Driver to drive?

[Ralph Barone]

Bill Sloman <bill.sloman@gmail.com> wrote:

On Thursday, 17 April 2014 17:09:15 UTC+10, whit3rd wrote:
On Wednesday, April 16, 2014 6:12:40 PM UTC-7, Ralph Barone wrote:
My suspicion is that the increased burden in the primary circuit from
adding a Hall effect sensor is probably unmeasurable, which is why I asked
about a theoretical solution. And it is just out of curiosity, but I
thought that somebody else with better analytical chops than me and better
knowledge of Hall effect device design might take up the challenge.

I'm pretty sure the Hall effect is due to electron drift velocity
and deflection of the moving electrons by a magnetic field. There's
no work whatever done by the magnetic field on such a moving charge.

IIRR Hall-effect sensors depend on having roughly equal currents being
carried by positive and negative charge-carriers.

The work being done by those charge carriers is supplied by your
measuring circuit and has no effect on the load.

Getting a magnetic field for the Hall effect sensor to detect could
involve forming the current carrying conductor into a loop, which would
add inductance to the circuit being measured, but you can just measure
the magnetic field being created around a straight wire, and the
inductance of a straight wire is around 5nH per cm, which isn't much.

http://www.k7mem.com/Electronic_Notebook/inductors/straight_wire.html

snip

Well, so much for my intuitive conclusion that Hall effect sensors have
burden. I suppose if that were true, one could cool a permanent magnet by
surrounding it with Hall effect sensors. Thanks all. That's one more small
piece of ignorance removed.

 

[Phil Hobbs]

On 4/18/2014 12:14 PM, Jim Thompson wrote:

On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

--
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

 

[Jim Thompson]

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
<hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

So you get something for nothing?

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[John Larkin]

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
<hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

Unless the acceleration on a charged particle radiates a photon.

If it does, the particle must slow down. Is that the reaction to the
momentum of the photon? Must be. So that is why the photon is emitted
in the direction of the particle motion?




--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[John Larkin]

On Fri, 18 Apr 2014 14:30:13 -0700, John Larkin
<jlarkin@highlandtechnology.com> wrote:

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

Unless the acceleration on a charged particle radiates a photon.

If it does, the particle must slow down. Is that the reaction to the
momentum of the photon? Must be. So that is why the photon is emitted
in the direction of the particle motion?

Which is why photons must have momentum.


--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[Martin Riddle]

On Fri, 18 Apr 2014 10:35:26 -0400, WangoTango
<Asgard24@mindspring.com> wrote:

In article <grl0l9hln3cqsr88vdaaqensjcpk5mkn8u@4ax.com>,
martin_rid@verizon.net says...
On Thu, 17 Apr 2014 13:56:01 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 17 Apr 2014 13:12:08 -0500, Tim Wescott
tim@seemywebsite.really> wrote:

On Thu, 17 Apr 2014 09:26:12 -0700, John Larkin wrote:

On Thu, 17 Apr 2014 00:03:04 -0500, Tim Wescott
tim@seemywebsite.really> wrote:

I have a customer who wants a USB-powered battery charger designed, with
certification -n- all. I figure the certification part will be harder
than the charger part, so I have to give it a pass.

Anyone do that and have spare cycles, or know someone? He wants someone
with a track record, or I'd talk him into using me!

What certs? UL/CSA/CE? FCC?

A test lab will do those, for a moderate pile of money.

Is there a USB certification standard?

You have to pass their compatibility tests if you want to use their logos
& such. I'm not sure whether you can even use "USB", but I suspect by
now that you can if you use the right wording.

You can use something like the FTDI chips and (maybe) inherit the
certs.

I think its $3K just for the PID and VID alone.

They have upped it to $5K.
We finally got into doing enough USB devices that I decided to pop out
the $2K they were asking for a VID (you make up the PID yourself), and
wouldn't you know it, they had increased the price to $5K just in time
for me to give them some money. That's my lot.
Naturally, they want you to sign on for the $4K a year subscription that
gets you logo use and so on.

I wonder if you go with the PID from microchip ( if they still offer
it) that you can get a cert from usb.org with it. Same for FTDI.

Cheers

 

[John Larkin]

On Thu, 17 Apr 2014 18:37:09 -0400, Martin Riddle
<martin_rid@verizon.net> wrote:

On Thu, 17 Apr 2014 13:56:01 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Thu, 17 Apr 2014 13:12:08 -0500, Tim Wescott
tim@seemywebsite.really> wrote:

On Thu, 17 Apr 2014 09:26:12 -0700, John Larkin wrote:

On Thu, 17 Apr 2014 00:03:04 -0500, Tim Wescott
tim@seemywebsite.really> wrote:

I have a customer who wants a USB-powered battery charger designed, with
certification -n- all. I figure the certification part will be harder
than the charger part, so I have to give it a pass.

Anyone do that and have spare cycles, or know someone? He wants someone
with a track record, or I'd talk him into using me!

What certs? UL/CSA/CE? FCC?

A test lab will do those, for a moderate pile of money.

Is there a USB certification standard?

You have to pass their compatibility tests if you want to use their logos
& such. I'm not sure whether you can even use "USB", but I suspect by
now that you can if you use the right wording.

You can use something like the FTDI chips and (maybe) inherit the
certs.

I think its $3K just for the PID and VID alone.

Cheers

We've done USB boxes with both the FTDI chips and using the USB
hardware inside an NXP uP. We use their VIDs. Seems to work fine.


--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[Phil Hobbs]

On 4/18/2014 4:50 PM, Jim Thompson wrote:

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

So you get something for nothing?

...Jim Thompson

No. If you try drawing current from a Hall sensor, it does work because
there's a component of the current in the direction of the E field.

Cheers

Phil Hobbs

--
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

 

[Jasen Betts]

On 2014-04-18, Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

So you get something for nothing?

No. Immagine a hall sensor with a permanent magnet.



--
umop apisdn


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[josephkk]

On Fri, 18 Apr 2014 13:03:27 -0700 (PDT), panfilero <panfilero@gmail.com>
wrote:

>How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?

Yes. And there are a handful or so of new versions tailored specifically
to that purpose. Search for "positive rail current monitor".


dn apisumop

?-)

 

[josephkk]

On Fri, 18 Apr 2014 11:00:00 -0700 (PDT), whit3rd <whit3rd@gmail.com>
wrote:

On Thursday, April 17, 2014 9:05:59 PM UTC-7, josephkk wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com

Magnetic force on a moving charge is perpendicular to velocity,

the power is zero because F-vector and V-vector are orthogonal.



Stuff and nonsense. If you change the path of a particle you have

accelerated it. That takes work.

Rethink that. The moon's straight path has been 'changed' into an ellipse by
Earth's gravity, but the continual acceleration does NO net work on
the moon. Similarly, a permanent magnet placed near a Hall sensor
will give a positive Hall effect indication for an indefinite period of time.

The magnet won't go flat. And, the moon isn't falling from the sky.
The word 'work' in physics has a specific energy-is-transferred meaning.

Thimking.

?-)

 

[Phil Hobbs]

On 4/18/2014 5:49 PM, John Larkin wrote:

On Fri, 18 Apr 2014 14:30:13 -0700, John Larkin
jlarkin@highlandtechnology.com> wrote:

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

Unless the acceleration on a charged particle radiates a photon.

If it does, the particle must slow down. Is that the reaction to the
momentum of the photon? Must be. So that is why the photon is emitted
in the direction of the particle motion?

Which is why photons must have momentum.


Classical fields have momentum too.

Cheers

Phil Hobbs

--
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

 

[Phil Hobbs]

On 4/18/2014 5:30 PM, John Larkin wrote:

On Fri, 18 Apr 2014 16:10:34 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:

On 4/18/2014 12:14 PM, Jim Thompson wrote:
On 18 Apr 2014 04:39:22 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-18, josephkk <joseph_barrett@sbcglobal.net> wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

Magnetic force on a moving charge is perpendicular to velocity,
the power is zero because F-vector and V-vector are orthogonal.

Stuff and nonsense. If you change the path of a particle you have
accelerated it.
That takes work.

Only if the applied force is in the direction of the motion, which is
never the case in Hall-effect cells.

?-)

Sounds like you need some remedial work in vector mathematics (and the
"right-hand rule"

...Jim Thompson

No, he's quite right. Work is a dot product, viz. force dot distance,
whereas Lorentz (magnetic) force goes as v cross B.

The dot product v dot (v cross B) is identically zero.

Otherwise, the Sun would be doing work on the Earth by bending its path
into an ellipse.

Cheers

Phil Hobbs

Unless the acceleration on a charged particle radiates a photon.

If it does, the particle must slow down. Is that the reaction to the
momentum of the photon? Must be. So that is why the photon is emitted
in the direction of the particle motion?

Nonrelativistic particles radiate like dipoles, i.e. the pattern has a
null along the direction of motion and a doughnut wrapped round it.
Relativistic particles do the same in their co-moving frame, but in the
lab frame, the radiation is concentrated near the direction of motion.
It's an effect of Lorentz transformation.

However, there's still a null in the actual direction of motion.

Cheers

Phil Hobbs



--
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

 

[panfilero]

On Friday, April 18, 2014 8:51:27 PM UTC-5, josephkk wrote:

On Fri, 18 Apr 2014 13:03:27 -0700 (PDT), panfilero <@gmail.com

wrote:



How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?



Yes. And there are a handful or so of new versions tailored specifically

to that purpose. Search for "positive rail current monitor".





dn apisumop



?-)

Positive rail current monitor? Like a high side shunt monitor? Does this apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?

 

On Sat, 19 Apr 2014 19:41:30 -0700 (PDT), panfilero
<panfilero@gmail.com> wrote:

On Friday, April 18, 2014 8:51:27 PM UTC-5, josephkk wrote:
On Fri, 18 Apr 2014 13:03:27 -0700 (PDT), panfilero <@gmail.com

wrote:

How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?


Yes. And there are a handful or so of new versions tailored specifically

to that purpose. Search for "positive rail current monitor".



dn apisumop


?-)

Positive rail current monitor? Like a high side shunt monitor? Does this apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?

Most likely just for DC.

The simples high side current monitor would be shunt between a PNP
transistor emitter and base and a high resistance from collector to
ground. Split the resistance into two resistors and get a nice voltage
(say 0..5 V) from that voltage divider proportional to the current
through the shunt.

For AC only, simply use a current transformer.

For AC+DC, use a isolated power supply to feed a V/f chip sitting on
the shunt and send down the frequency modulated signal through
capacitors and/or isolation transformers and then apply f/V
conversion.

For real high common mode voltages, say 400 kVac, use a laser injected
into one fiber to power the current measurement device sitting on the
EHT line, then use an other fiber to digitally transmit down the
measurement.

 

[Maynard A. Philbrook Jr.]

In article <vip3l9tnabqj3ovmrdn2v3rh64n5sts0i4@4ax.com>,
joseph_barrett@sbcglobal.net says...

On Fri, 18 Apr 2014 11:00:00 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

On Thursday, April 17, 2014 9:05:59 PM UTC-7, josephkk wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com

Magnetic force on a moving charge is perpendicular to velocity,

the power is zero because F-vector and V-vector are orthogonal.



Stuff and nonsense. If you change the path of a particle you have

accelerated it. That takes work.

Rethink that. The moon's straight path has been 'changed' into an ellipse by
Earth's gravity, but the continual acceleration does NO net work on
the moon. Similarly, a permanent magnet placed near a Hall sensor
will give a positive Hall effect indication for an indefinite period of time.

The magnet won't go flat. And, the moon isn't falling from the sky.
The word 'work' in physics has a specific energy-is-transferred meaning.


Thimking.

?-)

I think it's one of them quantum thingies!

Maybe zero point energy!

Jamie

 

On Sat, 19 Apr 2014 19:41:30 -0700 (PDT), panfilero
<panfilero@gmail.com> wrote:

On Friday, April 18, 2014 8:51:27 PM UTC-5, josephkk wrote:
On Fri, 18 Apr 2014 13:03:27 -0700 (PDT), panfilero <@gmail.com

wrote:



How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?



Yes. And there are a handful or so of new versions tailored specifically

to that purpose. Search for "positive rail current monitor".





dn apisumop



?-)

Positive rail current monitor? Like a high side shunt monitor? Does this apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?

Not the safest thing to do but with the appropriate gain, the high
common mode voltage is no problem. If you have a gain of 1/50, the
common mode voltage scales inside a single 5V rail. Of course the
differential signal is also divided by 50 and common mode rejection is
dependent on resistor tolerances so there may be other problems.

 

[pedro]

On Sun, 20 Apr 2014 17:57:33 -0700, Jeff Liebermann <jeffl@cruzio.com>
wrote:

When RoHS appeared, I expected to have Pb/Sn solder declared a
controlled or hazardous substance, making availability difficult. So,
I stocked up with four 1 lb rolls of various sizes of 63/37 activated
rosin core solder

You and a lot of others with a history of rework ....

 

[josephkk]

On Sat, 19 Apr 2014 22:40:31 -0400, "Maynard A. Philbrook Jr."
<jamie_ka1lpa@charter.net> wrote:

In article <vip3l9tnabqj3ovmrdn2v3rh64n5sts0i4@4ax.com>,
joseph_barrett@sbcglobal.net says...

On Fri, 18 Apr 2014 11:00:00 -0700 (PDT), whit3rd <whit3rd@gmail.com
wrote:

On Thursday, April 17, 2014 9:05:59 PM UTC-7, josephkk wrote:
On Thu, 17 Apr 2014 00:09:15 -0700 (PDT), whit3rd <whit3rd@gmail.com

Magnetic force on a moving charge is perpendicular to velocity,

the power is zero because F-vector and V-vector are orthogonal.



Stuff and nonsense. If you change the path of a particle you have

accelerated it. That takes work.

Rethink that. The moon's straight path has been 'changed' into an ellipse by
Earth's gravity, but the continual acceleration does NO net work on
the moon. Similarly, a permanent magnet placed near a Hall sensor
will give a positive Hall effect indication for an indefinite period of time.

The magnet won't go flat. And, the moon isn't falling from the sky.
The word 'work' in physics has a specific energy-is-transferred meaning.


Thimking.

?-)


I think it's one of them quantum thingies!

Maybe zero point energy!

Jamie

It actually works for classical physics as well. This is one of my
problem areas going through college.

?-)

 

[josephkk]

On Sat, 19 Apr 2014 19:41:30 -0700 (PDT), panfilero <panfilero@gmail.com>
wrote:

On Friday, April 18, 2014 8:51:27 PM UTC-5, josephkk wrote:
On Fri, 18 Apr 2014 13:03:27 -0700 (PDT), panfilero <@gmail.com

wrote:



How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?



Yes. And there are a handful or so of new versions tailored specifically

to that purpose. Search for "positive rail current monitor".





dn apisumop



?-)

Positive rail current monitor? Like a high side shunt monitor? Does this apply for AC? Both sides see high common mode voltage swings, +/- 110Vac ? What opamp can survive that?

Only for DC rails so far as i have seen.

?-)