Chip with simple program for Toy

When I was a kid my older brother had an "Electric Lab", the lab
contained a device that was a coil with contacts. The Idea was to
connect a small battery 1.5v to it and adjust the contacts to
"resonate" ( switch on/of continuosly ) . It had also 2 wires ending
in 2 separate pieces of pipe that I will hold and feel the
electricity.
I was used to connect it to an old train toy AC adapter and set it up
full blast.
I had to grab the pipes carefully ( more than letting them rest in the
palms of my hands and slowly grab them. I could feel the electricity
moving all the muscles of my hands and arms. So much that sometimes it
was difficult to open my hands to let go.

I also played with a neon sign transformer that make an arch 2 inches
long , and thru wood. When I turned off the light to do it I coud see
the cable and my fingers with a light blue color, I think it was 15000
volts. I got a shock once with it but the electricity whent in and out
in my hand , so nothing happened to me.

At school there was a cold water fountain that will shock you if you
touched it at the same time that the surrounding wall. I used to be in
line to drink water and touch the wall and the ears of the one
drinking water.....


On 5 Jun 2005 00:26:47 -0700, "Bill Bowden" <wrongaddress@att.net>
wrote:

Your experience notwithstanding, people die from 110 vac.
You're healthy and lucky. Not everybody is.

Actually, I'm not healthy, but I am lucky. I have coranary heart
disease and luckily had a triple bypass operation last year that
brought me back to normal. I had angina pains when walking uphill, so I
went to a cardiologist, who gave me a treadmill test, and then told me
to go to the hospital. I was lucky again to have insurance that paid
$85,000 for the angiogram and surgery. Otherwise, I would be bankrupt
today.

But I feel pretty good now. I can walk uphill for long distances and
climb 4 flights of stairs with no pain.

Haven't tried the 120VAC shock test yet, but I'm really more afraid of
falling down than getting shocked.

Last month, I fell down and hit my head on a open tool box and needed 5
stitches. I would have much prefered being shocked from 120VAC than
smashing my head on a open tool box.

-Bill
 
The problem is that you have a current source on one side. You only have two
independent equations to solve for the two unknown currents. The third
current is known. One approach is to replace the current source with a
voltage source equivalent then write the loop equations - you will then have
2 independent loops with two known voltage sources and two unknown currents.
This is the easy way
-----R1-----o--------R2-------o
| -->I1 | -->I3 |
V1 R3 V2 R2 in parallel
with current source Is becomes
|-------------|-------------------| V2 =R2*Is in series with
R2

V1=(R1+R3)I1 -R3*I3
-V2= -R3*I1 +(R2+R3)I3 or V2=R3*I1 -(R2+R3)I3

It can be solved using the current source but the "Ideal " current source is
known -eliminating the 3rd loop equation.
Put it this way with the original circuit as I see it.
-----R1-----o----------o---------
| -->I1 | --> | |
V1 R3 I3 R2 Is ^
|-------------|------------|---------|

Left loop V1=(R1+R3)I1 -R3*I3 where I3 is the loop current circulating
between the parallel resistors.
Center loop 0 =-R3I1 +(R3+R2)I3 +R2*Is
Right loop: V2 =R2*Is +R2*I3
V2 is unknown and Is is known. so the right loop equation simply gives the
unknown voltage V2 in terms of I3 and Is This leaves you with the two
equations, left and center in terms of the two unknown currents. . Knowing
I2 then allows you to solve the left and center loop equations
simultaneously- that is - there are only two independent loop equations in
fact as only two of the 3 current variables are unknown. The right side
equation can be used to solve for the voltage V2 but it isn't necessary once
you solve for I1, I3
Note that these become
center: R2*Is =V2 =R3*I1 -(R3+R2)I3 which is the same as above
left equation is unchanged from above.

Anytime you have a current source and are using loop equations, the known
current source eliminates a variable and hence eliminates a simultaneous
equation (making life easier)
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer


"ShamShoon" <mhelshou@yahoo.com> wrote in message
news:1117989523.499267.103060@g47g2000cwa.googlegroups.com...
Hello there,
In the book "Fundamentals of Electric Circuits" by Alexander/Sadiku,
the text says that "A loop is said to be independent if it contains at
least one branch which is not part of any other independent loop", I
find this definition misleading, since in the example on the previous
page there's a circuit that has a voltage source and a resistance in
series and those in parallel with a resistor, in parallel with another
resistor, and then in parallel with a current source.
If we apply the definition to the current source and voltage source, we
get two independent loops one on the right and one on the left, however
the middle loop shares its two elements (the two parallel resistors),
with both loops, yet it's still independent, which shows that the
definition is not accurate. I found it hard to find a definition of an
independent loop.

Any ideas? Am i missing something?
 
Don Kelly wrote:

The problem is that you have a current source on one side. You only
have two independent equations to solve for the two unknown currents.
The third current is known. One approach is to replace the current
source with a voltage source equivalent then write the loop equations
- you will then have 2 independent loops with two known voltage
sources and two unknown currents. This is the easy way
-----R1-----o--------R2-------o
| -->I1 | -->I3 |
V1 R3 V2
Use mono spaced fonts for this. Easier, and if someone uses a proportional
font, he/she can fix it by copy pasting it to an editor. The other way
around requires access to the same font :-(

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
 
aabid wrote:
I require a Single Op-Amp Bandpass Filter circuit with a gain of 100X
and the frequency range is 0.1 Hz to 45 Hz.

I found a circuit in the following website:

http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm

According to specs, the Q factor must be greater than the square root
of half the gain. But as you can see my requirements do not meet with
this condition. Hence, please help me out here.

Thanks.
This circuit will do what you want,but requires a big capacitor on the
input. If you use a dual opamp like the NE5532 you need the same space but
can reduce the cap by a factor of 100 or more. The noise will go up tho.

||
.----||--.
| 3n3|| |
\] ___ | ___ |
o---|]--|___|--o--|___|-o
/]+ 11k | 1M1 |
150u | |\ |
'--|-\ |
| >--o---o
.--|+/
| |/ NE5534
___ | ___
GND-|___|-o-|___|-VCC
22k | 22k
---
---0u47
|
===
GND

--
ciao Ban
Bordighera, Italy
 
aabid wrote:
I require a Single Op-Amp Bandpass Filter circuit with a gain of 100X
and the frequency range is 0.1 Hz to 45 Hz.

I found a circuit in the following website:

http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm

According to specs, the Q factor must be greater than the square root
of half the gain. But as you can see my requirements do not meet with
this condition. Hence, please help me out here.

Thanks.
This circuit will do what you want,but requires a big capacitor on the
input. If you use a dual opamp like the NE5532 you need the same space but
can reduce the cap by a factor of 100 or more. The noise will go up tho.

||
.----||--.
| 3n3|| |
\] ___ | ___ |
o---|]--|___|--o--|___|-o
/]+ 11k | 1M1 |
150u | |\ |
'--|-\ |
| >--o---o
.--|+/
| |/ NE5534
___ | ___
GND-|___|-o-|___|-VCC
22k | 22k
---
---0u47
|
===
GND

--
ciao Ban
Bordighera, Italy
 
On 6 Jun 2005 01:13:36 -0700, "aabid" <aabid.s@gmail.com> wrote:

I require a Single Op-Amp Bandpass Filter circuit with a gain of 100X
and the frequency range is 0.1 Hz to 45 Hz.

I found a circuit in the following website:

http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm

According to specs, the Q factor must be greater than the square root
of half the gain. But as you can see my requirements do not meet with
this condition. Hence, please help me out here.

Thanks.
A frequency range of 450:1 is way too large for a simple
bandpass filter. You should have separate low-pass
and high-pass stages for this.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Home of DaqGen, the FREEWARE signal generator
 
http://focus.ti.com/docs/prod/folders/print/tle2027.html
will run on an 8V supply and is very low noise.
cb

--
=================================
Some people have something to say... others have to say something!



"padmow" <padmow_69@yahoo.com> wrote in message
news:1118062073.069506.147970@z14g2000cwz.googlegroups.com...
Hi,

I wonder if there is any suggestion about which op amp I can use which
has Low Noise, and only need 9V from battery.

This is for amplifiying signal from electret microphone. The opamp
would then drive class AB transistor amplifier.

Thank you

Padmow
 
aabid wrote:
I require a Single Op-Amp Bandpass Filter circuit with a gain of 100X
and the frequency range is 0.1 Hz to 45 Hz.

I found a circuit in the following website:

http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm

According to specs, the Q factor must be greater than the square root
of half the gain. But as you can see my requirements do not meet with
this condition. Hence, please help me out here.
I think you should think in terms of a dual opamp and a 10X high pass
filter (2 or 3 poles) followed by a 10X low pass filter (2 or 3 poles).

You might download a free copy of this little filter design aid to
play around with variations.

http://www-s.ti.com/sc/techzip/slvc003.zip
 
padmow wrote:
Hi,

I wonder if there is any suggestion about which op amp I can use which
has Low Noise, and only need 9V from battery.

This is for amplifiying signal from electret microphone. The opamp
would then drive class AB transistor amplifier.

Thank you

Padmow
Inside the electret mike there is a FET amplifying or buffering the extremly
high impedance diaphragm signal. The output impedance is usually in the 2k2
to 6k8 region. You could take a OPA134, which will work nicely with only
4mA of supply current and from 5V on. It will not increase the noise and can
amplify a lot with very little distortion. It can also drive low impedance
loads. Very important for high gain amplifiers is that you know about layout
and associated issues, so you do not produce a generator instead of an
amplifier.
But the best choice would probably be the LT1677 Rail-to-Rail, low supply
current, high gain and CMRR if you pay the price for it($4.50 at Digikey).
maybe they can send you samples? On the first page of the datasheet is an
electret preamp (tho I do not think it is the best design possible with it).
--
ciao Ban
Bordighera, Italy
 
padmow wrote:
Hi,

I wonder if there is any suggestion about which op amp I can use which
has Low Noise, and only need 9V from battery.

This is for amplifiying signal from electret microphone. The opamp
would then drive class AB transistor amplifier.

Thank you

Padmow
Inside the electret mike there is a FET amplifying or buffering the extremly
high impedance diaphragm signal. The output impedance is usually in the 2k2
to 6k8 region. You could take a OPA134, which will work nicely with only
4mA of supply current and from 5V on. It will not increase the noise and can
amplify a lot with very little distortion. It can also drive low impedance
loads. Very important for high gain amplifiers is that you know about layout
and associated issues, so you do not produce a generator instead of an
amplifier.
But the best choice would probably be the LT1677 Rail-to-Rail, low supply
current, high gain and CMRR if you pay the price for it($4.50 at Digikey).
maybe they can send you samples? On the first page of the datasheet is an
electret preamp (tho I do not think it is the best design possible with it).
--
ciao Ban
Bordighera, Italy
 
Thanks. the dumn thing is that I knew this.

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
"John Bokma" <john@castleamber.com> wrote in message
news:Xns966CC05E7B1A0castleamber@130.133.1.4...
Don Kelly wrote:

The problem is that you have a current source on one side. You only
have two independent equations to solve for the two unknown currents.
The third current is known. One approach is to replace the current
source with a voltage source equivalent then write the loop equations
- you will then have 2 independent loops with two known voltage
sources and two unknown currents. This is the easy way
-----R1-----o--------R2-------o
| -->I1 | -->I3 |
V1 R3 V2

Use mono spaced fonts for this. Easier, and if someone uses a proportional
font, he/she can fix it by copy pasting it to an editor. The other way
around requires access to the same font :-(

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
 
On 7 Jun 2005 02:07:36 -0700, pie_garnishment@lycos.com (Winnie
Oakbob) wrote:

I think this task is too advanced for me, so please, can you help me ?
Just had to post light bulb joke :)

Q: How many Newsgroup members does it take to change a light bulb?

A: 1,331 Here's how:

1 to change the light bulb and to post to the newsgroup that the light
bulb has been changed.

14 to share similar experiences of changing light bulbs and how the
light bulb could have been changed differently.

7 to caution about the dangers of changing light bulbs.

27 to point out spelling/grammar errors in posts about changing light
bulbs.

53 to flame the spell checkers.

156 to write to the newsgroup complaining about the light bulb
discussion and its inappropriateness to this newsgroup.

41 to correct spelling in the spelling/grammar flames.

109 to post that the newsgroup is not about light bulbs and to please
take this posting to alt.lite.bulb

203 to demand that cross posting to alt.grammar, alt.spelling and
alt.punctuation about changing light bulbs be stopped.

111 to defend the posting to this list saying that we all use light
bulbs and therefore the posts **are** relevant to this mail list.

306 to debate which method of changing light bulbs is superior, where
to buy the best light bulbs, what brand of light bulbs work best for
this technique, and what brands are faulty.

27 to post URLs where one can see examples of different light bulbs.

14 to post that the URLs were posted incorrectly, and to post
corrected URLs.

3 to post about links they found from the URLs that are relevant to
this list which makes light bulbs relevant to this list.

33 to merge all posts to date, then quote them including all headers
and footers, and then add "Me Too."

12 to post to the list that they are unsubscribing because they cannot
handle the light bulb controversy.

19 to quote the "Me Too's" to say, "Me Three."

4 to suggest that posters request the light bulb FAQ.

1 to propose new alt.change.lite.bulb newsgroup.

47 to say this is just what alt.physic.cold_fusion was meant for,
leave it here.

143 votes for alt.lite.bulb.
--
When you hear the toilet flush, and hear the words "uh oh", it's already
too late. - by anonymous Mother in Austin, TX
To reply, replace digi.mon with phreaker.net
 
Kevin Aylward wrote:

Indeed. The explanation was why hfe matching mattered despite the
fact that the bipolar transistor is a voltage controlled device. To
wit, its the voltage drop across rbb' that is the killer.
Most of us know how to design a simple bipolar transistor stage using
the current model.
The design equations and explanations need just a few rows of text.

How would you design the same transistor stage using the
voltage-controlled model?

Can you show us how your way is faster, better, etc..?

How does it work in real transistor stage design?



--
Roger J.
 
"aabid" <aabid.s@gmail.com> wrote in message
news:1118045616.134854.136730@g43g2000cwa.googlegroups.com...
I require a Single Op-Amp Bandpass Filter circuit with a gain of 100X
and the frequency range is 0.1 Hz to 45 Hz.

I found a circuit in the following website:

http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm

According to specs, the Q factor must be greater than the square root
of half the gain. But as you can see my requirements do not meet with
this condition. Hence, please help me out here.

Thanks.
that page doesnt give the single op amp sallen key bandpass filter, wich u
might look at as an alternative, the one shown is a multiple feedback type.

try this calculator ....

http://www.analog.com/analog_root/static/techSupport/designTools/interactiveTools/filter/filter.html

Colin =^.^=
 
John Fields <jfields@austininstruments.com> wrote:

On Thu, 02 Jun 2005 23:02:00 +0100, Terry Pinnell
terrypinDELETE@THESEdial.pipex.com> wrote:

John Fields <jfields@austininstruments.com> wrote:

I've posted a schematic for you at alt.binaries.schematics.electronic
under "Random ON/OFF Timer Circuit".

Nice stuff! I'm hoping the OP will ask you for some narrative, as it
would save me a lot of time <g>.

---
You're in luck! He did, so I posted the circuit description and a
corrected schematic (sigh...) to abse.
Thanks. Looks good. Only wish I had time to either simulate or
breadboard it <g>.

--
Terry Pinnell
Hobbyist, West Sussex, UK
 
Roger Johansson wrote:
Kevin Aylward wrote:

Indeed. The explanation was why hfe matching mattered despite the
fact that the bipolar transistor is a voltage controlled device. To
wit, its the voltage drop across rbb' that is the killer.

Most of us know how to design a simple bipolar transistor stage using
the current model.
People believe they do.

The design equations and explanations need just a few rows of text.
As do the voltage controlled equations.

How would you design the same transistor stage using the
voltage-controlled model?
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

Can you show us how your way is faster, better, etc..?
It is the correct way. Once you get to really designing complex
transistor level circuitry, it will become obvious. All basic ac
understanding is based on gm, that is, vo=gm.vb

How does it work in real transistor stage design?
Its what real professional analogue designers do that actually
understand transistor level design. Unfortunately, they are many that
only believe they understand. They dont.

I have given the basics in the above link.

Kevin Aylward
informationEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
Kevin Aylward wrote:

How would you design the same transistor stage using the
voltage-controlled model?

http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html
There is no practical example on that web page, and it surely does not
look like a simpler or faster way to design a transistor stage.


--
Roger J.
 
<steve.parson@gmail.com> wrote in message
news:1118169966.758359.232540@g49g2000cwa.googlegroups.com...
Hi everybody. I'm an aspiring electrical engineer (going into first
year university in the fall) and have a problem with an old motherboard
(AOpen AK-73Pro(A)). Two capacitors (near the ZIFF slot on the
motherboard) are leaking electrolyte, and the computer experienced
completely random hard shutdowns. They are labelled in the diagram
provided by the manufacturer as "Low ESR condensers".
I assume the 2 caps are TC5 and TC6 and have a gold -ve stripe. Anyway look
carefully at all the others of the same manufacturer. (7, 16, 19, 26, 27,
29, 31, 39, 42) if they are not bulged or leaking now they will in the
future.

They are Lelon 10v 3300uF capacitors. I was wondering if there was a
distributor in Canada that doesn't ask for a minimum order so I can get
these 2 capacitors cheap. If anyone can get me a part number of the
equivalent and a distributor I would be forever grateful :) There seems
to be so many parts to choose from that I don't know what to pick.
Just go to your local parts place and get what he has on the shelf. I
replaced mine with cheap 105°C caps, TC5 & 6 with 3300uf 16V and the others
with 1000uf 16v even though TC16 & 19 & 27 were 1200uf. It works fine.

Also, I only have basic soldering experience. Is the process of
removing the old capacitors and putting the new ones in overly
difficult? If I approach it with extreme caution, should I have a
problem? Any insight would be appreciated.
I don't know what your skill level is when you say basic. I use an iron at
330°C and a solder sucker. Get on and off as soon as the solder is melted.
New cap in. Mind the orientation. Apply solder to the pad and cap lead,
apply iron to the solder so that solder runs to cover pad and lead and get
off. 1 to 2 seconds if the cap leads are clean, if not clean them with an
eraser first. Cut off leads and your done.


I understand the cost of replacing the capacitors is near the cost of
replacing the entire motherboard, but it is fitted with 1024mb of PC133
SDRAM that I don't want to replace with DDR. (cash-strapped).
No it's not. 11 caps = less than 6 $AU. Cheap fix.

Thank you in advance.
Steve ( steve.parson at gmail.com )
One other thing. Clean off any leakage from the caps.

Another one other thing. Static electricity. Ground yourself before you
start. But then you knew that didn't you :eek:)

Gordon
 
On 7 Jun 2005 11:46:06 -0700, steve.parson@gmail.com wrote:

problem with an old motherboard
(AOpen AK-73Pro(A)). Two capacitors are leaking electrolyte labelled
as "Low ESR condensers".
They are Lelon 10v 3300uF capacitors.
get otherway dead other MoBo with good caps somewhere & first practice
a bit soldering/desoldering on that. Try to find some that has on it
min.same capacity, voltage & size cap for spare parts ...
--
Regards , SPAJKY ÂŽ
mail addr. @ my site @ http://www.spajky.vze.com
3rd Ann.: - "Tualatin OC-ed / BX-Slot1 / inaudible setup!"
 
"Winnie Oakbob" <pie_garnishment@lycos.com> wrote in message
news:2d5631c0.0506070107.26f14f4d@posting.google.com...
I think this task is too advanced for me, so please, can you help me ?
You place your left hand on the light bulb and sit on the middle finger of
your right hand and rotate.
 

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