Chip with simple program for Toy

[Phil Allison]

"Puppet_Sock"

The situation where DC transmission is better is where the
transmission line is long compared to a 1/4 wavelength of
the AC transmission frequency. The typical transmission is
60 cps. So 3E8 m/s divided by 60/s, then divided by 4, is
1,250 km. So when you get transmission distances in
that range, radiative loss starts to be significant, and it
gets tough to match the impedance of the line to the load.
Basically what you get is a very large dipole antenna.

Anyway, long distance transmission is sometimes done
on DC lines to reduce radiative loss.


** I think this is entirely false.

There is no comparison with a 60 Hz dipole antenna.

AC power transmission is done with 3-phases ( spaced by exactly 120
degrees ) running a few metres apart, all in exact parallel across the
country - so the net 60Hz eclectic field at a distance always cancels to
ZERO !!.

Ergo, there simply is no "radiation loss" .

The real losses encountered are due to resistive heating, voltage drop and
phase angle changes on very long lines.

http://en.wikipedia.org/wiki/Electric_power_transmission#Limitations



...... Phil

 

[Dan Coby]

<lionelgreenstreet@gmail.com> wrote in message
news:abcb5809-a004-4fa6-a744-44f0512dd88c@a23g2000hsc.googlegroups.com...

Set Rs to 0 and then do your calculations.

ok, i tried but my procerure is wrong....Can you say me how obtain
Pout?

Yes, I also get a different result.

If Rs is 0 then it is a simple short. Note: The problem stated that Rs is much
less that RL. To really assume that Rs is negligible then we also need to
assume that Rs is less than 1 / (w * CL) for range of frequencies that are being
considered. At very high frequencies, this may not be true.

Then the circuit reduces to a simple parallel combination of the current source
Ip(w) with Rj, RL, Cj, and CL. Define Req = Rj * RL / (Rj + RL) and Ct = Cj = CL.
Then the impedance of the capacitors is Zc = 1 / s (Cj + CL).
(Where s is the Laplacian operator which is commonly treated as sqrt(-1) * w.)


The impedance of the parallel combination of Req and Zc is:

Zt = Req * Zc / (Req + Zc)

or Zt = Req / (Req/Zc + 1)

The voltage across this impedance (which is also the voltage across RL) is:

Vt = VL = Ip(w) * Zt

The power in RL is: Pout(w) = VL^2 / RL

or Pout(w) = Ip(w)^2 *Zt^2 / RL

It looks like they are simply using the magnitude of the impedance, i.e.

mag(Zt) = sqrt((Req^2 / (w * Ct))^2 / (Req^2 + 1 / (w * Ct)^2))

or mag(Zt) = sqrt(Req^2 / (Req^2 * (w * Ct)^2 + 1))

Using the magnitude of Zt we get the following for Pout:

or Pout(w) = Ip(w)^2 * Req^2 / (RL * (Req * (w * CT)^2 + 1)

Which differs from their equation by a factor of (Req^2 / RL)

I think that my result is correct. If Rj is infinity then the two equations are
equal. If Rj is zero then Ip is shorted and there should be no power in RL
however their equation gives Ip^2 * RL.

 

[Phil Allison]

<spamfree@spam.heaven>
Bret Cahill = TROLL

you came up with.

Never end sentences with prepositions.


Why?

** Cos it is a preposterous thing to do.....



...... Phil

 

[Claude]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:YqydncYrT6pQwJLVnZ2dnUVZ_j2dnZ2d@earthlink.com...

Claude wrote:

It's just for fun amongst friends, not to antagonize the general public!
Touchy touchy aren't we? I've done the spook the neighbour on the FM band
trick, it's time for the next fun packed spook.


It is illegal in every country in the world, and your little 'prank'
could cost someone their life.


--
http://improve-usenet.org/index.html


Use any search engine other than Google till they stop polluting USENET
with porn and junk commercial SPAM

If you have broadband, your ISP may have a NNTP news server included in
your account: http://www.usenettools.net/ISP.htm

Ok last post on the subject. They are NOT illegal everywhere, many
businesses and restaurants use them. I just don't know to how many uV/m they
are restricted.

What follows is in no way directed at others in this group who I find
helpful and friendly. I am just going to blow my steam valve off at this
particular responder

Aw crap I just saw the name on the reply, forget everything, I don't exist.
You are right the woman next door who stops mid stride in her driveway while
taking the garbage out to check her emails on her crackberry will surely
need an ambulance. Just to make sure all of the airwaves are safe I will
wrap tin foil around all of my electronic devices right now. I hope that
you are a minuscule exception to this groups usual civility.

Notwithstanding your possibly strange reaction I have been convinced to back
off the project.

Claude
Canada.

 

[Bob Eld]

"The Flavored Coffee Guy" <elgersmad@rock.com> wrote in message
news:cf4dadc5-7e3f-4731-a479-d24a866b4f05@h1g2000prh.googlegroups.com...

Did you know an effective electrostatic generator can produce plasma?
Well, if you get one of these electrostatic dusters, and place it
inside a 3 inch diameter PVC pipe you can generate quite the static
charge by allowing the air to flow through it as you drive. The PVC
should be 12 inches longer than the duster's fibers.

http://www.ettorecleaning.com/product-1385-Poly-(Static)-Duster.aspx

At that fine point, all you need is some aluminum foil and glue that
to the outside of the PVC pipe and strap it to the roof of your car or
truck. It may take 4 or 5 of these before your car will glow in the
dark or actually produce a plasma dense enough to absorb Police
Radar. In theory, it should work.

If not the alternative is a 300KV voltage muliplier and venting the
charge into the air as you drive. A much more complicated circuit.

I think it would work better if you took the aluminum foil, folded it into a
triangle shape then placed it on your head!

 

[Bob Eld]

"Boki" <boki.digital@gmail.com> wrote in message
news:9626a25e-24a6-4a7f-8d84-ba0d23c5fcba@1g2000prg.googlegroups.com...

Hi All,
What's the major difference between weak pull and strong pull ?

Does that means different bias voltage ?


Thanks.
Boki.

You come faster with a strong pull! Also faster helps.

 

[Bob Eld]

"Chris W" <1qazse4@cox.net> wrote in message
news:aYLPj.30338$KJ1.2115@newsfe19.lga...

The device listed at the link below automatically switch the output
between either a 12V (really 13.8V) power supply or a battery depending
of if the power supply is on. It says it "uses two 80 ampere Schottky
diodes connected as an OR-Gate to isolate the battery and power supply
from each other." My question is what is the or gate used for? I
understand to diodes to stop the Power supply from sucking power out of
the batteries when it is off, but have on idea what advantage an or gate
would give.


http://www.powerwerx.com/product.asp?ProdID=2047&CtgID=3574


--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

It is just two diodes in series with the two power sources, one from the
battery and one from the power supply. The cathodes of the diodes are
connected together forming the output. This way current flows from the
source with the highest voltage. Logically that is an "or" function. It is
one "or" the other. If the two sources are the same voltage, both would
supply current.

 

[Bob Eld]

<Derphy@gmail.com> wrote in message
news:70883867-70df-4d42-8e4e-a0804093b549@e39g2000hsf.googlegroups.com...

Before I go and start plugging things in I need to know this one
thing. I have a Philips VOIP841, the power adaptors that came with it
are 110VAC 60Hz only. I am now living in Belgium and of coarse cannot
plug my phone in without using the transformers. It seems easy enough,
and cheaper for me to just go out and purchase new AC to DC converters
for 220VAC 50Hz.

Original input: Proposed Input:
110VAC 60Hz 220VAC 50Hz
Original output: Proposed output:
7.5VDC 150mA ----handset charger------ 7.5VDC 300mA
7.5VDC 500mA ----base station----------- 7.5VDC 500mA

So here is my issue. I have two types of power outputs from my old AC/
DC converter. One is (7.5VDC 150mA) for the handset chargers, and the
other is (7.5VDC 500mA) for the base. I have found a universal AC/DC
220VAC converter for the 500mA base. But what I'd like to know is if
it's safe to use a (7.5VDC 300mA) converter on the handset charger
that originally requires (7.5VDC 150mA)?

Yes it is safe. You can ALWAYS use a power source with a higher current
rating than what you need. You cannot go the other way, however. That is you
cannot ues a 150mA device where you need 300mA.

 

[Bob Eld]

<Derphy@gmail.com> wrote in message
news:70883867-70df-4d42-8e4e-a0804093b549@e39g2000hsf.googlegroups.com...

Before I go and start plugging things in I need to know this one
thing. I have a Philips VOIP841, the power adaptors that came with it
are 110VAC 60Hz only. I am now living in Belgium and of coarse cannot
plug my phone in without using the transformers. It seems easy enough,
and cheaper for me to just go out and purchase new AC to DC converters
for 220VAC 50Hz.

Original input: Proposed Input:
110VAC 60Hz 220VAC 50Hz
Original output: Proposed output:
7.5VDC 150mA ----handset charger------ 7.5VDC 300mA
7.5VDC 500mA ----base station----------- 7.5VDC 500mA

So here is my issue. I have two types of power outputs from my old AC/
DC converter. One is (7.5VDC 150mA) for the handset chargers, and the
other is (7.5VDC 500mA) for the base. I have found a universal AC/DC
220VAC converter for the 500mA base. But what I'd like to know is if
it's safe to use a (7.5VDC 300mA) converter on the handset charger
that originally requires (7.5VDC 150mA)?

Yes it is safe. You can ALWAYS use a power source with a higher current
rating than what you need. You cannot go the other way, however. That is you
cannot ues a 150mA device where you need 300mA.

 

[Paul E. Schoen]

"John Popelish" <jpopelish@rica.net> wrote in message
news:I5KdnXOl8_a2Oo3VnZ2dnUVZ_r6rnZ2d@comcast.com...

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @ www.dse.com.au


For the two doubler capacitors, I would use something like 5,000 to
10,000 microfarads per amp of output load and rated for at least 16 volts
DC. Your diodes and transformer will allow about 300 to 400 mA load
current. If your load current is at or below 400 mA, that implies
something around 2,200 to 4,700 uF 16 or 25 volt capacitors for the
doubler.

Of course, if you use the full 30 volt winding, you can get pretty close
to 45 volts output with a bridge rectifier made with 4 of your diodes and
a single capacitor, say, 2,200 to 4,700 uF, rated for 50 volts. This
will give you a voltage around 40 to 45 volts, depending on load current
that can be up to about 0.7 amps.

You need to use capacitors rated at least as high as the maximum peak to
peak unloaded output voltage of the transformer, so for 24 VAC that would
be about 75 volts. Here is an LTspice ASCII showing a 24 VAC source boosted
to 68 VDC with two diodes and two capacitors. The first capacitor can be
rated at half the voltage of the second one. This circuit has poor
regulation, but is inherently current limiting, so you can just add a zener
(as shown) to the output for regulation. You need to adjust the value of C1
for the wattage of the zeners and load current. This is set for 45 VDC and
about 50 mA max.

Paul

=================================================================

Version 4
SHEET 1 880 680
WIRE 64 144 -48 144
WIRE 240 144 128 144
WIRE 288 144 240 144
WIRE 416 144 352 144
WIRE 512 144 416 144
WIRE 528 144 512 144
WIRE 640 144 528 144
WIRE 240 192 240 144
WIRE 416 192 416 144
WIRE -48 224 -48 144
WIRE 528 224 528 144
WIRE 640 224 640 208
WIRE 640 304 640 288
WIRE -48 384 -48 304
WIRE 240 384 240 256
WIRE 240 384 -48 384
WIRE 416 384 416 256
WIRE 416 384 240 384
WIRE 528 384 528 304
WIRE 528 384 416 384
WIRE 640 384 640 368
WIRE 640 384 528 384
WIRE 640 432 640 384
FLAG 640 432 0
FLAG 512 144 V+
SYMBOL diode 288 160 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 1000ľ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1
SYMBOL voltage -48 208 R0
WINDOW 3 -14 227 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 24 44 Left 0
SYMATTR Value SINE(0 35 60 0 0 0 120)
SYMATTR SpiceLine Rser=.1
SYMATTR InstName V1
SYMBOL res 512 208 R0
SYMATTR InstName R1
SYMATTR Value 2000
SYMBOL polcap 128 128 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 47ľ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1
SYMBOL diode 224 256 M180
WINDOW 0 24 72 Left 0
WINDOW 3 24 0 Left 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL zener 656 288 R180
WINDOW 0 40 32 Left 0
WINDOW 3 -137 34 Left 0
SYMATTR InstName D4
SYMATTR Value BZX84C15L
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL zener 656 208 R180
WINDOW 0 41 31 Left 0
WINDOW 3 -138 34 Left 0
SYMATTR InstName D3
SYMATTR Value BZX84C15L
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL zener 656 368 R180
WINDOW 0 41 47 Left 0
WINDOW 3 -144 33 Left 0
SYMATTR InstName D5
SYMATTR Value BZX84C15L
SYMATTR Description Diode
SYMATTR Type diode
TEXT 184 528 Left 0 !.tran 2

 

[Phil Allison]

"Tim Wescott"

Voltage doublers give poor regulation,

** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

and regenerative sets want a good, steady input voltage.

** Only way to get that is with a regulated supply.

You'll get better performance off of you pair of
wall warts, no matter how ugly things look.

** Not true.

Line voltage and load variations will affect them just like any unregulated
supply.

The actual percentage load regulation depends on the size of the transformer
and how much load is applied - ie light loads = good regulation.



...... Phil

 

[Phil Allison]

"Dave.H"

Actually the wallwarts are regulated,

I think they're M9926 @ www.dse.com.au

** That is a switch mode power supply ( SMPS ).




.... Phil

 

[Phil Allison]

"Mike Silva

"Phil Allison"

** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier
and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that?


** Yes - so fuck off.


Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.


** So what - shithead ???

Ripple voltage depends on cap value, so can be made as small as you like.



Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm


** The claims made are complete BOLLOCKS.

A "standard voltage doubler" generates 120 Hz ( or 100Hz ) ripple and
utilises the transformer just as fully as a bridge rectifier.

The page was written by a ham -

so you KNOW it is TOTAL SHIT !!!




........ Phil

 

[Phil Allison]

"Mike Silva

"Phil Allison"

** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier
and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that?


** Yes - so fuck off.


Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.


** So what - shithead ???

Ripple voltage depends on cap value, so can be made as small as you like.



Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm


** The claims made are complete BOLLOCKS.

A "standard voltage doubler" generates 120 Hz ( or 100Hz ) ripple and
utilises the transformer just as fully as a bridge rectifier.

The page was written by a ham -

so you KNOW it is TOTAL SHIT !!!




........ Phil

 

[Phil Allison]

"Mike Silva the fucking Cunt Head "



** Why don't you go eat a box of rat poison ???

Do the planet a favour by bleeding to death.




.... Phil

 

[Phil Allison]

"Mike Silva the fucking Cunt Head "



** Why don't you go eat a box of rat poison ???

Do the planet a favour by bleeding to death.




.... Phil

 

[Frnak McKenney]

On Thu, 24 Apr 2008 14:21:37 -0500, Chris W <1qazse4@cox.net> wrote:

Here is the problem. A client company has about 100 pieces of equipment
that they want to track when and how long they are down for service.
The equipment can be a few hundred feet from the nearest computer. The
requirement is for one computer to track every piece of equipment. The
interface will simply be the user equipment operator hitting a button
when the equipment goes down and then hitting another button when it
goes back in service or a toggle switch.
--snip--

Chris,

If (and I realize it's a big IF that assumes facts not in evidence
<grin!&gt all of the equipment is electrically powered, and if
(another big one) all are powered from a common power source, would
it be sufficient to monitor when each piece of equipment was drawing
power and when it wasn't?

I know that inductive current-detecting modules exist (think
clamp-on Ammeters); the hairy part would be locating the electrical
power lines and attaching the modules.

I admit I'm a little unclear on your statement about placement:

The equipment can be a few hundred feet from the nearest computer.

Does this mean that computers already exist, but none are close to
any given piece of equipment? That is, that a 100m RF link _could_
be set up between each piece of equipment and at least one computer,
and that those computers are networked?

Along the same lines, should we be picturing 100 pieces of
equipment, all within 50' of each other, but a hundred-yeard-dash
away from the nearest computer? Or 100 pieces of equipment scattered
across 20 acres (or at different locations across the country) which
are (say) in the middle of National Forests, well away from any
computer?


Frank McKenney
--
"If men were angels, no government would be necessary. If angels
were to govern men, neither external nor internal controls on
government would be necessary. In framing a government which is
to be administered by men over men, the great difficulty lies in
this: you must first enable the government to control the
governed; and in the next place oblige it to control itself. A
dependence on the people is, no doubt, the primary control on
the government; but experience has taught mankind the necessity
of auxiliary precautions."
-- James Madison / The Federalist Papers
--

 

[Frnak McKenney]

On Thu, 24 Apr 2008 14:21:37 -0500, Chris W <1qazse4@cox.net> wrote:

Here is the problem. A client company has about 100 pieces of equipment
that they want to track when and how long they are down for service.
The equipment can be a few hundred feet from the nearest computer. The
requirement is for one computer to track every piece of equipment. The
interface will simply be the user equipment operator hitting a button
when the equipment goes down and then hitting another button when it
goes back in service or a toggle switch.
--snip--

Chris,

If (and I realize it's a big IF that assumes facts not in evidence
<grin!&gt all of the equipment is electrically powered, and if
(another big one) all are powered from a common power source, would
it be sufficient to monitor when each piece of equipment was drawing
power and when it wasn't?

I know that inductive current-detecting modules exist (think
clamp-on Ammeters); the hairy part would be locating the electrical
power lines and attaching the modules.

I admit I'm a little unclear on your statement about placement:

The equipment can be a few hundred feet from the nearest computer.

Does this mean that computers already exist, but none are close to
any given piece of equipment? That is, that a 100m RF link _could_
be set up between each piece of equipment and at least one computer,
and that those computers are networked?

Along the same lines, should we be picturing 100 pieces of
equipment, all within 50' of each other, but a hundred-yeard-dash
away from the nearest computer? Or 100 pieces of equipment scattered
across 20 acres (or at different locations across the country) which
are (say) in the middle of National Forests, well away from any
computer?


Frank McKenney
--
"If men were angels, no government would be necessary. If angels
were to govern men, neither external nor internal controls on
government would be necessary. In framing a government which is
to be administered by men over men, the great difficulty lies in
this: you must first enable the government to control the
governed; and in the next place oblige it to control itself. A
dependence on the people is, no doubt, the primary control on
the government; but experience has taught mankind the necessity
of auxiliary precautions."
-- James Madison / The Federalist Papers
--

 

[Paul E. Schoen]

"Mike Silva" <snarflemike@yahoo.com> wrote in message
news:81006d7a-65d7-411a-946b-e8184dfde0ee@p25g2000hsf.googlegroups.com...
On Apr 24, 10:01 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

** The standard voltage doubler has exactly the *same* ( load )
percentage
regulation as the same transformer would provide with a bridge rectifier
and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that? Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.

Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm

==============================================================================

Two of those diodes are superfluous, and otherwise it is a standard dual
output voltage doubler, similar to what I posted with an LTspice ASCII
schematic. If your load is fairly constant and no more than a couple of
watts, the string of zeners (or a single 48V zener) in that circuit will
give you decent line and load regulation. It also has the advantage of
current limiting, so a short circuit on the output can be tolerated. It's
dirt simple, cheap, and meets your requirements. You don't need a fancy
voltage regulator, and you would need an LM317H to get a 48 VDC output, and
you might also exceed its 60 VDC input maximum.

Paul

 

[Paul E. Schoen]

"Mike Silva" <snarflemike@yahoo.com> wrote in message
news:81006d7a-65d7-411a-946b-e8184dfde0ee@p25g2000hsf.googlegroups.com...
On Apr 24, 10:01 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

** The standard voltage doubler has exactly the *same* ( load )
percentage
regulation as the same transformer would provide with a bridge rectifier
and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that? Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.

Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm

==============================================================================

Two of those diodes are superfluous, and otherwise it is a standard dual
output voltage doubler, similar to what I posted with an LTspice ASCII
schematic. If your load is fairly constant and no more than a couple of
watts, the string of zeners (or a single 48V zener) in that circuit will
give you decent line and load regulation. It also has the advantage of
current limiting, so a short circuit on the output can be tolerated. It's
dirt simple, cheap, and meets your requirements. You don't need a fancy
voltage regulator, and you would need an LM317H to get a 48 VDC output, and
you might also exceed its 60 VDC input maximum.

Paul