
> Yup, or just connect a high value resistor (eg 10 or 100K) between the 
> input and the other power rail - so you have your switch arranged like 
> this:
> 
> http://www.seattlerobotics.org/encoder/mar97/basics4.gif
> 
> That way, when the switch is open, the input is connected to 0V, and 
> when it's closed it is connected to V+.

Now, see, I've never been able to comprehend stuff like this. To me, 
this diagram just looks like when you close the switch, all the current 
will flow straight from one rail to the other, shorting out the battery 
and not providing any current at all to the input of the gate.

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> Presumably the resistence of the LED is finite and fixed.

No.  Here's a graph of voltage against current for a typical LED:

http://www.cq.cx/pics/int-led-vi.png

Note that for a fixed value resistor this would be a straight line starting 
at the origin.

Say you want to operate the LED at 20 mA (brightness/lifetime etc of LEDs is 
all related to current, not voltage).  That means you need to drive it at 
exactly 2.0 V.  You need to add a resistor in series to connect it to eg a 
5V supply.  So you need 3V across the resistor with 20 mA flowing through 
it, R=V/I=3.0/0.02= 150 ohms.

Of course there are more complex LED driving circuits that actively keep the 
current constant over a wide range of operating conditions, some can even 
drive chains of LEDs in series and flag up when there are failures etc.

> Now, see, I've never been able to comprehend stuff like this. To me, this 
> diagram just looks like when you close the switch, all the current will 
> flow straight from one rail to the other, shorting out the battery and not 
> providing any current at all to the input of the gate.

Yes, *some* current will flow from one rail to the other, which is why it is 
important to use a *high* value resistor.  If you use a 10kOhm resistor 
across 5 Volts, that's I=V/R=0.5mA which is hardly going to cause a problem. 
But the important thing is, that point in the middle that is directly 
connected to the input will be at 5V when the switch is closed.

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scott wrote:
>> Presumably the resistence of the LED is finite and fixed.
> 
> No.  Here's a graph of voltage against current for a typical LED:
> 
> http://www.cq.cx/pics/int-led-vi.png

Where do you *find* this stuff??

> Note that for a fixed value resistor this would be a straight line 
> starting at the origin.

OK, now I'm confused. A graph of current against voltage? But current is 
completely *determined by* voltage! o_O

>> Now, see, I've never been able to comprehend stuff like this. To me, 
>> this diagram just looks like when you close the switch, all the 
>> current will flow straight from one rail to the other, shorting out 
>> the battery and not providing any current at all to the input of the 
>> gate.
> 
> Yes, *some* current will flow from one rail to the other, which is why 
> it is important to use a *high* value resistor.  If you use a 10kOhm 
> resistor across 5 Volts, that's I=V/R=0.5mA which is hardly going to 
> cause a problem. But the important thing is, that point in the middle 
> that is directly connected to the input will be at 5V when the switch is 
> closed.

OK... so... why not an open circuit then? That would have infinite 
resistence?

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>> http://www.cq.cx/pics/int-led-vi.png
>
> Where do you *find* this stuff??

I did a google image search for "LED VI graph" IIRC.  Easier than trying to 
explain it.

>> Note that for a fixed value resistor this would be a straight line 
>> starting at the origin.
>
> OK, now I'm confused. A graph of current against voltage? But current is 
> completely *determined by* voltage! o_O

Of course it is determined, but it's only a linear relationship (V=IR) for 
resistors.  LEDs are diodes and are certainly not linear, so you need a 
graph.  You could draw a graph for a resistor if you wanted like I said, but 
it would be a bit pointless because we know V=IR for a resistor.

> OK... so... why not an open circuit then? That would have infinite 
> resistence?

Then the input would be equally connected to 0V and V+ with near infinite 
resistances, that's called a "floating" input, because it can be persuaded 
very easily to either go to 0V or V+, depending on the exact details inside 
the IC (and maybe some state of the other inputs, who knows?!).  Best to 
ensure you have at most around 100K of resistance to one of the rails to fix 
the input reliably.

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>> OK, now I'm confused. A graph of current against voltage? But current 
>> is completely *determined by* voltage! o_O
> 
> Of course it is determined, but it's only a linear relationship (V=IR) 
> for resistors.  LEDs are diodes and are certainly not linear, so you 
> need a graph.  You could draw a graph for a resistor if you wanted like 
> I said, but it would be a bit pointless because we know V=IR for a 
> resistor.

Wait a sec - so you're saying there are devices which actually violate 
Ohm's law?

>> OK... so... why not an open circuit then? That would have infinite 
>> resistence?
> 
> Then the input would be equally connected to 0V and V+ with near 
> infinite resistances, that's called a "floating" input, because it can 
> be persuaded very easily to either go to 0V or V+, depending on the 
> exact details inside the IC (and maybe some state of the other inputs, 
> who knows?!).  Best to ensure you have at most around 100K of resistance 
> to one of the rails to fix the input reliably.

Uh... so have it always connected to one rail, but through such a high 
resistance that the rail can't affect it?

Well, you're the one with the engineering degree, but it *still* doesn't 
make any sense to me...

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> Wait a sec - so you're saying there are devices which actually violate 
> Ohm's law?

Yes, more exactly, there are devices which do not have constant 
resistance/impedance.

> Uh... so have it always connected to one rail, but through such a high 
> resistance that the rail can't affect it?
>
> Well, you're the one with the engineering degree, but it *still* doesn't 
> make any sense to me...

Maybe it would help if you imagined that inside the IC there is a big 
resistor (eg 1Mohm) connecting the input to one of the supply rails (but you 
don't know which one).  In order to ensure this resistor cannot affect the 
input voltage, you need to make sure the input is always connected to the 
voltage you want through a smaller resistor (electricity likes the lowest 
resistance path!).

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scott wrote:
>> Wait a sec - so you're saying there are devices which actually violate 
>> Ohm's law?
> 
> Yes, more exactly, there are devices which do not have constant 
> resistance/impedance.

Are you saying the resistence doesn't determine the current? Or are you 
just saying that the current can cause the resistence to change?

>> Well, you're the one with the engineering degree, but it *still* 
>> doesn't make any sense to me...
> 
> Maybe it would help if you imagined that inside the IC there is a big 
> resistor (eg 1Mohm) connecting the input to one of the supply rails (but 
> you don't know which one).  In order to ensure this resistor cannot 
> affect the input voltage, you need to make sure the input is always 
> connected to the voltage you want through a smaller resistor 

Hmm, OK.

> (electricity likes the lowest resistance path!).

Now, see, I always thought electricity just takes *every* path, with the 
current being determined by Ohm's law.

So, if you have a high-resistence path from A to B, and then you add a 
second, lower-resistence path from A to B, does the current flowing 
through the first path change? Or does it remain the same?

(Clearly the *potential* at A and B will be changed - there's more 
current flowing now...)

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