Driver to drive?

[Maynard A. Philbrook Jr.]

In article <4i0l8914kgrirntgbdehi1odq24avfpe61@4ax.com>,
jlarkin@highlandtechnology.com says...

On Mon, 18 Nov 2013 12:25:28 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Mon, 18 Nov 2013 10:30:05 -0800, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Mon, 18 Nov 2013 11:14:39 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Mon, 18 Nov 2013 09:34:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 18 Nov 2013 09:05:54 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Mon, 18 Nov 2013 06:59:03 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 14 Nov 2013 08:13:54 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Wed, 13 Nov 2013 19:30:07 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 11 Nov 2013 10:49:05 -0600, "Tim Williams" <tmoranwms@charter.net
wrote:

Is J2 the socket for the PIC, or is it going through cables to the PIC
somewhere else?

What are the trace lengths on the loop from Q1 source (ground), to the
nearest bypass capacitor on +12VR, through D5, back to Q1? (Hint: D5 must
be placed at Q1, not at the coil.)

Why? It will clamp a flyback overshoot wherever it is in the circuit.

If the wiring from FET to coil is significantly inductive, then what?

...Jim Thompson


Since you can't think this through, you'll have to Spice it.

No. It's _you_ who'll have to LTspice it... and you'll get it wrong
:-}

...Jim Thompson

Having fun, John?

Jamie

 

[Anthony Stewart]

e = L*di/dt

Then the stored energy E=1/2 L*I^2 if dumped in a Cap = 1/2 C*V^2
assuming no loss in the transfer... does V = I * sq.rt.(L/C) ?

for
L=1uH
I=100mA
C=10pF
then V= 32V

The root (L/C) ratio is one of the factors that determines their "Q"
In this case ignoring the relay coil capacitance, the Q is 316 and only using the FET drain capacitance.

 

[Paul E Bennett]

bitrex wrote:

Use 4 optoisolated current sources/sinks under microprocessor control,
and sense the terminal voltage/current at each of the two ports. Apply
the appropriate passive device defining equations in software to
simulate a floating inductor, capacitor, negative impedance, etc.

Something like this maybe:

http://i.imgur.com/i8nSruE.png

PWM output goes to the NPN current sources, op amps scale and bias the
sense voltages to be in the microprocessor's ADC range.

Your circuit is not fully isolated, which may affect the results you are
after. If you were able to isolate the op-amps as well (isolated amplifiers
are quite expensive I know) then it would not interfere with the circuit
into which you were placing it. Otherwise quite a neat idea for the lower
frequency range (definitely not RF).

--
********************************************************************
Paul E. Bennett IEng MIET.....<email://Paul_E.Bennett@topmail.co.uk>
Forth based HIDECS Consultancy.............<http://www.hidecs.co.uk>
Mob: +44 (0)7811-639972
Tel: +44 (0)1235-510979
Going Forth Safely ..... EBA. www.electric-boat-association.org.uk..
********************************************************************

 

[RobertMacy]

On Thu, 21 Nov 2013 19:32:34 -0700, Jamie M <jmorken@shaw.ca> wrote:

Hi,

I was reading about topological insulators which can super-conduct
electrons on their edges, I was thinking about this in relation to
the high frequency skin effect. At a high enough frequency, the
skin depth should be 1 atom deep on the conductor, and then the electron
may have to travel on the "edge" only of the conductor, so
there could be zero losses from resistance and supercondivity.

The losses would be 100% from emission from the vibrating AC
electrons. At an infinite frequency the vibration amplitude of
the electrons would approach zero, and radiation losses would
approach zero. Does this make sense that losses from skin effect
could start to decrease or am I thinking about it completely backwards?!
Would it be possible to approach a high enough
frequency where the radiation losses could be recovered?

cheers,
Jaime

No. The skin effect you describe requires the current down inside to force
the current to the outside and the current down inside will eat power.

To test your idea, use free simulation tool, femm 4.2, and try it out,
you'll better understand where skin effect comes from. And learn a lot
about magnetics.

 

On Thu, 21 Nov 2013 19:45:17 -0800, DecadentLinuxUserNumeroUno
<DLU1@DecadentLinuxUser.org> wrote:

On Thu, 21 Nov 2013 19:42:24 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> Gave us:


My former boss's daughter came up with that one.(is that the right way
to pluralize 'boss'?). And she started out wanting to do Cosmology (not
cosmetology).


Not 'pluralize'! Possessive form! Doh!

The possessive form of boss is boss'. When the last letter is an 's'
or a 'z', whether it's plural or not, the possessive form has a
trailing apostrophe and no additional 's'.

 

[Phil Hobbs]

On 11/22/2013 01:32 PM, krw@attt.bizz wrote:

On Thu, 21 Nov 2013 19:45:17 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> wrote:

On Thu, 21 Nov 2013 19:42:24 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> Gave us:


My former boss's daughter came up with that one.(is that the right way
to pluralize 'boss'?). And she started out wanting to do Cosmology (not
cosmetology).


Not 'pluralize'! Possessive form! Doh!

The possessive form of boss is boss'. When the last letter is an 's'
or a 'z', whether it's plural or not, the possessive form has a
trailing apostrophe and no additional 's'.

Depends on whom you ask. Strunk & White says always to add "'s"
regardless. Chicago Manual of Style says to add "'s" if the noun is
singular, and just "'" if it's plural. The one constant in all this
confusion is RWWATP.(*)

Cheers

Phil Hobbs

(*) Real Writers Write Around The Problem.

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Jamie M]

On 11/22/2013 6:28 AM, RobertMacy wrote:

On Thu, 21 Nov 2013 19:32:34 -0700, Jamie M <jmorken@shaw.ca> wrote:

Hi,

I was reading about topological insulators which can super-conduct
electrons on their edges, I was thinking about this in relation to
the high frequency skin effect. At a high enough frequency, the
skin depth should be 1 atom deep on the conductor, and then the
electron may have to travel on the "edge" only of the conductor, so
there could be zero losses from resistance and supercondivity.

The losses would be 100% from emission from the vibrating AC
electrons. At an infinite frequency the vibration amplitude of
the electrons would approach zero, and radiation losses would
approach zero. Does this make sense that losses from skin effect
could start to decrease or am I thinking about it completely
backwards?! Would it be possible to approach a high enough
frequency where the radiation losses could be recovered?

cheers,
Jaime

No. The skin effect you describe requires the current down inside to
force the current to the outside and the current down inside will eat
power.

To test your idea, use free simulation tool, femm 4.2, and try it out,
you'll better understand where skin effect comes from. And learn a lot
about magnetics.

Hi,

Thanks I checked wiki: http://en.wikipedia.org/wiki/Skin_effect

The current down inside that forces the current to the outside, is the
AC induced eddy currents, and these eddy currents also move towards the
skin at higher frequency, so at some high enough frequency even the
eddy currents approach the outer layer of atoms on the conductor. Not
sure what would happen in this case, but probably the AC eddy currents
would create a plasma discharge or something off the surface?!

cheers,
Jamie

 

[DecadentLinuxUserNumeroUn]

On Fri, 22 Nov 2013 13:39:21 -0500, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> Gave us:

On 11/22/2013 01:32 PM, krw@attt.bizz wrote:
On Thu, 21 Nov 2013 19:45:17 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> wrote:

On Thu, 21 Nov 2013 19:42:24 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> Gave us:


My former boss's daughter came up with that one.(is that the right way
to pluralize 'boss'?). And she started out wanting to do Cosmology (not
cosmetology).


Not 'pluralize'! Possessive form! Doh!

The possessive form of boss is boss'. When the last letter is an 's'
or a 'z', whether it's plural or not, the possessive form has a
trailing apostrophe and no additional 's'.


Depends on whom you ask. Strunk & White says always to add "'s"
regardless. Chicago Manual of Style says to add "'s" if the noun is
singular, and just "'" if it's plural. The one constant in all this
confusion is RWWATP.(*)

Cheers

Phil Hobbs

(*) Real Writers Write Around The Problem.

By saying "The daughter of my former boss..."?

 

On Saturday, November 23, 2013 5:22:04 PM UTC+2, Robert Macy wrote:

I thought about that just after posting. Should check using FEA and a
model with a VERY thin surface layer of almost infinite conductivity and
see what happens. Something must not work here, else a single layer of
superconductivity would ALWAYS steal all the carriers so is not an easy
thing to even understand.

Current does flow on the surfaces in superconductors, even at DC.
The current layer has finite thickness, however, it's called the
London penetration depth.

The current tries to distribute itself across the cross-section
of the wire such that the stored energy is minimized. Each current
filament can lower its energy by moving away from other filaments.
One would then expect that all the filaments would move as far away
from each other as they can, i.e. to the surface of the conductor.

Anyway, there is only a finite density of superconducting charge
carriers available in a material. The smaller fraction of the
cross-sectional area the current filaments occupy, the smaller
number of carriers must carry the current, and the faster each
carrier has to move. Because the carriers have mass, there is
kinetic energy stored in their motion. This kinetic energy storage
looks to the driving electric circuit exactly like inductance,
hence it's called "kinetic inductance". However, the energy is stored
as kinetic energy, not in the magnetic field. Therefore the kinetic
inductance does not couple magnetically to other nearby inductors,
either.

So, there is a tendency of the current filaments to move farther
from each other, and thus reduce the stored magnetic energy. But
by doing so they induce a larger and larger kinetic energy storage
when the current gets packed to the surfaces. The two effects
balance when the current distribution reaches (in the case of an
infinite slab geometry IIRC) and exponentially decaying shape,
with decay length equaling the London penetration depth. For our
Nb thin films at LHe temperature this is roughly 90 nm.

Regards,
Mikko

 

[RobertMacy]

On Fri, 22 Nov 2013 16:30:44 -0700, Jamie M <jmorken@shaw.ca> wrote:

...snip...
Hi,

Thanks I checked wiki: http://en.wikipedia.org/wiki/Skin_effect

The current down inside that forces the current to the outside, is the
AC induced eddy currents, and these eddy currents also move towards the
skin at higher frequency, so at some high enough frequency even the
eddy currents approach the outer layer of atoms on the conductor. Not
sure what would happen in this case, but probably the AC eddy currents
would create a plasma discharge or something off the surface?!

cheers,
Jamie

I thought about that just after posting. Should check using FEA and a
model with a VERY thin surface layer of almost infinite conductivity and
see what happens. Something must not work here, else a single layer of
superconductivity would ALWAYS steal all the carriers so is not an easy
thing to even understand.

 

On Fri, 22 Nov 2013 13:39:21 -0500, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 11/22/2013 01:32 PM, krw@attt.bizz wrote:
On Thu, 21 Nov 2013 19:45:17 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> wrote:

On Thu, 21 Nov 2013 19:42:24 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> Gave us:


My former boss's daughter came up with that one.(is that the right way
to pluralize 'boss'?). And she started out wanting to do Cosmology (not
cosmetology).


Not 'pluralize'! Possessive form! Doh!

The possessive form of boss is boss'. When the last letter is an 's'
or a 'z', whether it's plural or not, the possessive form has a
trailing apostrophe and no additional 's'.


Depends on whom you ask. Strunk & White says always to add "'s"
regardless. Chicago Manual of Style says to add "'s" if the noun is
singular, and just "'" if it's plural. The one constant in all this
confusion is RWWATP.(*)

Cheers

Phil Hobbs

(*) Real Writers Write Around The Problem.

If there are standards on every side of the argument, it sounds like a
"real writer" can do it any way sheit pleases. "That's the nice thing
about standards..."

 

[Phil Hobbs]

On 11/23/2013 12:25 AM, DecadentLinuxUserNumeroUno wrote:

On Fri, 22 Nov 2013 13:39:21 -0500, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> Gave us:

On 11/22/2013 01:32 PM, krw@attt.bizz wrote:
On Thu, 21 Nov 2013 19:45:17 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> wrote:

On Thu, 21 Nov 2013 19:42:24 -0800, DecadentLinuxUserNumeroUno
DLU1@DecadentLinuxUser.org> Gave us:


My former boss's daughter came up with that one.(is that the right way
to pluralize 'boss'?). And she started out wanting to do Cosmology (not
cosmetology).


Not 'pluralize'! Possessive form! Doh!

The possessive form of boss is boss'. When the last letter is an 's'
or a 'z', whether it's plural or not, the possessive form has a
trailing apostrophe and no additional 's'.


Depends on whom you ask. Strunk & White says always to add "'s"
regardless. Chicago Manual of Style says to add "'s" if the noun is
singular, and just "'" if it's plural. The one constant in all this
confusion is RWWATP.(*)

Cheers

Phil Hobbs

(*) Real Writers Write Around The Problem.


By saying "The daughter of my former boss..."?

Sure, something like that. Usenet posts aren't important enough for the
bother, but in an article or a book, it's an essential skill.

Cheers

Phil Hobbs

 

[Jamie M]

On 11/23/2013 8:16 AM, reg@wmail.fi wrote:

On Saturday, November 23, 2013 5:22:04 PM UTC+2, Robert Macy wrote:
I thought about that just after posting. Should check using FEA and a
model with a VERY thin surface layer of almost infinite conductivity and
see what happens. Something must not work here, else a single layer of
superconductivity would ALWAYS steal all the carriers so is not an easy
thing to even understand.

Current does flow on the surfaces in superconductors, even at DC.
The current layer has finite thickness, however, it's called the
London penetration depth.

The current tries to distribute itself across the cross-section
of the wire such that the stored energy is minimized. Each current
filament can lower its energy by moving away from other filaments.
One would then expect that all the filaments would move as far away
from each other as they can, i.e. to the surface of the conductor.

Anyway, there is only a finite density of superconducting charge
carriers available in a material. The smaller fraction of the
cross-sectional area the current filaments occupy, the smaller
number of carriers must carry the current, and the faster each
carrier has to move. Because the carriers have mass, there is
kinetic energy stored in their motion. This kinetic energy storage
looks to the driving electric circuit exactly like inductance,
hence it's called "kinetic inductance". However, the energy is stored
as kinetic energy, not in the magnetic field. Therefore the kinetic
inductance does not couple magnetically to other nearby inductors,
either.

So, there is a tendency of the current filaments to move farther
from each other, and thus reduce the stored magnetic energy. But
by doing so they induce a larger and larger kinetic energy storage
when the current gets packed to the surfaces. The two effects
balance when the current distribution reaches (in the case of an
infinite slab geometry IIRC) and exponentially decaying shape,
with decay length equaling the London penetration depth. For our
Nb thin films at LHe temperature this is roughly 90 nm.

Regards,
Mikko

Hi,

I think a microwave waveguide is acting kind of like a topological
insulator too, the microwaves reflect off the surface and don't
penetrate, but still there is some interaction with the surface
atoms that is near superconducting type.

cheers,
Jamie

 

On Sunday, November 24, 2013 7:39:33 AM UTC-8, dave wrote:

On 11/24/2013 05:51 AM, William Sommerwerck wrote:

You might want to look at X10 modules. X10 has two-way switches.
It's not so much whether a switch system will work with CFLs, but
whether CFLs will work with switch systems. Most X10 modules can switch
CFLs, and dim some * models.
* Not to be confused with the food.

Get an appliance module if you want hard on/off

DIY if you want better lightning control. For example, put in a RGB LED light night. Turn it blue at night, unless you (or your wife) wants it in red.

 

Ok, I don't have any interest in dimming. What I see is most of the
simple remote switches are for incandescent bulbs or resistive loads.

So, where does this leave me?
Also, I was posting two way switch, i just realized what I want is a
three way switch system.

All you need is 2 relays. For example, one for 50W and one for 100W, and 150W together. So, you only need 2 bits. Plus another 3 bits for RGB LED.

I wired up an RF remote light for someone. They are more interested in the color LED. They don't seem to care if the main relays are working or not, but the color LED has to work.

 

[whit3rd]

On Thursday, November 21, 2013 9:20:54 PM UTC-8, bitrex wrote:

Use 4 optoisolated current sources/sinks under microprocessor control,

and sense the terminal voltage/current at each of the two ports. Apply

the appropriate passive device defining equations in software...

A Howland type current source can implement the voltage-to-current
function mainly effectively; you only need a couple of those (one dual
op amp) to make your voltage-output DAC do two bipolar current sources.
If op amp inputs can sense the voltages, the compliance range of the
current sources isn't so large an op amp cannot handle it.

 

[Rajdeep shaktawat]

There are also so many training centers who teaches to make such industrial things. And many manufacturer to such industrial instruments like temperature sensors and Thermal Imaging Camera such as, http://accuratesensors.com/thermal-imager-thermeye160.html

 

[Rajdeep shaktawat]

There are also so many training centers who teaches to make such industrial things. And many manufacturer produce such industrial instruments like temperature sensors and Thermal Imaging Camera such as, http://accuratesensors..com/thermal-imager-thermeye160.html

 

[Rajdeep shaktawat]

On Monday, November 25, 2013 4:35:28 PM UTC+5:30, John S wrote:

On 11/25/2013 4:16 AM, Rajdeep shaktawat wrote:

There are also so many training centers who teaches to make such industrial things. And many manufacturer to such industrial instruments like temperature sensors and Thermal Imaging Camera such as, http://accuratesensors.com/thermal-imager-thermeye160.html





This is an ad by deep tawat?

No, its just a recommendation John, nothing more.

 

[Jan Panteltje]

On a sunny day (Sun, 24 Nov 2013 19:30:52 -0700) it happened hamilton
<hamilton@nothere.com> wrote in <l6ucpa$s87$1@dont-email.me>:

http://www.washingtonpost.com/business/technology/new-research-aims-to-teach-computers-common-sense/2013/11/24/c7806b98-552e-11e3
-bdbf-097ab2a3dc2b_story.html

Warning
Washington post was bought by Amazon founder Jeff Bezos
He works for NSA, never makes profit, just wants an influence channel.

I have stopped reading it the day before he bought it.
Maybe try reuters.com