Chip with simple program for Toy

[Bob Eldred]

"Don Bruder" <dakidd@sonic.net> wrote in message
news:YeT3e.13180$m31.131165@typhoon.sonic.net...

In article <424ffb65$0$43995$14726298@news.sunsite.dk>,
"Roger Johansson" <no-email@no.invalid> wrote:

Well, the signal is actually an electromagnetic wave already when it
travels through the antenna wire. The antenna is just a bridge between
electromagnetic waves in conductors and electromagnetic waves in space.

Sounds like a variation on the explanation a Ham operator once gave me -
Boiled down from a conversation that lasted about half an hour, I came
away with the understanding that a transmitter and receiver operate in a
similar-in-result, although very different in actual mechanism, way to a
physically huge, ultra-tiny-rating capacitor, with each set's antenna
acting as one plate of the cap, and the "universe between them" as the
dielectric. It makes sense on one level, but on others, it kinda falls
apart. I've still never managed to wrap my head around the details of
"why" radioo works, even though I like to think I have a reasonable
grasp on the "how" part.

--
Don Bruder - dakidd@sonic.net - .html> for full details.

"a transmitter and receiver operate in a similar-in-result, although very
different in actual mechanism, way to a physically huge, ultra-tiny-rating
capacitor, with each set's antenna
acting as one plate of the cap, and the "universe between them" as the
dielectric."

Wrong, Wrong, Wrong! Radio energy is electomagnetic energy like light, just
lower in frequency. Physics tells us that any charged particle will radiate
electomagnetic energy when it accelerates or decellerates. Electrons in a
wire or antenna moving at high frequencies are under acceleration by their
motion they therfore radiate EMR. Likewise a EMR field will cause a current
in a wire of its same frequency. You can't get away from it. It's a matter
of physical principles. Look up Maxwell, Maxwell's equations and Hertz.
These guys figured this stuff out in the 19th century. It does not require a
return circuit, capacitance between antennas, eather or any other such
things, just empty space.
Bob

 

[Andrew Howard]

Thanks for all the responses! Now I have the general idea, so I can find
more info on my own.

Thanks,
Andrew Howard

 

[Bob]

<grug2005@hotmail.com> wrote in message
news:1112588882.530241.32760@l41g2000cwc.googlegroups.com...

Hi guys,

I'm a CS major, but an EE major at heart

I know how to build circuits, I use SPICE, but I am now trying to
figure out the more physical properties of devices.

As for capacitors, I know how to charge them, discharge them, use them
for timing circuits, etc.

However, I'd like to know what it actually means at the physical level.

For example:

Let's take a Cap rated at 35V, 100uf.

Let's apply a voltage source to it, say 10V - in other words, charge it
up

Taking another Cap that is exactly the same, and charge it with 20V.

Now, let's take a Cap rated at 35V, but has 1000uf capacity.

We charge with 10V as before to fully charge.

What I would like to know now is what is the difference between all
three Caps at the physical property level?

For example, the two 100uf caps are charged with different voltage
levels. What does that mean exactly? Does the one charged with 20V
have more electrons stored than the one charged with 10V? Or are they
at different energy levels? What gives?

The one with 1000uf certainly has more 'charge', but again, does the
mean more electrons?

Please help shed some light on this.

Thanks,
Grug

Grug,

The energy (E) stored in a capacitor is:

E=1/2 * C * V^2

where C is the capacitance (in Farads)
V is the voltage

You can do the math for yourself to see which cap has more or less energy
stored in it.


One thing to remember is that a capacitor always has zero net charge in it.
I know this seems strange, but it's true. The deletion of every electron
(from the more positive plate(s) of the capacitor) means an addition of
another electron on the other plate(s) (its more negative side). This
"separation" of charge is due to the transfer of energy from the external
source (in your case, your power supply). The more the charge is separated
then the more voltage there is across it (and vice versa).

Bob

 

[Michael Redmann]

kell schrieb:

I was experimenting driving various mosfets and igbts with a NE555 at
a low freq around one per sec turning a small bulb on and off just to
check how well the 555 works before I try to build the actual circuit
I have in mind, which will have an inductor operating at a few KHz in
a topology similar to a boost converter. (The inductor is intended to
reach a peak current of several amps.) I used no gate resistor, just
connected pin 3 directly to the gate.
Anyway, when I tried the IRF740 and a IGBT designed for car ignitions
(HGTP14N...), which both come in the TO-220 case, the bulb turned on
and off cleanly. But when I tried a couple of devices that come in
the big TO-247 case, the bulb didn't turn off cleanly. With the
IRG4PC30F IGBT, the bulb dimmed before it turned off. With the
STW18NB40 mosfet, the bulb didn't even turn off, it just flickered a
little.
I used a sealed lead acid battery and connected the Vcc pin of the
555 through a 75 ohm resistor to B+. I had a 100uF electrolytic
across the power pins.
I was powering the 555 through a resistor because that is the way I
intend to use it in the final circuit, as a way of protecting it from
the inductive spiking.
What is it about these big mosfets and IGBTs, that a 555 can't turn
them off?

Perhaps the 555 does not really switch off. What's the voltage at pin 3
when load should be off? The 555 can sink about 100 mA so the gate
should be grounded very well. What else do you have connected to the
555? (pins 2,5,6,7). Maybe something is ringing or even oszillating
which prevents the 555 from switching off correctly.

Regards
--
Michael Redmann
"It's life, Jim, but not as we know it." (Spock)

 

[Rheilly Phoull]

"padmow" <padmow_69@yahoo.com> wrote in message
news:1112614548.573885.12320@l41g2000cwc.googlegroups.com...

I would like to have opinion about Solid State Harddisk.
It would be great if my notebook computer is using this.

thank you
Padmow

Yes it would be great, I hope your notebook does use one.

--
Regards ..... Rheilly Phoull

 

[Bob Masta]

On 3 Apr 2005 14:12:23 -0700, ymustuknow@gmail.com wrote:

I am wondering if anybody knows of any power source(s) that produces or
contains LOW VOLTAGE and HIGH AMPERES?

One approach I've read about is to take the power transformer
from a microwave oven, saw off the secondary winding, and
wind on several turns of heavy-gage wire instead. People
use this to build welders; Google on microwave + welder
for more info.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

 

[Larry Brasfield]

"PeteS" <ps@fleetwoodmobile.com> wrote in message news:1112617041.370157.209520@g14g2000cwa.googlegroups.com...

The efficiency will eventually go down for a number of reasons.

1. Copper losses will increase due to P = I^2R. Increase the current
and you increase the losses in the windings (notably by a square
factor).

2. The input and output windings will heat up, thereby increasing their
resistance (copper has a positive temperature coefficient). With a
higher resistance, you exacerbate the problem in (1)

Agree with 1 and 2.

3. Core and eddy current losses in the magnetic core will increase.
Increased output current (which implies increased input current) will
increase the magnetic flux density. As you increase it, the losses in
the core will increase, up to magnetic saturation, where you can
effectively get no more current (the maximum energy through a
transformer is limited by the magnetics as well as the winding limits).

It is not true that magnetizing losses increase with
output current. They actually go down a little.
This is because the increased IR drop in the primary
reduces the amount of flux change necessary in the
core to provide enough induced voltage to equal the
applied primary voltage adjusted by the IR drop.

For magnetization, you can think of the primary as a
more or less pure inductor [1] in series with the primary
resistance and in parallel with some real impedance
representing the eddy current, hysteresis, (and radiated)
losses. As the voltage across that inductor drops, so do
its losses.

[1. The inductor is usually non-linear, but its inductance
is a monotonic function of current.]

....
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Rheilly Phoull]

"Ron Hubbard" <notat@hotmail.com> wrote in message
news:1152fhm95rpg380@corp.supernews.com...

I just soldered in parallel a couple of leads to the speaker in my clock
radio so I can tape the music without worrying about background noise.
But connected to the remote jack, there's no sound at all while
connected to the mic jack I get loud and distorted sound.

What should I do; place a small cap in line or use a 8ohm/1K audio
transducer and connect it to the mic input?

-rh

Try a 100ohm pot across the leads from the speakers, adjusted to the level
you need.

--
Regards ..... Rheilly Phoull

 

[mike]

padmow wrote:

I would like to have opinion about Solid State Harddisk.
It would be great if my notebook computer is using this.

thank you
Padmow

I've got a 10MB solid state hard drive in a 2.5" IDE format.
Make an offer.
mike

--
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
..
Wanted, PCMCIA SCSI Card for HP m820 CDRW.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Wanted, 12.1" LCD for Gateway Solo 5300. Samsung LT121SU-121
Wanted 13" LCD for Mitac 6133 Samsung HT13X13-201
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/

 

[Robert Monsen]

kell wrote:

I was experimenting driving various mosfets and igbts with a NE555 at a
low freq around one per sec turning a small bulb on and off just to
check how well the 555 works before I try to build the actual circuit I
have in mind, which will have an inductor operating at a few KHz in a
topology similar to a boost converter. (The inductor is intended to
reach a peak current of several amps.) I used no gate resistor, just
connected pin 3 directly to the gate.
Anyway, when I tried the IRF740 and a IGBT designed for car ignitions
(HGTP14N...), which both come in the TO-220 case, the bulb turned on
and off cleanly. But when I tried a couple of devices that come in the
big TO-247 case, the bulb didn't turn off cleanly. With the IRG4PC30F
IGBT, the bulb dimmed before it turned off. With the STW18NB40 mosfet,
the bulb didn't even turn off, it just flickered a little.
I used a sealed lead acid battery and connected the Vcc pin of the 555
through a 75 ohm resistor to B+. I had a 100uF electrolytic across the
power pins.
I was powering the 555 through a resistor because that is the way I
intend to use it in the final circuit, as a way of protecting it from
the inductive spiking.
What is it about these big mosfets and IGBTs, that a 555 can't turn
them off?

The 555 should really be able to shut them off. Test the MOSFETs that
fail to turn off by disconnecting them from the 555, and grounding the
gate. Does it turn off?

The gate threshold can be as high as 5V. It may take 10V Vgs to turn it
on completely.

Some mosfets may take even more voltage to turn completely on. The
IRF740 has a slightly lower Vgs(th) than the other.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Fritz Schlunder]

"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message
news:9eb4e.6$Sd3.222@news.uswest.net...

"PeteS" <ps@fleetwoodmobile.com> wrote in message
news:1112617041.370157.209520@g14g2000cwa.googlegroups.com...
The efficiency will eventually go down for a number of reasons.

1. Copper losses will increase due to P = I^2R. Increase the current
and you increase the losses in the windings (notably by a square
factor).

2. The input and output windings will heat up, thereby increasing their
resistance (copper has a positive temperature coefficient). With a
higher resistance, you exacerbate the problem in (1)

Agree with 1 and 2.

Also agree with 1 and 2. Item 1 is very real and makes a very tangible and
signicant impact and should not be ignored. The influence of item 2 however
is usually quite small for a normal transformer operated with a normal
temperature range. Since most loss and thermal rise calculations are
somewhat approximate anyway, item #2 can often be ignored.

3. Core and eddy current losses in the magnetic core will increase.
Increased output current (which implies increased input current) will
increase the magnetic flux density. As you increase it, the losses in
the core will increase, up to magnetic saturation, where you can
effectively get no more current (the maximum energy through a
transformer is limited by the magnetics as well as the winding limits).

It is not true that magnetizing losses increase with
output current. They actually go down a little.
This is because the increased IR drop in the primary
reduces the amount of flux change necessary in the
core to provide enough induced voltage to equal the
applied primary voltage adjusted by the IR drop.

Agreed.

For magnetization, you can think of the primary as a
more or less pure inductor [1] in series with the primary
resistance and in parallel with some real impedance
representing the eddy current, hysteresis, (and radiated)
losses. As the voltage across that inductor drops, so do
its losses.

[1. The inductor is usually non-linear, but its inductance
is a monotonic function of current.]

I suppose that is one way to think about it. I think of it a little
differently. Ampere's Law would have you believe the magnetic flux density
B is proportional to the number of turns times the current flowing through
those turns in any given magnetic device. Since a transformer is a magnetic
device, it would seem logical that as the output current increases the
magnetic flux density in the core also increases. This would suggest at
some current level the transformer's core would saturate.

This is not the case however for a regular transformer (IE: one not a
flyback transformer, they are different). What one must realize is that a
transformer has two or more independent windings on a single core. Ampere's
Law applies to the primary winding, and it also applies for the secondary.
As the load current on the secondary increases you would tend to get more
flux generated in the core. However, as the primary current increases to
supply that secondary current, the primary winding also generates it's own
flux. If you studiously apply the right hand rule, you will find that the
flux will be in different directions for primary and secondary, and so they
both serve to cancel each other out. As a consequence the flux density in
the core is essentially independent of the load current of the transformer.
In effect the maximum power output rating of a transformer is limited by the
winding resistances, or by total thermal dissipation limits resulting in a
given temperature rise. Practical transformers are thermal dissipation
limited long before winding resistance limited. In theory a transformer
made with superconducting windings could be made very small and output an
outrageously huge amount of power (efficiently at that).

Flyback transformers are different, and have properties more like inductors
than do regular transformers. In the flyback transformer current does not
normally flow simultaneously through primary and secondary windings. At any
given time only one of them conducts. As a consequence you don't get the
flux canceling effect mentioned above for regular transformers. If you keep
increasing the load on a flyback transformer it will eventually saturate the
core.

As for the OP's original question, does the efficiency improve with
increasing or decreasing load current... Obviously at zero output current
the efficiency is zero since any transformer will waste some idle power
primarily due to core hysteresis and eddy current loss. As you apply a
heavier and heavier load the efficiency continues to improve, up to a point.
At some point the I^2*R loss effect will start to dominate and thus the
efficiency will start to decrease again. In practice, where this peak of
efficiency occurs depends on the design of the transformer. Typical
transformers will often be designed to have maximum efficiency somewhat near
(though often a little below) their maximum rated continuous output current.
The efficiency peak is relatively broad.

 

[Bob Eldred]

<charles.macleod@gmail.com> wrote in message
news:1112615498.839444.320040@z14g2000cwz.googlegroups.com...

In a standard (low voltage eg 40V)transformer - same number of input
turns as output turns, should the effiency go up with increasing the
output current or down?

All transformers consume magnetizing current to energize the core. Some of
this is reactive while some is resistive. The resistive portion involve
copper and core losses, mostly core when the current is low. These losses
occur whether there is a load on the transformer or not. If there is no load
on the transformer, no output, there is still magnetizing loses so the
efficiency is zero. Pout/Pin = 0/(small number) = 0. As you increase the
load, the output becomes higher and the copper losses also become higher but
the efficiency increases because there is now an output. The efficiency
continues to increase with load until it reaches some maximum where the
transformer is most efficient. Once that point is reached, the efficiency
will decrease with increasing load because heating will exacerbate copper
losses (higher resistance with temperature). Any transformer can give
several times it rated output current for short durations. Heat is what
limits it. The flux density in the core and the related magnetic losses are
NOT a function of load. They are a function of the primary voltage and not
load, by faraday's law. The secondary load current balances and is in the
opposite direction of the primary current. The flux density stays the same.
When the primary goes positive, current flows into the primary, by
convention. The same polarity winding on the secondary also goes positive
but the current flows out of that winding. The two currents times their
number of turns (amp-turns) balance, one in the other out. This is reversed
every half cycle. That balance means that there is no net flux density
caused by the load.

To sum it up, the efficiency is maximum when the load is at some rated value
usually at or near the nameplate values and is zero with no load and becomes
low again when the transformer is smoking.
Bob

 

[Active8]

On Mon, 4 Apr 2005 06:17:43 -0700, Ron Hubbard wrote:

I just soldered in parallel a couple of leads to the speaker in my clock
radio so I can tape the music without worrying about background noise.
But connected to the remote jack, there's no sound at all while
connected to the mic jack I get loud and distorted sound.

What should I do; place a small cap in line or use a 8ohm/1K audio
transducer and connect it to the mic input?
^^^^^^^^^^^^

You mean transformer? Use the pot suggestion.

The remote jack is supposed to connect to a mic switch to turn the
tape transport on and off.
--
Best Regards,
Mike

 

[Tom Biasi]

"Ron Hubbard" <notat@hotmail.com> wrote in message
news:1152fhm95rpg380@corp.supernews.com...

I just soldered in parallel a couple of leads to the speaker in my clock
radio so I can tape the music without worrying about background noise.
But connected to the remote jack, there's no sound at all while
connected to the mic jack I get loud and distorted sound.

What should I do; place a small cap in line or use a 8ohm/1K audio
transducer and connect it to the mic input?

-rh


In addition to what has already been posted:

You may be able to get suitable audio directly from the radio's volume
control.
Tom

 

[Ron Hubbard]

"Active8" <reply2group@ndbbm.net> wrote in message
news:d2rgj8.3qo.1@active8.fqdn.th-h.de...

On Mon, 4 Apr 2005 06:17:43 -0700, Ron Hubbard wrote:

I just soldered in parallel a couple of leads to the speaker in my
clock
radio so I can tape the music without worrying about background
noise.
But connected to the remote jack, there's no sound at all while
connected to the mic jack I get loud and distorted sound.

What should I do; place a small cap in line or use a 8ohm/1K audio
transducer and connect it to the mic input?
^^^^^^^^^^^^
You mean transformer?

Yeah, I did mean transformer. It's funny the things you have on your
mind when you type at 3:00 in the morning....

Ron

 

[Michael A. Terrell]

padmow wrote:

I would like to have opinion about Solid State Harddisk.
It would be great if my notebook computer is using this.

thank you
Padmow

Disk-on-chip from M-Systems:

http://www.m-systems.com/Content/Products/Product.asp?pid=2

http://www.m-systems.com/

You better have a lot money if you want to buy anything from them.

--
Former professional electron wrangler.

Michael A. Terrell
Central Florida

 

[Lord Garth]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:425207F2.CA7C7961@earthlink.net...

padmow wrote:

I would like to have opinion about Solid State Harddisk.
It would be great if my notebook computer is using this.

thank you
Padmow


Disk-on-chip from M-Systems:

http://www.m-systems.com/Content/Products/Product.asp?pid=2

http://www.m-systems.com/

You better have a lot money if you want to buy anything from them.

I have two 16MB M-Disk ICs plugged into two ICOP PC-104 cards.
They work very well on these essentially 386-40 computers. I've
even put Win 3.11 on them with a ram disk but later returned to a
DOS only environment.

What I want to know is if any of you guys ever connected a CF card
to the IDE port and used DOS to format it into a proper HD structure?
The result is easily loads more storage for the dollar than an M-Disk.
Note the M-Disk is in a 600 mil DIP and plugs in as any JEDEC byte wide
device would. The BIOS on the ICOP is set to boot this socket so
having a CF on the IDE port means no special BIOS options are needed.

 

[Lord Garth]

"Eric R Snow" <etpm@whidbey.com> wrote in message
news:fci351h3ba686hujfbb4spdkmaq6hs42mb@4ax.com...

On Mon, 04 Apr 2005 19:00:29 GMT, "CWatters"
colin.watters@pandoraBOX.be> wrote:


"KILOWATT" <kilowatt"nospam"@softhome.net> wrote in message
news:Ew34e.3293$Fy3.280205@news20.bellglobal.com...
Hi everyones...thanks for your time. I wish to make a large ozone
generator
and i'm wondering if one of those two methods (see subject line)
produces
less byproduct (like nitrous oxyde) than the other ? Maybe i'm wrong
and
pure oxygen MUST be used to not have any byproducts...not shure. Any
comments? TIA for any replies.

Just be aware that ozone is a lung irritant and levels used to clean air
(for example) are a hazard to health. Very bad for you if you have asthma
etc. I believe it also effects some rubber and plastic insulation.

Ozone is indeed a major cause of rubber deterioration. Latex is
especially affected.
ERS

Makes me wonder what sort of rubber is used on TV remotes...
I assumed it was silicone rubber but there is usually an oily
residue under them, probably from deterioration of the rubber.

 

[me]

Anyway, when I tried the IRF740 and a IGBT designed for car ignitions
(HGTP14N...), which both come in the TO-220 case, the bulb turned on
and off cleanly. But when I tried a couple of devices that come in
the big TO-247 case, the bulb didn't turn off cleanly. With the
IRG4PC30F IGBT, the bulb dimmed before it turned off. With the
STW18NB40 mosfet, the bulb didn't even turn off, it just flickered a
little.
I used a sealed lead acid battery and connected the Vcc pin of the
555 through a 75 ohm resistor to B+. I had a 100uF electrolytic
across the power pins.
I was powering the 555 through a resistor because that is the way I
intend to use it in the final circuit, as a way of protecting it from
the inductive spiking.
What is it about these big mosfets and IGBTs, that a 555 can't turn
them off?

Could be the gate capacitance, try a low value resistor (maybe 200 ohm)
from gate to ground in the circuit...

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[Michael Redmann]

PeteS schrieb:

For an interesting experiment, charge two capacitors in parallel,
disconnect the supply and reconnect them in series. if the caps were
equal, you have a voltage doubler (because you now have half the
capacitance). Then do it with one cap 10 times the capacitance of the
other. Measure the new voltage. You'll be quite surprised. This trick
is used in Parametric amplifiers (common in Radars, for example).

Remarkable! If my computation is right, the factor is 11/sqrt(10). So
the total voltage will be 34.8 volts, which is more than the sum of
every single voltage. I'm ashamed but up to now I thought resonance is
the only effect to get voltages up.

Regards
--
Michael Redmann
"It's life, Jim, but not as we know it." (Spock)