Marriage is under fire!!

On 5 Sep 2004 20:47:00 -0700, tatto0_2000@yahoo.com (Wong) wrote:

Hi,
Based on this circuit, I have some strange results from simulations.

5V
___
|
|
\
/ 2K
\
|
|--------------------------o C
| |
| |
\ |
/ 2K |
\ |
| |
| |
| |
o o
A B
----------------------------
A B C
0v 0v 0v
5v 0v 0v
x 5v 5v (something wrong here)
----------------------------
I think that when point B is 5V when point A either 5 or 0v, point C
should not get 5v as well. But I just cant figure out why the
simulator behave like this.
I am using Electronics Workbench 5.12. Thanks in advance
Aren't B and C the same node?

John
 
You might want to check out

http://www.streamlight.com

http://www.flashlights.com

Streamlight makes a 2 AA Luxeon flashlight
and a dimmable 3 AAA Luxeon headlight. I believe the flashlight
has a DC-DC voltage converter, so it runs well on NiMH rechargeables.


"BruceW..1" <sorry@noDirectEmail.com> writes:
I'd like to power a 1-watt Luxeon Star
http://www.lumileds.com/products/family.cfm?familyId=2
with two AA batteries.

And I'd like to be able to dim the light. I've found that the Luxeon
Star puts out useable light down to 50mA, or up to about 350 mA.

This is for a headlamp to be used for camping and backpacking. Most
manufacturers of these headlamps use 3 batteries and a voltage dropping
resistor. IMHO, this is pretty cheesy and besides I'd like to mod one
of my existing AA headlamps.

Actually I'd like to make it run off of a single AA battery because I'd
rarely use full brightness.

Has Maxim come out with any new IC'c in the last year that might be
right for the job? I just ran across this, don't know what's inside,
but it doesn't dim:
http://www.luxeonstar.com/item.php?id=415&link_str=89&partno=02009A

I'm no electronics expert but I'm smarter than the average bear in this
respect and I can solder.

Can anyone recommend a circuit?

Thanks for your help.
 
Wong wrote:
But with point A equal to 0v. The node between 2 resistors should have
2.5v according to the voltage divider rule. So this is something that
I am not so sure with the 'mix' of 2.5v and 5v (point B).
Is point C still 5v under this circumstances ?
If the ASCII drawing is correct, B and C will always be the same; they are
directly connected.

--
John Miller
Email address: domain, n4vu.com; username, jsm

A diplomat is a man who can convince his wife she'd look stout in a fur
coat.
 
"BruceW..1" <sorry@noDirectEmail.com> wrote:

I'd like to power a 1-watt Luxeon Star
http://www.lumileds.com/products/family.cfm?familyId=2
with two AA batteries.

And I'd like to be able to dim the light. I've found that the Luxeon
Star puts out useable light down to 50mA, or up to about 350 mA.
The Linear Technology LT1618

http://www.linear.com/prod/datasheet.html?datasheet=736

Looks like a useful part for this application. Unfortunately like most
switchers for portable applications it is only available in tiny surface
mount packages.
 
Do you have a connection to ground in your circuit? Electronics Workbench
(Multisim) and similar simulators need one, in order to calculate voltages.
 
Wong wrote:

But with point A equal to 0v. The node between 2 resistors should have
2.5v according to the voltage divider rule. So this is something that
This is correct, as long as *no voltage* is being
applied at point B. The resistance between B and C
is 0 ohms, so Vc will always equal Vb.

I am not so sure with the 'mix' of 2.5v and 5v (point B).
Is point C still 5v under this circumstances ?
The voltage chart is correct for the circuit as drawn.

If you connect B to 5v, you're putting a 0 (that's zero)
ohm resistance in parallel with the upper 2k resistor.
What does your 'voltage divider rule' say about this?

If you connect B to A (0v), you're putting a 0 (that's zero)
ohm resistance in parallel with the *lower* 2k resistor.
What does your 'voltage divider rule' say about this?


John Larkin <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in message news:<7conj0lfr533flgsufs1u5p0s18kg6p6qd@4ax.com>...

On 5 Sep 2004 20:47:00 -0700, tatto0_2000@yahoo.com (Wong) wrote:


Hi,
Based on this circuit, I have some strange results from simulations.

5V
___
|
|
\
/ 2K
\
|
|--------------------------o C
| |
| |
\ |
/ 2K |
\ |
| |
| |
| |
o o
A B
----------------------------
A B C
0v 0v 0v
5v 0v 0v
x 5v 5v (something wrong here)
----------------------------
I think that when point B is 5V when point A either 5 or 0v, point C
should not get 5v as well. But I just cant figure out why the
simulator behave like this.
I am using Electronics Workbench 5.12. Thanks in advance

Aren't B and C the same node?

John
 
On 6 Sep 2004 03:04:46 -0700, tatto0_2000@yahoo.com (Wong) wrote:

But with point A equal to 0v. The node between 2 resistors should have
2.5v according to the voltage divider rule.

It would if nothing else is connected to it. But when you apply a hard
voltage source to the junction, the source wins.

Please don't top post.

John
 
eh?
Point B and C are connected! why shouldn't they be the same?
just think about that.
if Point B is applied voltage from a source its ovbious that
Point C will look the same.



Wong wrote:
Hi,
Based on this circuit, I have some strange results from simulations.

5V
___
|
|
\
/ 2K
\
|
|--------------------------o C
| |
| |
\ |
/ 2K |
\ |
| |
| |
| |
o o
A B
----------------------------
A B C
0v 0v 0v
5v 0v 0v
x 5v 5v (something wrong here)
----------------------------
I think that when point B is 5V when point A either 5 or 0v, point C
should not get 5v as well. But I just cant figure out why the
simulator behave like this.
I am using Electronics Workbench 5.12. Thanks in advance
 
Hi,

The screw only touches the ground pad at first layer.
No full ground plane is used. Insulation of screw from ground pad or
removal of screw can solve distortion problem. It's still a mystery to
me.


I agree with John, this sure sounds like a ground loop. It can be very
hard to make a RF/controller combination work reliably without a full
ground plane. I would probably spring for a relayout with a full ground
plane even if that means one or two more layers on the boards. The other
issue with this unit might be EMC, especially with a plastic enclosure.

Regards, Joerg

http://www.analogconsultants.com
 
Navid Naghdi wrote:
Some tips will be appreciated here from the following question:

A rocket being prepared for launch has two crewmembers, each of whom
has a two-position switch, one position marked 'hold' and the other
marked 'go'. Draw a circuit so that:

(a) - A hold light comes on when either crew member has their
switch in the 'hold' position.
(b) - A 'go' light comes on only if both crewmembers have their
switch in the 'go' position.
<sigh> Is it that time of year already?

This ain't alt.get.your.homework.done.for.free, son.

Crack open your textbook, show us *your* design and
we'll critique it...
 
In sci.electronics.design BruceW..1 <sorry@nodirectemail.com> wrote:
I'd like to power a 1-watt Luxeon Star
http://www.lumileds.com/products/family.cfm?familyId=2
with two AA batteries.

And I'd like to be able to dim the light. I've found that the Luxeon
Star puts out useable light down to 50mA, or up to about 350 mA.
It'll put out a lot more light at low average currents if you keep the
peak current at 350ma.
This is for a headlamp to be used for camping and backpacking. Most
manufacturers of these headlamps use 3 batteries and a voltage dropping
resistor. IMHO, this is pretty cheesy and besides I'd like to mod one
of my existing AA headlamps.

Actually I'd like to make it run off of a single AA battery because I'd
rarely use full brightness.
For most of the SMPS chips available, they are significantly more efficient
at a battery voltage of 2.4V, rather than 1.2V.
You can probably get away with a step-up only device, 2 fresh alkaline AA
cells won't provide enough current at ~3.5V to kill the luxeon device, even
neglecting diode drops.
Synch rectification will buy you a little bit, but not much.
 
In sci.electronics.design Ian Stirling <root@mauve.demon.co.uk> wrote:
In sci.electronics.design BruceW..1 <sorry@nodirectemail.com> wrote:
I'd like to power a 1-watt Luxeon Star
http://www.lumileds.com/products/family.cfm?familyId=2
with two AA batteries.

And I'd like to be able to dim the light. I've found that the Luxeon
Star puts out useable light down to 50mA, or up to about 350 mA.

It'll put out a lot more light at low average currents if you keep the
peak current at 350ma.

This is for a headlamp to be used for camping and backpacking. Most
manufacturers of these headlamps use 3 batteries and a voltage dropping
resistor. IMHO, this is pretty cheesy and besides I'd like to mod one
of my existing AA headlamps.
What may also be useful would be a 3 cell device.
2 AA 1.5V cells, and 1AA 3V cell, for emergencies.
 
zoid999@excite.com (Navid Naghdi) wrote:

Some tips will be appreciated here from the following question:

A rocket being prepared for launch has two crewmembers, each of whom
has a two-position switch, one position marked 'hold' and the other
marked 'go'. Draw a circuit so that:

(a) - A hold light comes on when either crew member has their
switch in the 'hold' position.
(b) - A 'go' light comes on only if both crewmembers have their
switch in the 'go' position.
Sounds like you need to review the concepts of series and parallel...

More about me: http://www.jecarter.com/
VB3/VB6/C/PowerBasic source code: http://www.jecarter.com/programs.html
Freeware for the Palm with NS Basic source code: http://nsb.jecarter.com
Drivers for Pablo graphics tablet and JamCam cameras: http://home.earthlink.net/~mwbt/
Email here: http://www.jecarter.com/contactme.htm
 
In article <413ce387$0$24441$ed2619ec@ptn-nntp-reader03.plus.net>, Ian
Stirling wrote in part:

It'll put out a lot more light at low average currents if you keep the
peak current at 350ma.
I have found the efficiency of red, orange and yellow Luxeons to be
maximized at instantaneous currents somewhat less, maybe around 200 mA.
For white and blue ones, the efficiency is maximized at even lower
instantaneous currents around 100 mA or less.
If your average current is around or over 50 mA and the Luxeon is a
white one, I recommend steady rather than pulsating current. But if the
average current is only a few milliamps, then you are better off using
pulses with a higher instantaneous current - perhaps 50-100 mA.

- Don Klipstein (don@misty.com)
 
Which professor assigned this as your homework?

This is not a very difficult problem. I'm not sure why you're asking for
help.

"Navid Naghdi" <zoid999@excite.com> wrote in message
news:26278ab4.0409061301.43bc2d8f@posting.google.com...
Some tips will be appreciated here from the following question:

A rocket being prepared for launch has two crewmembers, each of whom
has a two-position switch, one position marked 'hold' and the other
marked 'go'. Draw a circuit so that:

(a) - A hold light comes on when either crew member has their
switch in the 'hold' position.
(b) - A 'go' light comes on only if both crewmembers have their
switch in the 'go' position.
 
Rich Webb wrote:
On Sun, 5 Sep 2004 16:12:27 -0400, "Rodney Kelp"
Maybe this is off topic but does anybody know how you convert the gps
position coordinace to distance? For instance If I am currently at
44 degrees 18 minutes 35.3 seconds lattitude and 69 46 52.4 longtitude
and I move to 44 18 35.3 by 69 46 54.0 how many feet have I moved
easterly? I can't seem to find any conversion tables that convert degrees,
minutes, seconds to feet. Am I expecting too much?
For a local approximation, consider 1 minute of latitude to be exactly
1852 meters, which is 1 nautical mile or about 6076 feet. Then 1 minute
of longitude is 1852 meters times the cosine of your latitude. Neither
is strictly true (we live on a lumpy planet) but close enough.
So, you moved 1.6 seconds of arc (0.027 minutes) to the east (assuming
your positions above are N and W) or about 116 feet.

A "flat earth" approximation like this fine as long as you stay in
about the same general area. In the example above, if the starting
points had been 1 nm (1 minute) further north then the change in
distance traveled to the east would be less than an inch.
A profile view of the earth, exaggerated, would look like the
3-lobed rotor of a Wankel engine, but rounded, of course. I remember
when I was quite young that there was an announcement that scientists
had discovered that the earth bulged, but that the bulge is south of
the equator. And it's kind of elongated at the north pole. One thing
that makes this kinda cool for me is that one time in the USAF I was
BSing with some guy who turned out to be in on the early missile tests
where they discovered the bulge. He said that there was this one series
of flights that kept landing short and to the south of where they had
calculated. (or short and north, whatever). He also said that in one
of their experiments they put a whole rocket in orbit, like, one stage.
At the time, I had thought that that was impossible, like everybody
else. Presumably, it's still up there. :)

Cheers!
Rich
 
"Rodney Kelp" <Rodneykelp605@hotmail.com> wrote in message
news:U_ydnQoH4se47qbcRVn-gw@adelphia.com...
Maybe this is off topic but does anybody know how you convert the gps
position coordinace to distance? For instance If I am currently at
44 degrees 18 minutes 35.3 seconds lattitude and 69 46 52.4 longtitude
and
I move to 44 18 35.3 by 69 46 54.0 how many feet have I moved easterly?
I
can't seem to find any conversion tables that convert degrees, minutes,
seconds to feet. Am I expecting too much?
My GPS receiver will give me coordinance and speed in miles per hour
and
direction but not the distance traveled. What's up with that?
I'm trrying to measure some acerage. I feel like I am going to feel
really
stupid when someone tells me.
Rodney,

For great circle distance computations it goes like this:

Distance = 69.1 x (180/pi) x arccos | sin(lat1) x sin(lat2) + cos(lat1) x
cos(lat2) x cos(long2 - long1) |
 
"billy_evil" <billy_evil@yahoo.com> wrote in message
news:bbd2a5b4.0409061853.1cf01d6b@posting.google.com...
What are the differences between 4-bit and 8-bit MCU?
Size of the registers.
 
[snip]

You might try my ascii schematic drawing program. I think you will find it
easier to use than Andy's and it has a few more features.

It's at www.fidalgo.net/~garyr/pyascii/

The executable and associated files are in PyAscii_04_Win9x.zip (~ 2 MB)

The source files are in PyAscii_04_source.zip

Just click on the filename to start the download.

Gary Richardson
 

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