Chip with simple program for Toy

[Paul Burke]

Kryten wrote:

Of course you can verb nouns, and noun verbs,


No you can't, unless a word specifically does have two forms.

Look at the everyday word 'knife'. A noun, unless I knife you (verb), or
I'm in a knife fight or on a knife edge (adjective). The point about
English is that almost any part of speech is potentially almost any
other part. The rules for transfer are very simple, embedded in the
language since the year dotcom, and one of its glories. Even the title
of the infamous preskriptionfest 'Eats shoots and leaves' or however she
punctuates it advertises this.

Paul Burke

 

[Mike]

"John Fields" <jfields@austininstruments.com> wrote in message
news:8fqpp099lq18pm0qsc6ifi597dqnnlekjv@4ax.com...

On Wed, 17 Nov 2004 08:54:52 -0600, John Fields
jfields@austininstruments.com> wrote:


Ooops...

If all you're using the supply for is to charge a cap, then you could
do something like this:

+HV
+------+ +-----+ |
MAINS>----|P1 S1|--|~ +|---[16K]--+--[1N4007>]---<<---+
| | | | | |
| | | | [40K] [CAP]
| | | | | |
MAINS>----|P2 S2|--|~ -|----------+--------------<<---+
+------+ +-----+
XFMR FWB


Assuming no copper losses in the transformer and no leakage current
through the cap and based on 120VRMS nominal mains, 10mA out of the
^^^^^^^^^^^^^^^^^^^^^
a 10VRMS input to the transformer

secondary, and the resistances shown, +HV will be about 400V. the 16k
resistor will dissipate about 1.6 watts and the 40k will dissipate
about 4 watts after the cap is charged, so a 5 watter and a 10 watter
would probably be OK.

and the schematic should look like this:
+HV
+------+ +-----+ |
10VRMS>----|P1 S1|--|~ +|---[16K]--+--[1N4007>]---<<---+
| | | | | |
| | | | [40K] [CAP]
| | | | | |
10VRMS-----|P2 S2|--|~ -|----------+--------------<<---+
+------+ +-----+
XFMR FWB


--
John Fields

That looks simple enough, but what if the 10VRMS has some noise? Wouldn't
that cause more than 400V to be put across the capacitor?

 

[Randy Gross]

Robert Monsen <rcsurname@comcast.net> wrote in message news:<Q3bnd.60292$5K2.32791@attbi_s03>...

<snip>

Select the 'edit'
button, and an editor comes up that allows you to create a new model.
Get the values out of the datasheet, and rename the model, and then you
can pick it just like any other transistor.

Same goes for diodes.

Some of the models will just bring up a netlist editor. In that case,
just replace the appropriate values in the netlist.

This approach will suit my needs perfectly. Thanks Robert! I have one
more question concerning the IN457 diode listed in the LM317/LM301A
circuit on page 16 of the datasheet. I can't find any reference to
this diode (National, Motorola, DigiKey, datasheet search, etc.) so
there's no data to sub the device if it proves outdated. I'm starting
to think the number may be incomplete or in err.
Can anyone help.

Randy

 

[Terry Pinnell]

aaawelder@yahoo.com (Randy Gross) wrote:

Robert Monsen <rcsurname@comcast.net> wrote in message news:<Q3bnd.60292$5K2.32791@attbi_s03>...

snip

Select the 'edit'
button, and an editor comes up that allows you to create a new model.
Get the values out of the datasheet, and rename the model, and then you
can pick it just like any other transistor.

Same goes for diodes.

Some of the models will just bring up a netlist editor. In that case,
just replace the appropriate values in the netlist.

This approach will suit my needs perfectly. Thanks Robert! I have one
more question concerning the IN457 diode listed in the LM317/LM301A
circuit on page 16 of the datasheet. I can't find any reference to
this diode (National, Motorola, DigiKey, datasheet search, etc.) so
there's no data to sub the device if it proves outdated. I'm starting
to think the number may be incomplete or in err.
Can anyone help.

Randy

CircuitMaker 1N457 (not IN457!) diode model:

IS: Saturation current 28.40p
RS: Ohmic resistance [0,] 2.100
N: Emission coefficient 1.700
TT: Transit-time [0,] 4.320u
CJO: Zero-bias junction capacitance [0,] 10.60p
VJ: Junction potential 750.0m
M: Grading coefficient 333.0m
EG: Activation energy 1.110
XTI: Saturation-current temperature exponent 3.000
KF: Flicker-noise coefficient 0.000
AF: Flicker-noise exponent 1.000
FC: Foward-bias depletion coefficient 500.0m
BV: Reverse breakdown voltage 60.00
IBV: Current at breakdown voltage 66.00n
TNOMaramameter measurement temperature 27.00

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[CBarn24050]

Subject: Resonance
From: daksoy@gmail.com (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <2502fe80.0411201404.3fc06c24@posting.google.com


My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.

2)If we want voltage across R to reach its maximum, we calculate
complex expression for Vr and if we let Vr's imaginary part to be 0,
we come up with a new resonance frequency w1 = (-R/L)^0.5, which is
meaningless. So can we immediately say or predict that voltage accross
R will reach its maximum when we set w = wo = [1/LC - (R/L)^2]^0.5
(which will reveal that finding resonance frequency has nothing to do
with letting the imaginary parts equal to 0)? If wo is making both the
voltage across Zo and R maximum, what is the reason for this?

Maximum voltage accros R is at DC.

(calculating the frequency which will make abs(Vr) maximum seemed to
be impossible, so i made a prediction)

Maximun impedence for a parallel tuned circuit is at resonance.

Yes, but here the minimum impedence occurs at resonance.

 

[Randy Gross]

Thanks to every one of you for solutions to my problems. Now I can get
this circuit up and running to study what's going on in it before I
build.

Randy

 

[Spajky]

On 20 Nov 2004 14:04:51 -0800, daksoy@gmail.com (Deniz) wrote:

-->------L--
I | |
C R
| |
= =

Circuit seen above (series RL and a parallel C) is supposed to
resonate somehow. There is a current source driving the circuit (since
I thought that a voltage source driving the circuit would be useless).

could be driven also with voltage IMHO & rappresents a impedance step
down transformation @ resonant frequency; R is just a load ...
--
Regards, SPAJKY ÂŽ
& visit my site @ http://www.spajky.vze.com
"Tualatin OC-ed / BX-Slot1 / inaudible setup!"
E-mail AntiSpam: remove ##

 

[The Phantom]

On Sun, 21 Nov 2004 19:50:44 GMT, Robert Monsen
<rcsurname@comcast.net> wrote:

The Phantom wrote:
On 20 Nov 2004 22:48:45 GMT, cbarn24050@aol.com (CBarn24050) wrote:


Subject: Resonance
From: daksoy@gmail.com (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <2502fe80.0411201404.3fc06c24@posting.google.com

My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.


What does it mean for the capacitor and inductor to have opposite
energies?



At any given time, they don't have 'opposite' energy

CBarn24050 says they do. My question was directed at him; I was
hoping he would explain about these 'opposite' energies.

He also denies that the energy in the cap is max when the energy in
the inductor is 0. I was hoping he could give some details showing
just how this comes about.

, because energy is

a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

 

[Steve Evans]

On Sun, 21 Nov 2004 19:50:44 GMT, Robert Monsen
<rcsurname@comcast.net> wrote:

At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

It is a fact that in a parallel resonant circuit, the impedance is at
a maximum. Is this due to the fact that - at resonance - the energy
flows between the cap and coil are so large as to be able to repell
any current from the external energy source, thereby rendering its
path effectively blocked?
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[CBarn24050]

Subject: Re: Resonance
From: Robert Monsen rcsurname@comcast.net
Date: 21/11/04 19:50 GMT Standard Time
Message-id: <ok6od.649997$8_6.899@attbi_s04

The Phantom wrote:
On 20 Nov 2004 22:48:45 GMT, cbarn24050@aol.com (CBarn24050) wrote:


Subject: Resonance
From: daksoy@gmail.com (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <2502fe80.0411201404.3fc06c24@posting.google.com

My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.


What does it mean for the capacitor and inductor to have opposite
energies?



At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

--
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.

Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does not go
from 1 to the other and back again. When I said that they equal but opposite,
what I meant was that they allways add up to zero at any time in the cycle.

 

[Steve Evans]

On 22 Nov 2004 00:21:38 GMT, cbarn24050@aol.com (CBarn24050) wrote:

Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does not go
from 1 to the other and back again. When I said that they equal but opposite,
what I meant was that they allways add up to zero at any time in the cycle.

Zero over an aeveage *whole* cycle, I think you mean.
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[CBarn24050]

Subject: Resonance
From: The Phantom phantom@aol.com
Date: 22/11/2004 13:04 GMT Standard Time
Message-id: <knm3q0dfe1ui7s0715dhghs1ar0ccs0593@4ax.com

When the voltage across a capacitor is greater than zero, is there
not an energy of .5*C*V^2 associated with that capacitor? And
likewise given a current greater than zero in an inductor, is there
not an energy of .5*L*I^2 stored in the inductor? Are these not
*real* energies?

Just because the energy is not real doesn't mean that it does not exist, it's
just the confusing terms (real & imaginary) that apply to complex (but not
complicated) maths. You would think that they would have come up with something
a bit better by now.

It does not go
from 1 to the other and back again. When I said that they equal but
opposite,
what I meant was that they allways add up to zero at any time in the cycle.

What I should have said was that there is no CHANGE in the total energy level

during the cycle.

 

[Steve Evans]

On Mon, 22 Nov 2004 01:19:16 GMT, Robert Monsen
<rcsurname@comcast.net> wrote:

I dont' want to put you on the spot, but I think there is energy stored
in resonant systems. Look at the film of the tacoma narrows bridge being
torn apart due to resonant oscillations. Each little nudge from the wind
stores more energy in the resonant system, until it collapses.

Yeah, but5 the energy has to come at the *right* moment each time.
it's a bit like pushing a child on a swing. you have to impart the
force at the rpecise time in each arc to get the swing moving with
minimal effort. i think the equivalent term in elctronics is the
'flywheel effect'; get the timing spot-on and you can keep the
oscillation moving with minimal energy input/maximal efficitency.
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Steve Evans]

On 22 Nov 2004 07:57:18 -0600, The Phantom <phantom@aol.com> wrote:

tnx, phantom! Thats the clearest explanation i've come across so far.
there are still a couple of outstandingpoints i need to clear up....
I'll get back in a shrot while!

Steve
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Rich Grise]

On Mon, 22 Nov 2004 16:58:20 +0000, CBarn24050 wrote:

Subject: Resonance
From: The Phantom phantom@aol.com
Date: 22/11/2004 13:04 GMT Standard Time
Message-id: <knm3q0dfe1ui7s0715dhghs1ar0ccs0593@4ax.com

When the voltage across a capacitor is greater than zero, is there
not an energy of .5*C*V^2 associated with that capacitor? And
likewise given a current greater than zero in an inductor, is there
not an energy of .5*L*I^2 stored in the inductor? Are these not
*real* energies?

Just because the energy is not real doesn't mean that it does not exist, it's
just the confusing terms (real & imaginary) that apply to complex (but not
complicated) maths. You would think that they would have come up with something
a bit better by now.

It's definitely real - it's just that it stays in the resonant circuit,
rather than doing work or being thrown away as heat, which is the part
that you pay for. This is why power companies like power factor
correction.

It does not go
from 1 to the other and back again. When I said that they equal but
opposite,
what I meant was that they allways add up to zero at any time in the
cycle.

What I should have said was that there is no CHANGE in the total energy
level
during the cycle.

Yeah.

Cheers!
Rich

 

[Rich Grise]

On Mon, 22 Nov 2004 01:19:16 +0000, Robert Monsen wrote:

CBarn24050 wrote:

I dont' want to put you on the spot, but I think there is energy stored
in resonant systems. Look at the film of the tacoma narrows bridge being
torn apart due to resonant oscillations. Each little nudge from the wind
stores more energy in the resonant system, until it collapses.

They've determined that the bridge wasn't resonating - it just got blown
down:

http://www.math.umbc.edu/~gobbert/teaching/math101.s2003/reports/Group1Tacoma.doc

Cheers!
Rich

 

[Robert Monsen]

The Phantom wrote:

On 20 Nov 2004 22:48:45 GMT, cbarn24050@aol.com (CBarn24050) wrote:


Subject: Resonance
From: daksoy@gmail.com (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <2502fe80.0411201404.3fc06c24@posting.google.com

My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.


What does it mean for the capacitor and inductor to have opposite
energies?

At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

--
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.

 

[Steve Evans]

On Tue, 23 Nov 2004 04:32:52 GMT, Rich Grise <rich@example.net> wrote:

They're 90 degrees out of phase, because the capacitor stores energy as
charge, and the inductor stores energy as current. The current hits its
maximum as charge is going through zero, and vice-versa.

Well, they're *each* 90 degrees out of phase, so they're in complete
antiphase (180') WRT each other. The cap's eneryg is stored in an
electric field; the inductor's is stored in a magnetic field and when
one's at a maximum, the other's at a minimum and vice versa.
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Don Kelly]

"CBarn24050" <cbarn24050@aol.com> wrote in message
news:20041121192138.21610.00000937@mb-m12.aol.com...

Subject: Re: Resonance
From: Robert Monsen rcsurname@comcast.net
Date: 21/11/04 19:50 GMT Standard Time
Message-id: <ok6od.649997$8_6.899@attbi_s04

The Phantom wrote:
On 20 Nov 2004 22:48:45 GMT, cbarn24050@aol.com (CBarn24050) wrote:


Subject: Resonance
From: daksoy@gmail.com (Deniz)
Date: 20/11/2004 22:04 GMT Standard Time
Message-id: <2502fe80.0411201404.3fc06c24@posting.google.com

My first question is: At resonance frequency wo, can we immediately
(without calculating time dependent expressions) say that when the
stored energy in the capacitor reaches maximum, the stored energy in
the inductor becomes 0 ?

NO, the enrgy in both components are equal but opposite.


What does it mean for the capacitor and inductor to have opposite
energies?



At any given time, they don't have 'opposite' energy, because energy is
a scalar value.

However, the energy that they have is passed back and forth between the
two elements; thus, when the capacitor has maximal energy, the inductor
has minimal energy.

You could say that the energy for each is a sinusoidal wave above the x
axis, 180' out of phase, and that the sum of the two sine waves is equal
to the total energy in the system. That energy can be increasing if
there is an impulse, decreasing if the oscillation is damped, or
'constant' if the damping and impulse balance out.

--
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.







Perhaps not the best choice of words, i'll try again. These components are
reactive so there is no energy as such, ie it's not real energy. It does
not go
from 1 to the other and back again. When I said that they equal but
opposite,
what I meant was that they allways add up to zero at any time in the
cycle.

----------
There is real energy but it is being shuttled back and forth. The average
power over a cycle is 0 so that the total energy input during the cycle is
also 0. At any instant in time the power is not 0 nor is the sum of the
energies stored in the L and C.

Where does the energy come from?
The conditions when operating at steady state are not the same as when the
circuit is first energised. There is a transient period in which energy is
initially stored in the capacitor and inductor (not necessarily the same in
each). You can't handle this period with the concepts of AC steady state
analysis (phasors, reactive, etc) but need to consider the differential
equations involved.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[The Phantom]

On Tue, 23 Nov 2004 19:17:43 GMT, Steve Evans
<smevans@jif-lemon.co.mars> wrote:

On Tue, 23 Nov 2004 04:32:52 GMT, Rich Grise <rich@example.net> wrote:

They're 90 degrees out of phase, because the capacitor stores energy as
charge, and the inductor stores energy as current. The current hits its
maximum as charge is going through zero, and vice-versa.

Well, they're *each* 90 degrees out of phase, so they're in complete
antiphase (180') WRT each other. The cap's eneryg is stored in an
electric field; the inductor's is stored in a magnetic field and when
one's at a maximum, the other's at a minimum and vice versa.

I think we are talking about energy here, and it is true that the
time functions of the two energies are sinusoids, 180 out of phase,
but the reason is slightly more complicated, Steve. The current in L
and C are each 90 degrees out of phase with the reference (voltage),
in opposite directions (so to speak), so the two currents are indeed
180 out. But one might expect that since the energy in C is a
function of voltage and the energy in L is a function of current, that
the energies might be 90 degrees out of phase, since the voltage
across the C is only 90 degrees out phase with the current in the L.

The detail Rich is missing is that if you plot the energy in L and
C separately, you will see that the energy vs. time plot is a *double*
frequency function, compared to the voltage or current. This is
because the energy involves the *square* of the voltage or current
(for C or L), which is always positive regardless of whether the
voltage (or current) is in the positive or negative direction.
Remember your trigonometry, specifically the formula: SIN^2(x) = (1 -
COS(2x))/2 When you square a sinusoid, you get a double frequency
sinusoid plus a constant (the constant is the *average* energy). When
you look at the squares of two sinusoids that are 90 deg out of phase,
you get a couple of double frequency sinusoids that are *180* out of
phase with each other. When you add these two double frequency
sinusoids (plus their constant terms), the sinusoidal portions cancel
and the contants add to give a constant equal to the total energy in L
and C.