Chip with simple program for Toy

[Claude]

Here is what I am using, it doesn't get any simpler. It is a course book of
57 lessons and you buy the component kit that comes with it in a nice little
tool box complete with breadboard and jumpers. The book and kit are 100%
compatible. The kit is exclusively sold by www.abra-electronics.com The kit
is loaded with all of the required components and cost approximately 35%
less than if bought separately. The book covers some theory then immediately
has you breadboarding the lesson. They start with the schematic and pictures
of the breadboard then wane you off of the breadboard pictures. The book has
a few errors but the author maintains the required corrections on his site.
The course takes about a 100 hours to get through if you are passionate.
There was an error in the NOR and NAND gates lesson that threw me out for 2
days until I checked the site ( schematics where reversed)

Check out the author's site for more questions, answers and examples.
www.elxevilgenius.com

There is an Abra in Montreal Canada and one in the state of New-York, same
owner and they do most of their business on-line. Very reputable and cater
mostly to educational institutions.

The way the Navy course suggested in the other posts is outstanding!!!!! If
there is one thing where the military excel in it is course design. It is
thousands of pages so when I get confused on a concept in my Evil Genius
course I check out the Navy course. It is a theory course, no practical
experiments at all.

Are you lazy? I know I am so there is a 40 hour electronics course on
Youtube given by a university in India. The teacher Mr Natarajan has a bit
of an accent but the course is awesome. He power points the theory then
breadboards everything in front of you. Those 40 hours are worth 2 years of
school!!!!! The reason they did this very professional course is to be able
to train people in remote areas of India who can't get to a proper
institution. I would have paid good coin if they had charged for this
course.

If nothing above helps consider knitting )

Claude
Montreal

<danikar@gmail.com> wrote in message
news:217a24f3-3235-4c06-a004-bc7b6d847b1b@q27g2000prf.googlegroups.com...

I am looking to learn more about electronics and what not. I have
looked at a few books and purchased one. However, I didn't like it
that much.

I got Electronic Projects for Dummies. The book started off like I was
going to understand it, then I got to chapter 5 and it asked me to buy
a bunch of components to put together this project. Some of the items
on the list I couldn't find even in the online stores that they
suggested in the book. The project was a little above me anyway.

What I am looking for is a book that will explain a concept to me,
like capacitors. Then give a few circuits that demonstrate what they
just explained to me. Is there anything like that? Or, do you
generally have to read through a book that is all explanation then get
another book that has circuits you can play with?

Thanks in advanced!

 

[Ken Fowler]

On 2-Jun-2008, Freelance Embedded Systems Engineer <g9u5dd43@yahoo.com>
wrote:

danikar@gmail.com wrote:
I am looking to learn more about electronics and what not. I have
looked at a few books and purchased one. However, I didn't like it
that much.

I got Electronic Projects for Dummies. The book started off like I was
going to understand it, then I got to chapter 5 and it asked me to buy
a bunch of components to put together this project. Some of the items
on the list I couldn't find even in the online stores that they
suggested in the book. The project was a little above me anyway.

What I am looking for is a book that will explain a concept to me,
like capacitors. Then give a few circuits that demonstrate what they
just explained to me. Is there anything like that? Or, do you
generally have to read through a book that is all explanation then get
another book that has circuits you can play with?

Thanks in advanced!

I reccomend reading several editions of "The ARRL Handbook for Radio
Amateurs". A new edition appears every year and has for about 75 years.
Most big libraries have a collection of issues and old issues show up at
swap meets and garage sales. It's oriented towards equipment for Amateur
Radio stations but has a good basic electronics section and lots of
construction projects. As someone else pointed out, the US Navy NEETS
manuals are very comprehensive and written at a level appropriate for
beginning Electronics Technicians. You can buy the NEETS set on a CD Rom.

Ken Fowler, KO6NO

 

[Joel Koltner]

"David L. Jones" <altzone@gmail.com> wrote in message
news:407f6b41-5953-45db-b24b-c8803c3a3193@a32g2000prf.googlegroups.com...

The Talking Electronics books are quite eclectic but do a pretty good
job for the beginner:

Be sure to look for Colin Mitchell's rant in one of them about how compact
fluorescent lightbulbs are never going to catch on.

I like your description -- definitely accurate.

 

[Tom Biasi]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

Hi Lee,
Without knowing what you monitor circuit is looking for we can only make
some assumptions based on the circuits that work and the problems you have
with your circuit.
It would probably be easier to start with a high value load that works for
the LEDs and reduce it until it "fools" the monitor and the go a little
lower.

Tom

 

[Michael Robinson]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

Check your wiring.

If you connect the "running light" load resistor to the wrong end of the 100
ohm resistor, it will make the lights go out.

 

[Michael Robinson]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

Check your wiring.

If you connect the "running light" load resistor to the wrong end of the 100
ohm resistor, it will make the lights go out.

 

[Michael Robinson]

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CC395.744%thepanda@btconnect.com...

On 04/06/2008 21:46, in article g26uv8$8aa$1@aioe.org, "Michael Robinson"
kellrobinson@yahoo.com> wrote:


"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find
some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web
to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket
lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running
light
and full level brake light and looks great. I guessed on the 100 ohms
for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs
are
already dim on the running light circuit, that they get too little
current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee


Check your wiring.

If you connect the "running light" load resistor to the wrong end of the
100
ohm resistor, it will make the lights go out.


Hi Michael,

I was connecting it the same way as CA connected theirs - earth to the
feed
side of the 'running light' 100 ohm reduction resistor not the LED/Brake
light feed side.

If I connect it to the LED side next to the brake feed I'll effectively be
adding another load resistor in parallel to the brake load won't I, and
halve the load resistance? I'm not sure how it would be 'seen' on the
running light monitor?

Cheers,
Lee


Well, something doesn't jibe.

If I understand you correctly,

when the brake is not applied, and you don't have a load resistor installed
on the running light feed, the running light works;

but then, when you insert a load resistor from the "+14.4V Running Light"
node in your circuit to ground, the running light dies.

However, "+14.4 Running Light" is a low impedance power source, and
diverting some small fraction of an amp to ground from that power source
should have negligible effect; the voltage at that node should hardly change
at all with the addition of some small load. Either something's out of
kilter with your circuitry or else I misunderstood your post.

The points on your diagram labeled in red "+14.4V" signify a connection to
your bike's positive power source, do they not?

 

[Michael Robinson]

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CC395.744%thepanda@btconnect.com...

On 04/06/2008 21:46, in article g26uv8$8aa$1@aioe.org, "Michael Robinson"
kellrobinson@yahoo.com> wrote:


"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find
some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web
to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket
lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running
light
and full level brake light and looks great. I guessed on the 100 ohms
for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs
are
already dim on the running light circuit, that they get too little
current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee


Check your wiring.

If you connect the "running light" load resistor to the wrong end of the
100
ohm resistor, it will make the lights go out.


Hi Michael,

I was connecting it the same way as CA connected theirs - earth to the
feed
side of the 'running light' 100 ohm reduction resistor not the LED/Brake
light feed side.

If I connect it to the LED side next to the brake feed I'll effectively be
adding another load resistor in parallel to the brake load won't I, and
halve the load resistance? I'm not sure how it would be 'seen' on the
running light monitor?

Cheers,
Lee


Well, something doesn't jibe.

If I understand you correctly,

when the brake is not applied, and you don't have a load resistor installed
on the running light feed, the running light works;

but then, when you insert a load resistor from the "+14.4V Running Light"
node in your circuit to ground, the running light dies.

However, "+14.4 Running Light" is a low impedance power source, and
diverting some small fraction of an amp to ground from that power source
should have negligible effect; the voltage at that node should hardly change
at all with the addition of some small load. Either something's out of
kilter with your circuitry or else I misunderstood your post.

The points on your diagram labeled in red "+14.4V" signify a connection to
your bike's positive power source, do they not?

 

[Dan Coby]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits - fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my eyes.

If I take one of the aftermarket board load resistors and connect it across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work together.

Even better, if someone were actually able to explain to me how it's worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

The short form of the answer is that you need to reduce the 100 ohm resistor.
If I were you, I would try 43 ohms (i.e. the value from the original circuit.)

Without more information about the monitor circuit, we can only guess about
the values for the green resistors. The 46 ohm value in the original circuit seems
to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5 watts).
Since this value works, you can continue to use it. If you want, you can try higher
values. Higher values will give you less power being wasted in these resistors.
However at some point with higher values the monitor circuit will decide that
the 'light bulb' is burned out.

You chose the 100 ohm resistor to give the correct brightness for the running
lights. However, I assume, that the second 46 ohm resistor was not connected
when you chose that value. Put the second 46 ohm resistor into your circuit and
try lower values for the current 100 ohm resistor until you get the brightness
desired.

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4 volts,
R1 is 100 ohms and R2 is 46 ohms.

For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
volts (4 x 1.8 volts) before then will produce light. So you need to change the
ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you probably
cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).

 

[Dan Coby]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits - fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my eyes.

If I take one of the aftermarket board load resistors and connect it across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work together.

Even better, if someone were actually able to explain to me how it's worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

The short form of the answer is that you need to reduce the 100 ohm resistor.
If I were you, I would try 43 ohms (i.e. the value from the original circuit.)

Without more information about the monitor circuit, we can only guess about
the values for the green resistors. The 46 ohm value in the original circuit seems
to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5 watts).
Since this value works, you can continue to use it. If you want, you can try higher
values. Higher values will give you less power being wasted in these resistors.
However at some point with higher values the monitor circuit will decide that
the 'light bulb' is burned out.

You chose the 100 ohm resistor to give the correct brightness for the running
lights. However, I assume, that the second 46 ohm resistor was not connected
when you chose that value. Put the second 46 ohm resistor into your circuit and
try lower values for the current 100 ohm resistor until you get the brightness
desired.

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4 volts,
R1 is 100 ohms and R2 is 46 ohms.

For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
volts (4 x 1.8 volts) before then will produce light. So you need to change the
ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you probably
cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).

 

[John G]

"panfilero" <panfilero@gmail.com> wrote in message
news:d77221c0-d315-4854-93e9-098bf42a0195@u12g2000prd.googlegroups.com...
On Jun 4, 4:33 pm, Mark <makol...@yahoo.com> wrote:

so would you say I can just multipy the instantaneous values of V and
I to get the real power?

Yes... but they have to be the V and I taken _at the same instnat in
time__, you multiply them and that gives you the instantaneous power
at that time....then repeat for many periods.. take all the power
results and average them and you will have the true average power...

but why do you want to measure the true average power of a small fan
motor?

lets get to the meat of the real problem you are trying to solve..

Mark

Thanks Mark,

well basically Im tryign to characterize a small fan, and I want to
know how much power this little blower is consuming... the blower
comes with a little controller that sends pulses out to each winding
in order to keep the fan spinning. That is the only power the fan
gets, the controller gets power from somewhere else but I'm just
interted in the fan... so I have set out to measure the power each
winding consumes, and that's how've I've ended up with all these
issues.

I don't want to criticize you at all, but as you do not seem to understand
the difference between real and apparent power you are Way way out of your
depth wanting to worry about the power used by a small fan.

John G.

 

[BobW]

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

Lee,

I should be able to figure this out since I'm an EE and I have a BMW RT. The
keyword is "should".

I sure love my RT. There's nothing on this planet like them. Great
acceleration, nimble in the twisties, smooth as silk at high speeds, great
gas mileage, comfortable, yada yada...

So, here's what I think is going on.

First, I'm making the assumption that the original circuit looks like this
for both the BRAKE and RUNNING circuits:

12V->SWITCH->LAMP->0V (aka GND), and there's also a high value "sense"
resistor across the SWITCH.

Assuming that this is true, then when the BRAKE SWITCH is off and the LAMP
is good then the monitor circuit will see a "low voltage" across the LAMP.
If the LAMP is burned out (i.e. it is open) then the monitor circuit will
12V across the LAMP even though the SWITCH is off (thanks to the "sense"
resistor that's across the SWITCH).

If the above operation is correctly described then you'll have a problem
with your circuit since you're trying to use the same LED string for both
the BRAKE and RUNNING circuits. Specifically, a problem will occur in your
circuit if the RUNNING circuit is activated (RUNNING LIGHTS on) and then the
BRAKE CIRCUIT is tested for a blown lamp. In this condition, even though the
BRAKE SWITCH is off the monitor circuit will see something much greater than
a "low voltage" and will thus declare a blown BRAKE LAMP.

The solution would be to put a diode between the BRAKE load resistor and the
lower LED drive resistor. The anode would connect to the BRAKE load
resistor. This way, the BRAKE monitor circuit will see its "low voltage"
even though the RUNNING circuit is already turned on, and when the BRAKE
switch is on then the LED string will brighter.

It may be necessary to add a diode from the RUNNING circuit, too, in the
case that you've depressed one of the brakes as it's doing its diagnostics.
You'll have to play with this scenario, but (assuming this is all correct) I
would add the addtional diode just in case.

I hope this makes sense and I hope it solves your problem. After all, we're
like BMW motobrothers.

Bob

--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

 

[John G]

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CD344.756%thepanda@btconnect.com...

On 04/06/2008 22:59, in article
__-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
adcoby@earthlink.net> wrote:

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find
some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
body).

I've made my own LED board and have found lots of resources on the web
to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
has a bulb failure warning circuit. To get round this the aftermarket
lamp
has a load resistor across each of the running and brake circuits -
fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running
light
and full level brake light and looks great. I guessed on the 100 ohms
for
the running light but it looks about the same as the 5W tungsten to my
eyes.

If I take one of the aftermarket board load resistors and connect it
across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs
are
already dim on the running light circuit, that they get too little
current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work
together.

Even better, if someone were actually able to explain to me how it's
worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

The short form of the answer is that you need to reduce the 100 ohm
resistor.
If I were you, I would try 43 ohms (i.e. the value from the original
circuit.)

Without more information about the monitor circuit, we can only guess
about
the values for the green resistors. The 46 ohm value in the original
circuit
seems
to have been chosen to give about a 5 watt load at 14.4 volts (actually
4.5
watts).
Since this value works, you can continue to use it. If you want, you can
try
higher
values. Higher values will give you less power being wasted in these
resistors.
However at some point with higher values the monitor circuit will decide
that
the 'light bulb' is burned out.

You chose the 100 ohm resistor to give the correct brightness for the
running
lights. However, I assume, that the second 46 ohm resistor was not
connected
when you chose that value. Put the second 46 ohm resistor into your
circuit
and
try lower values for the current 100 ohm resistor until you get the
brightness
desired.

The current combination of a 46 ohm resistor and a 100 ohm resistor
creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5
volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is
14.4
volts,
R1 is 100 ohms and R2 is 46 ohms.

For your values, this gives 4.5 volts. However your LEDs need a minimum
of 7.2
volts (4 x 1.8 volts) before then will produce light. So you need to
change
the
ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
probably
cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).



Dan,

Thanks ever so much for the comprehensive reply- I'm going to check all
again in the morning anyway and will study what you've put here to make
sure
I understand properly and make changes.

Cheers,
Lee

I agree completly with Dan Coby.

Surely the circuits should be the SAME except for a variation in the values
of the LED dropping resistors if your new LEDS are different to the original
circuit.

Why did you change anything beyond the geometery?
Did you change the TYPE of lead ie. for a brighter model?

The 2 loading resistor should still be 46 ohms as that is what works for the
check circuit which we can only guess about.
If the running lights go out then the 100 ohm resistor is the culprit and
should be lower.

John G.

 

[Dan Coby]

"Gyro" <thepanda@btconnect.com> wrote in message news:C46CD344.756%thepanda@btconnect.com...

On 04/06/2008 22:59, in article
__-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
adcoby@earthlink.net> wrote:

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits - fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my eyes.

If I take one of the aftermarket board load resistors and connect it across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work together.

Even better, if someone were actually able to explain to me how it's worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

The short form of the answer is that you need to reduce the 100 ohm resistor.
If I were you, I would try 43 ohms (i.e. the value from the original circuit.)

Without more information about the monitor circuit, we can only guess about
the values for the green resistors. The 46 ohm value in the original circuit
seems
to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5
watts).
Since this value works, you can continue to use it. If you want, you can try
higher
values. Higher values will give you less power being wasted in these
resistors.
However at some point with higher values the monitor circuit will decide that
the 'light bulb' is burned out.

You chose the 100 ohm resistor to give the correct brightness for the running
lights. However, I assume, that the second 46 ohm resistor was not connected
when you chose that value. Put the second 46 ohm resistor into your circuit
and
try lower values for the current 100 ohm resistor until you get the brightness
desired.

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4
volts,
R1 is 100 ohms and R2 is 46 ohms.

For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
volts (4 x 1.8 volts) before then will produce light. So you need to change
the
ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
probably
cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).



Dan,

Thanks ever so much for the comprehensive reply- I'm going to check all
again in the morning anyway and will study what you've put here to make sure
I understand properly and make changes.

Cheers,
Lee

A few more comments:

I suggested earlier that you replace your 100 ohm resistor with a 43 ohm resistor
(the same as the original circuit). However that value may still be a little large. With
a 43 ohm resistor, the output of the voltage divider is 14.4 * 46 / (43 + 46) = 7.44 volts.
That voltage is still probably too low to get the desired brightness. (I did not do the
math earlier, I simply assumed that the original circuit had reasonable values and
that the LEDs were similar.) My guess is that something in the 33 to 36 ohm range will
be better.

The 46 ohm resistors will dissipate about 4.5 watts. The other resistor will also dissipate
about that much heat. Obviously they need to be sized to handle that power level. I suggest
at least a 10 watt power rating. Even so they will get warm. Larger resistors with more
surface area (or a heat sink) will run cooler.

The brightness of the LEDs in the 'running light' mode will be pretty sensitive to the
battery voltage. (The original circuit had the same problem.) You could add a diode
into the connection between the 'brake light' 46 ohm resistor and your 100 ohm resistor.
This would reduce the voltage sensitivity by eliminating the voltage divider. Thus you
could use the 100 ohm value that you already have. (The voltage divider is eliminated
since the diode is reversed biased when the brake light switch is not active.) The
voltage drop across this diode will reduce the drive signal to the LEDs by about 0.7 volts.
You might want to reduce the 130 and 270 ohm resistors by 5% to compensate.

Both the original circuit and your version have groups of LEDs arranged as two strings
of four LEDS in parallel sharing a single current limiting resistor. This is not really a good
idea. (Its only advantage is minimizing the resistor count.) LEDs are very non-linear in
their current/voltage characteristics. It is likely that the groups of 4 LEDs will not share
the current equally between the two strings. A better design would have a separate
270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in strings of at
least 4 is good since it helps reduce wasted power.)

 

[Bob Eld]

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:4847F90A.C7A0333@hotmail.com...

lerameur wrote:

HI all,

I just build two electro magnet with the same Gauss output

Isn't it Teslas these days ?

Graham

Tesla is an MKS or SI unit. Gauss is a CGS unit. a Tesla is 10^4 Gauss.
What's 10,000 among friends?

 

[BobW]

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46DB058.775%thepanda@btconnect.com...

On 05/06/2008 08:51, in article
6JudnUt4Z5V7B9rVnZ2dnUVZ_qPinZ2d@earthlink.com, "Dan Coby"
adcoby@earthlink.net> wrote:

[snip]

The brightness of the LEDs in the 'running light' mode will be pretty
sensitive to the
battery voltage. (The original circuit had the same problem.) You could
add a
diode
into the connection between the 'brake light' 46 ohm resistor and your
100 ohm
resistor.
This would reduce the voltage sensitivity by eliminating the voltage
divider.
Thus you
could use the 100 ohm value that you already have. (The voltage divider
is
eliminated
since the diode is reversed biased when the brake light switch is not
active.) The
voltage drop across this diode will reduce the drive signal to the LEDs
by
about 0.7 volts.
You might want to reduce the 130 and 270 ohm resistors by 5% to
compensate.

Both the original circuit and your version have groups of LEDs arranged
as two
strings
of four LEDS in parallel sharing a single current limiting resistor. This
is
not really a good
idea. (Its only advantage is minimizing the resistor count.) LEDs are
very
non-linear in
their current/voltage characteristics. It is likely that the groups of 4
LEDs
will not share
the current equally between the two strings. A better design would have a
separate
270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in
strings
of at
least 4 is good since it helps reduce wasted power.)



My sincere thanks to everyone that replied. This has helped me out a lot
and
I now have it sorted as per Dan's recommendations - a 33ohm 10W resistor
rather than the 100 has everything functioning correctly.

I now understand how to work out stuff a little more too.

Will look to adding a diode as per Dan & Bob for 'belt and braces'.

Cheers, Lee

Lee,

If what Dan and I suspect is truly going on, if you add the diodes then the
only values that would matter are the two fool-the-monitor-circuit load
resistors. These load resistors should only need to be some maximum value
(i.e. minimum sense current). After that, all the other brightness-setting
resistors will have no effect on the proper operation.

I would be concernced that without doing a lot of characterization with the
current circuit, you may get into trouble (e.g. lights not coming on) as
your battery voltage varies and the monitor circuit characteristics vary.
Again - if it were my bike I would spend the extra effort to see if the
diodes allow you to vary resistor values without concern for the operational
stability of your lights.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

 

[BobW]

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46DB058.775%thepanda@btconnect.com...

On 05/06/2008 08:51, in article
6JudnUt4Z5V7B9rVnZ2dnUVZ_qPinZ2d@earthlink.com, "Dan Coby"
adcoby@earthlink.net> wrote:

[snip]

The brightness of the LEDs in the 'running light' mode will be pretty
sensitive to the
battery voltage. (The original circuit had the same problem.) You could
add a
diode
into the connection between the 'brake light' 46 ohm resistor and your
100 ohm
resistor.
This would reduce the voltage sensitivity by eliminating the voltage
divider.
Thus you
could use the 100 ohm value that you already have. (The voltage divider
is
eliminated
since the diode is reversed biased when the brake light switch is not
active.) The
voltage drop across this diode will reduce the drive signal to the LEDs
by
about 0.7 volts.
You might want to reduce the 130 and 270 ohm resistors by 5% to
compensate.

Both the original circuit and your version have groups of LEDs arranged
as two
strings
of four LEDS in parallel sharing a single current limiting resistor. This
is
not really a good
idea. (Its only advantage is minimizing the resistor count.) LEDs are
very
non-linear in
their current/voltage characteristics. It is likely that the groups of 4
LEDs
will not share
the current equally between the two strings. A better design would have a
separate
270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in
strings
of at
least 4 is good since it helps reduce wasted power.)



My sincere thanks to everyone that replied. This has helped me out a lot
and
I now have it sorted as per Dan's recommendations - a 33ohm 10W resistor
rather than the 100 has everything functioning correctly.

I now understand how to work out stuff a little more too.

Will look to adding a diode as per Dan & Bob for 'belt and braces'.

Cheers, Lee

Lee,

If what Dan and I suspect is truly going on, if you add the diodes then the
only values that would matter are the two fool-the-monitor-circuit load
resistors. These load resistors should only need to be some maximum value
(i.e. minimum sense current). After that, all the other brightness-setting
resistors will have no effect on the proper operation.

I would be concernced that without doing a lot of characterization with the
current circuit, you may get into trouble (e.g. lights not coming on) as
your battery voltage varies and the monitor circuit characteristics vary.
Again - if it were my bike I would spend the extra effort to see if the
diodes allow you to vary resistor values without concern for the operational
stability of your lights.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

 

[Dan Coby]

<google@woodall.me.uk> wrote in message
news:c0827f1c-dd15-409c-9772-ee73d7f6b7c6@x35g2000hsb.googlegroups.com...

On Jun 4, 10:59 pm, "Dan Coby" <adc...@earthlink.net> wrote:

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4 volts,
R1 is 100 ohms and R2 is 46 ohms.


Unless it's been wired up differently to the way it's been drawn then
this isn't the circuit the OP is using. He's using:

-----
------------------+--| 100 |------o
| | -----
+ ___
14.4 volts | 4 |
- | 6 |
| ---
| |
|__________________|________o


The only way I can see this failing is if there's a high resistance
somewhere (bad connection? Bad earth?). Occasionally you see a similar
problem in car rear light clusters where the indicator flashing causes
the brake light to flash in anti-phase.

Tim.

I did not include the 46 ohm resistor that is across the 'running light' switch
since it has no effect upon the LED brightness (unless as you noted there is a
bad connection.) The 46 ohm resistor that I did show is the one across the
'brake light' switch. When the 'running light' switch is on and the 'brake light'
switch is off then the circuit is:

_____
------------+----| 100 |-----+--------o
| | ----- |
+ ___ ___
14.4 volts | 4 | | 4 |
- | 6 | | 6 |
| --- ---
| | |
|____________|________________|________o

The 46 ohm resistor on the left does not affect (at least with ideal components)
the LED brightness.

 

[Dan Coby]

<google@woodall.me.uk> wrote in message
news:c0827f1c-dd15-409c-9772-ee73d7f6b7c6@x35g2000hsb.googlegroups.com...

On Jun 4, 10:59 pm, "Dan Coby" <adc...@earthlink.net> wrote:

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4 volts,
R1 is 100 ohms and R2 is 46 ohms.


Unless it's been wired up differently to the way it's been drawn then
this isn't the circuit the OP is using. He's using:

-----
------------------+--| 100 |------o
| | -----
+ ___
14.4 volts | 4 |
- | 6 |
| ---
| |
|__________________|________o


The only way I can see this failing is if there's a high resistance
somewhere (bad connection? Bad earth?). Occasionally you see a similar
problem in car rear light clusters where the indicator flashing causes
the brake light to flash in anti-phase.

Tim.

I did not include the 46 ohm resistor that is across the 'running light' switch
since it has no effect upon the LED brightness (unless as you noted there is a
bad connection.) The 46 ohm resistor that I did show is the one across the
'brake light' switch. When the 'running light' switch is on and the 'brake light'
switch is off then the circuit is:

_____
------------+----| 100 |-----+--------o
| | ----- |
+ ___ ___
14.4 volts | 4 | | 4 |
- | 6 | | 6 |
| --- ---
| | |
|____________|________________|________o

The 46 ohm resistor on the left does not affect (at least with ideal components)
the LED brightness.

 

[Tim Woodall]

On Thu, 5 Jun 2008 10:27:55 -0700,
Dan Coby <adcoby@earthlink.net> wrote:

I did not include the 46 ohm resistor that is across the 'running light' switch
since it has no effect upon the LED brightness (unless as you noted there is a
bad connection.) The 46 ohm resistor that I did show is the one across the
'brake light' switch. When the 'running light' switch is on and the 'brake light'
switch is off then the circuit is:

_____
------------+----| 100 |-----+--------o
| | ----- |
+ ___ ___
14.4 volts | 4 | | 4 |
- | 6 | | 6 |
| --- ---
| | |
|____________|________________|________o

The 46 ohm resistor on the left does not affect (at least with ideal components)
the LED brightness.

Ah, ok. Yes, you're right.

Tim.



--
God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t,"
and there was light.

http://tjw.hn.org/ http://www.locofungus.btinternet.co.uk/