Chip with simple program for Toy

[Fred Bloggs]

sert wrote:

Fred Bloggs <nospam@nospam.com> wrote in
news:484C6008.5070809@nospam.com:


View in a fixed-width font such as Courier.
.
--------------
. 9 2 | 2
. or t = - - x R x (v - 1) from which v = | 1 -
----- x t . 2
\| 9 x R .
.




Thanks. But there's something unusual going on; the capacitor is
discharging completely. A linear capacitor, as we know, never
discharges completely.

Also, for values of t over a certain threshold the formula is
undefined.

Let's assume that the capacitor will discharge completely as the
formula suggests. Surely the time t will continue (time will not
stop!) but the formula doesn't actually give v=0 for any t after
the discharging, it's simply undefined.

Why? Is there some other non-obvious solution to the DE?

View in a fixed-width font such as Courier.

.
.
.
.
. i
. ->
. .----------. v= i x R
. | ^ |
. | | | d 3
. +| | | and i= - -- (3 x v )
. --- / dt
. C --- v \ R
. | / 2 dv
. | | \ so v = -9 x R x v x --
. | | | dt
. | | |
. '----------.
.
.
. 2 dv dv
. this becomes v + 9 x R x v x -- = v x ( 1 + 9 x R x v x -- )=0
. dt dt
.
. at each instant of time because although C is nonlinear the basic
.
.
. laws of physics still apply to the circuit.
.
.
. The above equation has only two possible solutions:
.
.
. 1) v(t)=0 for all t, which is ruled out because we need v(0)=1
.
. --------------
. | 2
. 2) v(t)= | 1 - ----- x t
. \| 9 x R
.
.
.
. 9 x R dv
. at time t= ----- v=0, the v x ( 1 + 9 x R x v x -- )=0
. 2 dt
.
. 9 x R
. equation still applies where t is now measured from -----
. 2
.
. and since v(0)=0 the v(t)=0 solution does
.
. dv
. apply, and the 1 + 9 x R x v x -- = 0 solution leads to an
. dt
.
. an imaginary and extraneous result.
.
.
. All of this can be condensed into:
.
.
.
. ----------------------
. | 2
. v(t)= | MAX(0,1 - ----- x t) t>=0
. \| 9 x R
.
.

 

[Rich Grise]

On Tue, 10 Jun 2008 02:42:03 -0500, Stephen J. Rush wrote:

You can't win.
You can't break even.
You can't even quit the game.

Christianity is the belief that you can win.
Judaism/Islam is the belief that you can break even.
Mysticism is the belief that you can quit the game.

;-)

Cheers!
Rich

 

[JosephKK]

On Sun, 08 Jun 2008 12:52:59 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

On Sun, 8 Jun 2008 19:40:46 +0000 (UTC), sert <jerry@hotmail.com
wrote:

Jim Thompson <To-Email-Use-The-Envelope-Icon@My-Web-Site.com
wrote in news:85ao441vohrgf00boagq6lclq2ih99degm@4ax.com:

Too many twists and turns.

Fundamentals...

i = d/dt(c*v) =

i = c*dv/dt + v*dc/dt

Try manipulating from there, adding in c as a function of
v. I suspect you inserted too many artificial restraints.


(Your post came up a bit garbled so I have it fixed above.)

I believe what you wrote is wrong, it is simply:

i = c*dv/dt

That equation is in the book, too.

Nope! The general equation is... i = d/dt(C*v)

Likewise for inductors it's... v = d/dt(L*i)

Don't dispute the MIT grad who found this stuff fascinating ;-)

...Jim Thompson

Nice swat on some twit who clearly deserved it.

 

[JosephKK]

On Sun, 08 Jun 2008 15:56:31 -0400, John Popelish <jpopelish@rica.net>
wrote:

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)

While it is true that LTSpice supports it, i believe that it is
generic to all spice v.3

 

[JosephKK]

On Sun, 08 Jun 2008 16:07:04 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

sert wrote:
Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

The Euler method and its relatives (such as the Milne
predictor-corrector) have real trouble with decaying exponential-ish
things like this. The problem is that the difference scheme has an
extraneous solution (the oscillating one) that dominates the decaying
exponential.

You need a better solver--a simple one that works OK for this is the
classical fourth-order Runge-Kutta scheme, or with a bit more work you
can code up the Adams-Bashforth-Moulton predictor-corrector (my personal
fave). You can look them up on the web, or find them in most elementary
numerical analysis books.

Cheers,

Phil Hobbs

Haven't a clue if that is over OP's head or not, but now i know who to
call when the math is getting tough for me.

 

[redbelly]

On Jun 8, 4:27 pm, Tim Wescott <t...@seemywebsite.com> wrote:

So v = sqrt(1V^2 - (10/9 V^2/sec)t)

I get the same result, but have a problem that v is imaginary for
times greater than 0.9 sec. I guess the way to interpret this is that
this "capacitor" really discharges to 0 V in a finite time, and simply
remains 0 thereafter.

Mark

 

[Helmut Sennewald]

"redbelly" <redbelly98@yahoo.com> schrieb im Newsbeitrag
news:2b4e90a7-74f5-4843-830b-790d7b572f6e@c65g2000hsa.googlegroups.com...

On Jun 8, 4:27 pm, Tim Wescott <t...@seemywebsite.com> wrote:

So v = sqrt(1V^2 - (10/9 V^2/sec)t)

I get the same result, but have a problem that v is imaginary for
times greater than 0.9 sec. I guess the way to interpret this is that
this "capacitor" really discharges to 0 V in a finite time, and simply
remains 0 thereafter.

Mark

Hello Mark,

When the capacitor is discharged by the resistor,
its energy is changed to heat in the resistor.
That means that all the energy is lost.

E = 1/2*C*V^2

After V has become 0, E will be 0:

0 = 1/2*C*V^2

-> V remains 0V after the discharge.

Best regards,
Helmut

 

[Helmut Sennewald]

"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:e2tu44t1r59k0948mf43roo879s9rdlqo8@4ax.com...

On Sun, 08 Jun 2008 15:56:31 -0400, John Popelish <jpopelish@rica.net
wrote:

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)

While it is true that LTSpice supports it, i believe that it is
generic to all spice v.3

Hello,

A nice wish regarding SPICE3, but it's not available there!
I have the feeling there isn't much activity around SPICE3
since a decade.

Helmut

 

[Stephen J. Rush]

On Wed, 11 Jun 2008 17:28:03 -0700, BobW wrote:

For those of you that haven't seen this in action:

http://www.youtube.com/watch?v=pcVG6c_OvYU&NR=1

An eBay seller called magnet4less has a nice variety of neodymium
magnets. I bought some of his 1/2" N48 sphere magnets. They work very
well in 3/4" copper pipe.

The effect is amazing -- especially when you first demonstrate that the
magnet is not attracted to the copper.

I dropped one through a 10' piece of 3/4" copper pipe and it took about
13 seconds to make it all the way through. It works well in short
pieces, too.

Cool. I've seen several demonstrations of eddy-current braking, but this
is the best one that doesn't require a big electromagnet.

 

[Rich Grise]

On Wed, 11 Jun 2008 17:28:03 -0700, BobW wrote:

For those of you that haven't seen this in action:

http://www.youtube.com/watch?v=pcVG6c_OvYU&NR=1

Tis one's cool too:
http://www.youtube.com/watch?v=f43fXde2S78

I guess one utube leads to another! ;-)

Cheers!
Rich

 

[JosephKK]

On Wed, 11 Jun 2008 20:08:07 +0200, "Helmut Sennewald"
<helmutsennewald@t-online.de> wrote:

"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:e2tu44t1r59k0948mf43roo879s9rdlqo8@4ax.com...
On Sun, 08 Jun 2008 15:56:31 -0400, John Popelish <jpopelish@rica.net
wrote:

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)

While it is true that LTSpice supports it, i believe that it is
generic to all spice v.3


Hello,

A nice wish regarding SPICE3, but it's not available there!
I have the feeling there isn't much activity around SPICE3
since a decade.

Helmut

The current base version from Berkeley is v.3f5. The initial
development of v.3 included all kinds of added non-linear components
changes and a complete set of linear or non-linear dependant sources.

 

[Helmut Sennewald]

"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:4uk354l503c7a2gu4nk334dvba7jrehu5o@4ax.com...

On Wed, 11 Jun 2008 20:08:07 +0200, "Helmut Sennewald"
helmutsennewald@t-online.de> wrote:


"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:e2tu44t1r59k0948mf43roo879s9rdlqo8@4ax.com...
On Sun, 08 Jun 2008 15:56:31 -0400, John Popelish <jpopelish@rica.net
wrote:

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)

While it is true that LTSpice supports it, i believe that it is
generic to all spice v.3


Hello,

A nice wish regarding SPICE3, but it's not available there!
I have the feeling there isn't much activity around SPICE3
since a decade.

Helmut



The current base version from Berkeley is v.3f5. The initial
development of v.3 included all kinds of added non-linear components
changes and a complete set of linear or non-linear dependant sources.

Hello Joseph,

SPICE 3f5 is from 1997.

http://www.mes.tu-darmstadt.de/lectures/summer/gel/schmankerl/SpiceSchmankerl.pdf

Do you have any link about a manual-page with an
arbitrary capacitor model? I couldn't find one.

Best regards,
Helmut

 

[JosephKK]

On Fri, 13 Jun 2008 19:59:43 +0200, "Helmut Sennewald"
<helmutsennewald@t-online.de> wrote:

"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:4uk354l503c7a2gu4nk334dvba7jrehu5o@4ax.com...
On Wed, 11 Jun 2008 20:08:07 +0200, "Helmut Sennewald"
helmutsennewald@t-online.de> wrote:


"JosephKK" <quiettechblue@yahoo.com> schrieb im Newsbeitrag
news:e2tu44t1r59k0948mf43roo879s9rdlqo8@4ax.com...
On Sun, 08 Jun 2008 15:56:31 -0400, John Popelish <jpopelish@rica.net
wrote:

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)

While it is true that LTSpice supports it, i believe that it is
generic to all spice v.3


Hello,

A nice wish regarding SPICE3, but it's not available there!
I have the feeling there isn't much activity around SPICE3
since a decade.

Helmut



The current base version from Berkeley is v.3f5. The initial
development of v.3 included all kinds of added non-linear components
changes and a complete set of linear or non-linear dependant sources.

Hello Joseph,

SPICE 3f5 is from 1997.

http://www.mes.tu-darmstadt.de/lectures/summer/gel/schmankerl/SpiceSchmankerl.pdf

Do you have any link about a manual-page with an
arbitrary capacitor model? I couldn't find one.

Best regards,
Helmut

It does not seem to be listed on the Berkeley SPICE models page:

http://bwrc.eecs.berkeley.edu/Classes/icbook/SPICE/UserGuide/elements_fr.html

but with a pair of non-linear dependant sources is should be pretty
easy to make a macro for one. In fact they provide an annotated
subcircuit for a non-linear capacitor in the non-linear source
description.

 

[John Fields]

On Sat, 14 Jun 2008 03:15:26 -0700 (PDT), "j_slobo@hotmail.com"
<j_slobo@hotmail.com> wrote:

Is there some way to detect a 3-5volt say khz clock using an ordinary
instrument(no osciloscope)

---
AC or DC voltmeter.

JF

 

[redbelly]

On Jun 11, 1:59 pm, "Helmut Sennewald" <helmutsennew...@t-online.de>
wrote:

"redbelly" <redbell...@yahoo.com> schrieb im Newsbeitragnews:2b4e90a7-74f5-4843-830b-790d7b572f6e@c65g2000hsa.googlegroups.com...

On Jun 8, 4:27 pm, Tim Wescott <t...@seemywebsite.com> wrote:

So v = sqrt(1V^2 - (10/9 V^2/sec)t)

I get the same result, but have a problem that v is imaginary for
times greater than 0.9 sec. I guess the way to interpret this is that
this "capacitor" really discharges to 0 V in a finite time, and simply
remains 0 thereafter.

Mark

Hello Mark,

When the capacitor is discharged by the resistor,
its energy is changed to heat in the resistor.
That means that all the energy is lost.

E = 1/2*C*V^2

After V has become 0, E will be 0:

0 = 1/2*C*V^2

-> V remains 0V after the discharge.

Best regards,
Helmut

Hello Helmut,

Much of what you say is also true of standard linear capacitors, which
of course do not go to zero volts in a finite time.

After reviewing my derivation, I found that as V --> 0, the current
becomes infinitely large, thereby draining all the charge. At V=0,
the differential equation I had come up with is no longer valid, so I
guess it makes sense in a way that the solution is not valid for times
beyond when V reaches 0.

Regards,
Mark

 

[Frank Buss]

Talal Itani wrote:

My boss asked me to evaluate oscilloscopes and to recommend one. I know
Tektronix and LeCroy. Which is better? Are they any others? Ideally, we
would get a 4-channel scope at 300 MHz, with a 5 million sample record
length. Thanks for your advise.

Agilent scopes are nice, too. Some scope companies have local resellers,
which can demonstrate their products at your workplace and even lend you a
scope for some days, so you can evaluate it yourself, if it fits your
needs.

--
Frank Buss, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de

 

[Jim Granville]

Talal Itani wrote:

Hello,

My boss asked me to evaluate oscilloscopes and to recommend one. I know
Tektronix and LeCroy. Which is better? Are they any others? Ideally, we
would get a 4-channel scope at 300 MHz, with a 5 million sample record
length. Thanks for your advise.

Google is your friend.
Find a broker, and then pick brands from that.
eg

http://www.teknetelectronics.com/Search.asp?p_Var1=Oscilloscope&F=C

-jg

 

[Dirk Bruere at NeoPax]

Talal Itani wrote:

Hello,

My company wants to buy an oscilloscope. The scope that meets our needs is
around $7,000. I am sure we can save much money if we buy a PC-based unit,
like Picoscope. Do you have experience with PC based scopes? Do you
recommend them for serious work?

I don't really like them, no matter what the spec.
It ties up a PC, for one thing. It makes it super non-portable.
Just too many 'bits' floating about.

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
Remote Viewing classes in London

 

[donald]

Talal Itani wrote:

Hello,

My company wants to buy an oscilloscope. The scope that meets our needs is
around $7,000. I am sure we can save much money if we buy a PC-based unit,
like Picoscope. Do you have experience with PC based scopes? Do you
recommend them for serious work?

Thanks


Which Picoscope are you looking at:

http://www.picotech.com/oscilloscope-specifications.html

At $11,990.00 or $2,390.00, the price or specs are better with the
$7,000.00 scope, which ever scope that is.

donald

 

[John Larkin]

On Sat, 21 Jun 2008 06:19:05 GMT, "Talal Itani" <titani@verizon.net>
wrote:

Hello,

My boss asked me to evaluate oscilloscopes and to recommend one. I know
Tektronix and LeCroy. Which is better? Are they any others? Ideally, we
would get a 4-channel scope at 300 MHz, with a 5 million sample record
length. Thanks for your advise.

T.I.

LeCroy is evil. Weird, too.

John