Chip with simple program for Toy

[John Fields]

On 7 Mar 2006 05:49:45 -0800, "Chris" <cfoley1064@yahoo.com> wrote:

John Fields wrote:
On Tue, 7 Mar 2006 00:38:28 -0600, Jamie <jamie@gnulife.org> wrote:




I have a question about wire gauge ratings. Why do wire gauge charts
show how many amps that the wire is able to carry, and not the wattage
that the wire can dissipate safely? Resistors are rated in watts, not
amps.

---
OK, lets say that you've got a 100 foot extension cord and it's
rated to safely dissipate 10 watts. What are you going to do with
that information? More importantly, how do you know what you can
plug into it safely?

On the other hand, if it's rated to safely carry 10 amps all you
have to do is look at the nameplate of what you're going to plug
into it to know how much current the extension cord will have to
carry.
---


I would think that if I had a length of wire that was just barely
capable of conducting a charge at the rate of 10 amps at 12 volts, that if
I were to hook up 120 volts to the same piece of wire and attempted to run
THAT charge at the rate of 10 amps, that, since P=I*E, the wire would be
dissipating heat much more rapidly and could overheat.

---
You forget that the wire isn't supposed to be a significant portion
of the load. Using our 100 foot extension cord as an example, if it
can safely dissipate 10 watts with 10 amps going through it, then
the voltage dropped across it will be:

P 10W
E = --- = ----- = 1V,
I 10A

with the balance being dropped across the load, and its
(the extension cord's) resistance will be:


E 1V
R = --- = ---- = 1 ohm,
I 1A



and the circuit will look like this:


E1
|
[R1]
|
+----E2
|
[R2]
|
GND

Where E1 is the supply voltage, E2 is the voltage appearing across
the load, R1 is the resistance of the extension cord, and R2 is the
resistance of the load.


So, let's take a look at what actually happens with a load hooked up
to our extension cord.

First, if we have a 120V supply feeding a load which draws 10 amps,
the load will look like:


E 120V
R --- = ----- = 12 ohms
I 10A

and our circuit will look like:


120V
|
[R1] 1R
|
+----E2
|
[R2] 12R
|
GND

Now, since the extension cord and the load are in series, that's a
total of 13 ohms, so the current the extension cord will have to
carry will be:

E 120V
I = --- = ------ ~ 9.2 amperes
R 13R

and instead of 120V appearing across the load, there'll be about 119
there because of the ~ 1V drop in the cord.

On the other hand, if you wanted to treat the extension cord like a
resistor, all you'd have to do would be to short out the socket end
of it and plug it in.

Since the cord looks about like an ohm, when you first plugged it in
you'd get:

E 120V
I = --- = ------ = 120 amperes
R 1R

through it, and the power it would be dissipating would be:


P = IE = 120A * 120V = 14,400 watts!



Why isn't voltage taken into account in determining the right diameter
for a conductor in a circuit?

---
Because it's largely irrelevant since the bulk of the voltage is
supposed to appear across the load. The diameter is important
because that, in conjuction with the length of the wire will
determine its resistance and that, in turn, will determine how hot
the wire gets with a specified current running through it.

--
John Fields
Professional Circuit Designer

Good morning, Mr. Fields. If I might ask the master if he'd like a
second cup of coffee?

E 1V
R = --- = ---- = .1 ohm,
I 10A

and then through to:

E 120V
I = --- = ------ = 1200 amperes
R .1R

and

P = IE = 1200A * 120V = 144,000 watts!!!

(Extra exclamation points added -- it only makes your point to the OP
more relevant.)

I never post before the second cup of coffee anymore. I used to do
this almost every morning.

Cream or sugar, sir?

---
Neither, thanks, but now that you've roused me from my turpor, I
think I'll go pour that second cup, LOL!

--
John Fields
Professional Circuit Designer

 

What more can you tell me about a switching regulator? What would it
take to use one of those instead of the LM317T?

 

[w2aew]

The document is wrong, on two counts! The equation for Fc in a simple
RC circuit is 1/2piRC as you've stated (they left out the 1/ part, or a
(-1) exponent). Also, they calculated wrong - it should be 11.6KHz,
not 1.1KHz.

 

[BobG]

I found a spec that this thermopile is 2.89 ohms, so that is like the
source impedance right?

 

[Bill Bowden]

tysonclugy@gmail.com wrote:

What more can you tell me about a switching regulator? What would it
take to use one of those instead of the LM317T?

Switching regulators make a lot of noise. You might get better sound
with the LM317. The battery life will be a about 1/2 as much compared
to a switching regulator since the input is about twice the output.

Basically, a switching regulator charges a capacitor very quickly to
the desired output voltage and then waits for the capacitor voltage to
fall a small amount and then blast it again with a pulse of current to
bring the voltage back up. It sort of switches on and off as necessary
to keep the output fairly constant. But it won't be as smooth as the
linear LM317 and so needs more filtering on the output to get rid of
the noise and small variations in output voltage.

-Bill

 

[Phil Allison]

"w2aew"

The document is wrong, on two counts! The equation for Fc in a simple
RC circuit is 1/2piRC as you've stated (they left out the 1/ part, or a
(-1) exponent). Also, they calculated wrong - it should be 11.6KHz,
not 1.1KHz.

** Not to mention the danger and stupidity of connecting the chassis ground
of an Spectrum Analyser to the supply neutral.

I note the article was written by two frogs.

Beware - they are really toads.




......... Phil

 

[JimW52]

eden wrote:

I posted about this circuit
[http://www.captain.at/electronics/atmel-programmer/atmega16-programmer.png]
some months ago. Back then I thought I cleared all mysteries, but it
seems I still have some problems.

I have put together all parts, but using uisp program I can't see the
ATmega16 chip, so I guess I must have wired something wrong.

First, I realize that I used wrong capacitors for C3 and C4, instead
100n I took 10n. Could this be the problem? If I measure the voltage
after IC2 it seems correct.

Second, I wonder now, do I have to put some GNDs together, like the one
near C1 and C2 with the one near C3? And maybe also the one coming out
from pin 11 from IC1?

When I asked first time about the circuit, I was confused by the 3 VCC
symbols, and then most agreed that they all should be joint, and I did
that?

Do somebody see some catch here, what could I be doing wrong?
Thanks for any help.
Goran

You need to connect all grounds together.

Jim

 

[Ant_Magma]

D**n!

The blind leading the blind.

I wonder how many papers and ANs i have that has such mistakes. Don't
they ever check before they publish?

 

[eden]

JimW52 wrote:

eden wrote:
I posted about this circuit
[http://www.captain.at/electronics/atmel-programmer/atmega16-programmer.png]
...

You need to connect all grounds together.

Jim

By all you mean even the one coming after the LED? And maybe even the
two from the Parallel port?

Goran

 

[Joel Kolstad]

"Charles Schuler" <charleschuler@comcast.net> wrote in message
news:wpSdnVi8VOPRW5HZRVn-jw@comcast.com...

This site will attempt to install files on your computer!

Only tried to open up a PDF file on me... seems benign enough?

 

[Charles Schuler]

"rLet" <honey_arlet24@yahoo.com> wrote in message
news:1141693243.566737.304480@j33g2000cwa.googlegroups.com...

i need to design a small signal amplifier in a voltage divider biasing,
can anyone help??? pls???

http://www.montagar.com/~PATJ/electrowares.htm

Down about four lines is a free program for common emitter design. I have
no idea if it is of any value. You can try it and you can Google for
others.

 

[Charles Schuler]

Only tried to open up a PDF file on me... seems benign enough?

You could be right.

I have learned, however, to avoid sites of this type.

I got a different warning, by the way.

My computer has various condoms fitted over its silicon head.

Better safe than sorry is a good way to go these days.

 

[JimW52]

eden wrote:

JimW52 wrote:
eden wrote:
I posted about this circuit
[http://www.captain.at/electronics/atmel-programmer/atmega16-programmer.png]
...

You need to connect all grounds together.

Jim

By all you mean even the one coming after the LED? And maybe even the
two from the Parallel port?

Goran

Yes, ALL gnds must be joined together, also all vccs should be joined

together. If you think about it,. how would say the LED work if its
ground connection was disconnected. There would be no current flow and
so no light.

Note that VCC_OUT needs to be connected to the points marked VCC. This
is obviously an error in the diagram. Also AVCC needs to connect to VCC
and AGND needs to connect to GND.

For better programming options have a look at
http://www.lancos.com/prog.html

HTH

Jim

 

[eden]

JimW52 wrote:

eden wrote:
JimW52 wrote:
eden wrote:
I posted about this circuit
[http://www.captain.at/electronics/atmel-programmer/atmega16-programmer.png]
...

You need to connect all grounds together.

Jim

By all you mean even the one coming after the LED? And maybe even the
two from the Parallel port?

Goran

Yes, ALL gnds must be joined together, also all vccs should be joined
together. If you think about it,. how would say the LED work if its
ground connection was disconnected. There would be no current flow and
so no light.

Note that VCC_OUT needs to be connected to the points marked VCC. This
is obviously an error in the diagram. Also AVCC needs to connect to VCC
and AGND needs to connect to GND.

For better programming options have a look at
http://www.lancos.com/prog.html

HTH

Jim

Thank you very much Jim. I am still quite new and some things are still
confusing me.
Of course, if I can't make it work I will look for something like
ponyprog, but I wanted to try this by my self, like learning curve for
both electronics and embedded programming.

Thank you once more,
Goran

 

[ehsjr]

tysonclugy@gmail.com wrote:

I got the LM317T from Radio Shack and I am trying to wire up an
external battery pack to run an mp4 player which has an internal 3.7v
battery.

Can someone explain to me (in plain english, I'm new with this stuff)
what I need and how to adjust this thing?
I want to run this mp4 player on a rechargable 9.6v battery that I
have.

Thanks in advance.

Here's a "picture".


+-----------------------> Gnd To MP3 (-)
|
------ | ---------
+---|470ohm|---+---|-BATTERY+|---+
| ------ --------- |
| |
| ------ |
| | o | |
| |______| |
| /______/| |
| | || |
| |LM317 |/ |
| ------ |
| | | | |
+----------+ | +---------------+
| |
| |
| ------ |
+---|240ohm|--+------------------------> 3.7V To MP3 +
------

With the 240 and 470 resistors shown, you'll be within
..003 volts of target (3.7) per the computation. You do
not need any capacitors if you keep the LM317 relatively
close to the battery.

Ed

 

[ehsjr]

Jamie wrote:

I have a question about wire gauge ratings. Why do wire gauge charts
show how many amps that the wire is able to carry, and not the wattage
that the wire can dissipate safely? Resistors are rated in watts, not
amps.

I would think that if I had a length of wire that was just barely
capable of conducting a charge at the rate of 10 amps at 12 volts, that if
I were to hook up 120 volts to the same piece of wire and attempted to run
THAT charge at the rate of 10 amps, that, since P=I*E, the wire would be
dissipating heat much more rapidly and could overheat.

Why isn't voltage taken into account in determining the right diameter
for a conductor in a circuit?



- Jamie

P = I*E which is also equal to I^2*R

In the math, voltage is automatically taken into account:
E = I*R. Substitute I*R for E in your power equation and you get
P = I*(I*R), which is more commonly written P=I^2*R. That means
power is equal to current squared times resistance.

The neat thing about this is that they don't need to
know the source voltage when making a wire table.
They know the resistance of the wire and can determine
how many watts will be produced in it for a given
current through it. They know how many watts the wire
can safely dissipate, so it is simple to rate the
wire in amps.

Ed

 

[Ryan]

Why are ordinary DC power supplies so complicated? What's going on in
there that seems to take so many components and makes them so big and

All electronics is complex. That's why it's fun, but also why it's
hard to learn and understand. I could ask you "why is a car engine so
complicated?"

Then I would explain part of the complexity. There is a lubrication
system, and this needs a pump, and to be cooled. There is a cooling
system, which needs to circulate and needs a pump and a whole lot more
cooling. There is a timing system, which must be precise and many
things exist to make this happen. And so forth.


If I knew the equivalent for a power supply, I'd post about that.

 

[John Larkin]

On Fri, 10 Mar 2006 01:12:44 GMT, Ryan <quakeserver149@yahoo.com>
wrote:

Why are ordinary DC power supplies so complicated? What's going on in
there that seems to take so many components and makes them so big and


All electronics is complex. That's why it's fun, but also why it's
hard to learn and understand. I could ask you "why is a car engine so
complicated?"


Then I would explain part of the complexity. There is a lubrication
system, and this needs a pump, and to be cooled. There is a cooling
system, which needs to circulate and needs a pump and a whole lot more
cooling. There is a timing system, which must be precise and many
things exist to make this happen. And so forth.


If I knew the equivalent for a power supply, I'd post about that.

For a simple linear supply, there's

AC inlet, fuse, switch, EMI filtering

Transformer for isolation and voltage stepdown

Rectifier for ac-to-dc

Filter capacitors

Regulator, an IC (complex inside) or a discrete circuit with
reference, error amp, pass element, current limiter

Output capacitor

Bleeder resistor

Maybe remote sense circuits

Maybe overvoltage crowbar

Heatsinks, maybe a fan

PC board, metal chassis, in/out terminals

Switchers are more complex.

John

 

[Joel Kolstad]

"Ryan" <quakeserver149@yahoo.com> wrote in message
news:go4Qf.811083$x96.491817@attbi_s72...

If I knew the equivalent for a power supply, I'd post about that.

There is a good comparison between engines and power supplies in that
"environmental friendliness" drives some of the added complexity -- all the
EGR components, catalytic converter stuff, etc. in an engine syste, all the
power factor correction stuff in a power supply.

 

[Ant_Magma]

I'm building a powerline modem using Belfuse's powerline module:
http://www.belfuse.com/Data/DBObject/0804_5000_03_04.pdf

I've requested for reference design and they say no reference design
are available.

My question:
I have realized with the reference design of my ethernet transceiver
that most of the pins (unless specifically stated) have a 50ohm
resistor (i'm assuming its a terminating resistor?).

Since without a reference design, i'm guessing i just connect the pins
of the powerline module directly to the corresponding pins at my
ethernet transceiver. My question is, do i have to place any resistors
btwn these connections too? Is there like a conventional rule of thumb
etc when dealing with these kind of situations?