Chip with simple program for Toy

[Gordon Reeder]

pie_garnishment@lycos.com (Winnie Oakbob) wrote in
news:2d5631c0.0506070107.26f14f4d@posting.google.com:

I think this task is too advanced for me, so please, can you help me ?

Stand on a step ladder or stool and firmly grasp the
light bulb. Have four of your friends grab hold of the
ladder or stool and turn you about.

--
Just my $0.02 worth. Hope it helps
Gordon Reeder
greeder
at: myself.com

Hey EVERYBODY!
Unity means let's try to meet each other halfway

 

[John - KD5YI]

BobG wrote:

Is there a 'rule of thumb' for figuring out much torque it takes to put
out xx volts and yy amps from a PMA given the gauss of the magnet and
the N turns of the coils? Should spin free open circuit, and really
stall out short circuit, and be somewhere in the middle during
operation. I guess the 'max power' point is when the load drags the
volts down to half the open circuit value? I guess the current here
would be half the short circuit current too? I want to rig the pwm
controller to not load down the pma too much under low power/low
torque/ slow spinning conditions, like from a stirling or micro hydro.
In general volts (and current) is proportional to rpm... someone got
some formulas that are a little more specific?

Horsepower = Volts*Amps/746 = torque*RPM/5252 with torque in lb*ft.

John

 

[Ban]

padmow wrote:

You are very detail about explaining it. Thank you very much.

I hope you don't mind, how do I measure the "output impedance" of the
electret microphone.

Also how to choose the suitable resistor which connected to +, the
positive voltage suply.

About the LT1677, why it is not the best for the design.

Again Thank you very much
Padmow

The O/P impedance of the mike is mainly governed by the resistor to supply,
in the datasheet
http://www.linear.com/pc/downloadDocument.do;jsessionid=CnsD27wbAuj8MkHkAaal1sX2gMBrcaFtv2st2KIzyrQYFwZMNg7N!1727118076?navId=H0,C1,C1154,C1009,C1026,P1844,D2102
this is resistor R1 with 10k.
You can eliminate R2 all together, it only attenuates the signal and adds
noise. You can then reduce the feedback resistor R3 until the gain is right.
I would also add a 270R resistor from pin6 to output, to isolate the cable
capacitance. This preamp will work on 2AA batteries(better would be a 9V
battery, but require some additional circuitry) and consume very little
current (3mA).
Nice easy circuit.
--
ciao Ban
Bordighera, Italy

 

[John Bokma]

Dave wrote:

I would like to build a portable battery power supply for my portable
USB hard drive/CD/DVD enclosures.

Now that would have been something to put in your subject, don't you think?

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[John Fields]

On 9 Jun 2005 16:33:15 -0700, "davidd31415" <davidd31415@yahoo.com>
wrote:

I'm looking for methods to obtain positive and negative 12 volt signals
from a single positive 12 v signal (simultaneously, not simply
reversing the polarity).

I'm specifically looking for a circuit design and an explanation of how
the circuit works.

---
When you say 'signal', do you mean something which only has two
states; +12V and 0V, and that when it's at +12V the other signal is
at -12V and that when it's at 0V the other signal is at 0V?

--
John Fields
Professional Circuit Designer

 

[John Fields]

On 9 Jun 2005 17:37:43 -0700, wdflannery@aol.com wrote:

I'd like to hook up a flashlight lamp, 1.5 v battery, and capacitor in
series, close the circuit, and watch the lamp come on and slowly go off
as the capacitor charges up.

I tried it with a double electric layer capacitor (1 farad), but I now
think this type of cap has a big internal resistance, so, nothing
(observable) happened..

Now I'm thinking an electrolytic cap (0.1 farad) and maybe a 10 ohm
resistor in the circuit to get the time constant up to > 1 sec (R*C).

Is this going to work?

No.

Try this:

+------->\ <------+
| \ |
|+ O |
[BAT] |+ [LAMP]
| [1F] |
| | |
+----------+---------+

It ought to work with the 0.1F as well.








--
John Fields
Professional Circuit Designer

 

[John Popelish]

wdflannery@aol.com wrote:

I'd like to hook up a flashlight lamp, 1.5 v battery, and capacitor in
series, close the circuit, and watch the lamp come on and slowly go off
as the capacitor charges up.

I tried it with a double electric layer capacitor (1 farad), but I now
think this type of cap has a big internal resistance, so, nothing
(observable) happened..

Now I'm thinking an electrolytic cap (0.1 farad) and maybe a 10 ohm
resistor in the circuit to get the time constant up to > 1 sec (R*C).

Is this going to work?

It might produce a brief flash. There are a couple problems with
lamps in this experiment. The first is the very low resistance of a
cold filament. By the time the filament has gotten hot enough to see,
most of the energy has already been drained out pf the capacitor with
a much shorter time constant than you are expecting. The second
problem is that the light output from a filament is at all
proportional to the current through the lamp.

I suggest you go back to your original cap, but replace the lamp with
a red LED. it will produce light very nearly proportional to the
current through it, and you should see a nice decay each time you
reverse the capacitor connections. I say a red LED, because it takes
less voltage to light it than other colors.

 

[John Popelish]

Thaqalain wrote:

I want to apply DC safely to the primary of a transformer,how can i
do?by
*When limiting resistance is in series.
*When limiting resistance is in parallel.
*when neither of above

The windings of a transformer act just as any other piece of wire
where DC is involved, so think of how you could safely connect wire
across a DC supply.

If the supply is a voltage source (voltage essentially constant, no
matter how much current is drawn from it), than adding other loads in
parallel with the wire would only provide additional paths for more
current, without limiting or reducing the current through the wire.

Resistances connected in series (the wire is just a low value
resistance) add. So a series resistor will reduce the current that
passes through the transformer winding.

If, however, the DC supply is a current source (which means that the
current is nearly constant, regardless of how much resistance is
connected across it because it varies its voltage to whatever is
necessary to drive that current), then paralleling the wire with other
resistances would detour some of that total current to other paths,
lowering the total voltage the supply needed to produce to maintain
the fixed, total current. This lowers the fraction of that total
passing through the transformer.

Since your mention of "applying DC" does not specify whether the
source is more like a voltage source or a current source, or something
in between, the question cannot be answered, except to discard the
third choice.

Tell your teacher that you learned this on the Internet, not in their
class.

 

[John Popelish]

Thaqalain wrote:

This was test question for ems comapny,they don't describe
Volatge/Current Source?here is original question:
DC is safely applied to the primary of a transformer:
*When limiting resistance is in series with primary.
*When -------------------------Parallel------------.
*When neither of preceding applies

Does my answer make sense to you?

Since voltage sources are so much more common than current sources, I
suspect they are expecting the first answer.

 

[Lord Garth]

"Thaqalain" <saqlain92110@yahoo.com> wrote in message
news:1118372943.860328.274500@f14g2000cwb.googlegroups.com...

1)An 8 bit transistor register has output voltage of LHLHLHLH,What is
equivalent decimal number being stored?

55 hex or 85 decimal

(2) A 4 bit transistor registry has output voltage of HLHL,what binary
number and it's decimal equivalent is stored?

1010 binary or 10 decimal

I am new in ems cos want to qualify test for entry.

 

[John Popelish]

Thaqalain wrote:

1)An 8 bit transistor register has output voltage of LHLHLHLH,What is
equivalent decimal number being stored?
(2) A 4 bit transistor registry has output voltage of HLHL,what binary
number and it's decimal equivalent is stored?

I am new in ems cos want to qualify test for entry.

Yikes. more poorly constructed questions.

Binary logic can code a 1 as either a more positive (H) or more
negative (L) level. It is strictly a matter of convention. But a
logic high is more commonly used to represent a value of 1 than a
logic low is, so lets go with that convention.

Then there is a convention, not a hard rule whether the left most bit
represents the smaller or higher binary value. Based on the
conventions for decimal numbers let's assume that the right most bit
represents the smallest power of 2.

Bit there are still lots of ways that bits might represent value.
There are positive only values, sign and value, 2's compliment codes,
1's complement and then there are many ways to represent a floating
point value. There is also a decimal code commonly used called binary
coded decimal or BCD that uses each group of 4 binary bits to
represent a decimal digit. But let's guess that they are asking about
the simplest code for integer positive values.

Under those assumptions, LHLHLHLH would be the binary 01010101 and
would represent the binary value of
2^6 + 2^4 + 2^2 + 2^0 or 64 + 16 + 4 + 1 = 85 decimal.

Using these same assumptions for the 4 bit register holding HLHL, the
pattern would represent the binary number, 1010 which would represent
a value of
2^3 + 2^1 = 8 + 2 = 10 decimal.

This result rules out the possibility that this code is BCD, since 9
is the highest value used for a digit.

Have these examples made it clear how these conversions work (based on
the assumed but not stated conventions)?

 

[Lord Garth]

"John Popelish" <jpopelish@rica.net> wrote in message
news:9emdnZL1X6PjlDTfRVn-uw@adelphia.com...

Thaqalain wrote:

Yikes. more poorly constructed questions.

I suspect this is done on purpose...perhaps to cause the unsure to
second guess themselves. I did give the engineers answer to a
Ph.D. uncle of mine when he asked if the glass was half full or
half empty...that being that the container is twice the required volume.
He had to think on that one!

 

[Lord Garth]

"Thaqalain" <saqlain92110@yahoo.com> wrote in message
news:1118375502.107293.60650@g47g2000cwa.googlegroups.com...

i have posted same in Com.Arch group,here is one answer,which one I
believe?

"Thaqalain" <saqlain92...@yahoo.com> writes:
(1)An 8 bit transistor register has output voltage of LHLHLHLH,What is
equivalent decimal number being stored?



If the machine is using negative logic with excess-3 BCD arithmetic,
it's storing 77 decimal or 22 decimal, depending on which end is the
MSB.
Unless it's a signed tens-comlement number, in which case it is 23
decimal.

You are given no encoding scheme such as positive or negative logic
or what the column values represent, all of which John touch upon.
The answer given above is on the outer fringe.

 

[John Popelish]

Thaqalain wrote:

If the machine is using negative logic with excess-3 BCD arithmetic,
it's storing 77 decimal or 22 decimal, depending on which end is the
MSB.
Unless it's a signed tens-comlement number, in which case it is 23
decimal.

They are messing with you.

 

[Bob Monsen]

wdflannery@aol.com wrote:

I'd like to hook up a flashlight lamp, 1.5 v battery, and capacitor in
series, close the circuit, and watch the lamp come on and slowly go off
as the capacitor charges up.

I tried it with a double electric layer capacitor (1 farad), but I now
think this type of cap has a big internal resistance, so, nothing
(observable) happened..

Now I'm thinking an electrolytic cap (0.1 farad) and maybe a 10 ohm
resistor in the circuit to get the time constant up to > 1 sec (R*C).

Is this going to work?

Ideas?

Thanks,
Will

I think the cap is doomed to failure. The bulb probably draws more than
500mA. Thus, the voltage will change at 0.5V/s. If there is even 1 ohm
of resistance in the cap, it'll suck up 0.5V, probably preventing your
bulb from lighting at all.

If you have a darlington transistor, (or two NPN transistors,) you can
use it to do something similar to what you want, although you will have
to use a couple of batteries in series.

Use a 100uF capacitor, and a 100k resistor; connect both to ground, and
to each other. Connect the positive junction between them to the base of
the darlington. Then, connect your flashlight battery to the emitter of
the darlington, and the collector to 3V.

Now, momentarily touch the junction of the base, 100uF, and 100k
resistor to 3V, through a normally open pushbutton or something.

It'll charge up to 3V. When you release the button, it will decay down
to 0V using a nearly exponential decay. The bulb will thus have an
exponential decay of the voltage across it, and should dim accordingly,
over about 5 seconds.

I just built one using two 2N4401 transistors to simulate the
darlington. I only had a 3V flashlight, so I set my power supply to
4.5V, and gave it a try. It starts out quite bright, and decays over a
period of about 7 seconds to nothing. It is hard to determine if the
decay is exponential, but I'm guessing that the non-linearities in light
vs voltage are keeping that from happening.

The flashlight bulb passes 0.5A at 3V, so the 4401 is getting warm.
However, it's well within the dissipation limits of the device.


.-------------o---o----------.
| | | |
no switch |/ | |
o------o----| ' |
| | |> |/ ---
| | |-| - 3V minimum
| | |> |
'+ .-. | |
=== | | | ---
/-\ | | .-. -
| '-' ( X ) |
| | '-' |
| | | |
'------o----------o----------'
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

---
Regards,
Bob Monsen

 

[Terry Pinnell]

John Fields <jfields@austininstruments.com> wrote:

On 9 Jun 2005 16:33:15 -0700, "davidd31415" <davidd31415@yahoo.com
wrote:

I'm looking for methods to obtain positive and negative 12 volt signals
from a single positive 12 v signal (simultaneously, not simply
reversing the polarity).

I'm specifically looking for a circuit design and an explanation of how
the circuit works.

---
When you say 'signal', do you mean something which only has two
states; +12V and 0V, and that when it's at +12V the other signal is
at -12V and that when it's at 0V the other signal is at 0V?

Another possibility is that Dave means 'supply', not 'signal'.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[John Fields]

On Thu, 09 Jun 2005 19:51:06 -0500, John Fields
<jfields@austininstruments.com> wrote:

On 9 Jun 2005 17:37:43 -0700, wdflannery@aol.com wrote:

I'd like to hook up a flashlight lamp, 1.5 v battery, and capacitor in
series, close the circuit, and watch the lamp come on and slowly go off
as the capacitor charges up.

I tried it with a double electric layer capacitor (1 farad), but I now
think this type of cap has a big internal resistance, so, nothing
(observable) happened..

Now I'm thinking an electrolytic cap (0.1 farad) and maybe a 10 ohm
resistor in the circuit to get the time constant up to > 1 sec (R*C).

Is this going to work?

No.

Try this:

+------->\ <------+
| \ |
|+ O |
[BAT] |+ [LAMP]
| [1F] |
| | |
+----------+---------+

It ought to work with the 0.1F as well.

---

Or, better yet:

+-----> |
| |
| O
|+ |
[BAT] +------+
| |+ |
| [1F] [LAMP]
| | |
+--------+------+

That way, when you turn the switch ON the battery will get the lamp up
to operating temperature and charge the cap at the same time, then
when you turn the switch OFF the cap will discharge through the lamp
without having to use a lot of its charge to heat the filament from a
cold start.




--
John Fields
Professional Circuit Designer

 

[Lord Garth]

"Genome" <ilike_spam@yahoo.co.uk> wrote in message
news:eCfqe.16284$%21.13004@newsfe2-gui.ntli.net...

"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:M98qe.3156$%j7.1022@newssvr11.news.prodigy.com...

"John Popelish" <jpopelish@rica.net> wrote in message
news:9emdnZL1X6PjlDTfRVn-uw@adelphia.com...
Thaqalain wrote:

Yikes. more poorly constructed questions.

I suspect this is done on purpose...perhaps to cause the unsure to
second guess themselves. I did give the engineers answer to a
Ph.D. uncle of mine when he asked if the glass was half full or
half empty...that being that the container is twice the required volume.
He had to think on that one!



Cross posted to SED.

The proper answer is.

'How much is left in the bottle?'

Obviously, for non-engineers, this will lead to a discussion about the
relative levels of liquid left in the bottle.

However the engineer will consider factors such as,

'Have I got another one?'
'Is the shop still open?'
'Will this be enough?'
'Do I have the means required to aquire another one?'
'Perhaps I'll just go to bed.'

DNA

Those are good questions...

Another quip to doctor relatives is that there is a 'vas deferens' between
children and no children.

 

[Dr Engelbert Buxbaum]

Impmon wrote:

Just had to post light bulb joke

Wonderful, thanks for sharing that with us!

By the way: How many students does it take to change a light bulb? -
None, they call the landlord.

 

[Bob Eldred]

"pinpassion" <mjdrum@comcast.net> wrote in message
news:1118411406.528867.42800@g47g2000cwa.googlegroups.com...

Hi Gang,

I have a large and heavy power supply transformer
that is part of a high current power supply project
that was featured in 73 magazine back in 1973. It
was started by a ham radio operator and was never
finished. I am going to finish it, if I can. I am
trying to identify the transformer leads. This is
for a 12 volt, 40 to 60 amp output. Here are the markings:

This transformer was made by ADC and is marked 541-010 REV H.
There is a marking on it that says 3-19470 and what I think
is a date code of 7438. There are terminal connections on one
side that are numbered 1 - 6. On the other side the connections
are numbered 7 - 14. I need to know the connection scheme for
this transformer so I can put it to use. I contacted who I
thought was the manufacturer "ADC" for information, and that
did not help.

I don't have the issue of 73 magazine, and I don't even know
if this is the same transformer that may have been part of the
construction article. I can take pictures and post them if that
might help.

Anybody have any ideas?

Thanks a lot.

Mike

The first thing you do is make a table of all of the resistances between the
various connections to determine where each winding is. All of the
resistances will be quite low some less than an ohm. The primary will
probably be two windings maybe with taps so that it can be series'd or
paralleled with taps to adjust for various line voltages. On a transformer
like that it is likely to be useable on line voltages from 90 Volts to 260
Volts depending on how the windings and taps are connected.

Once you think you have found the primary, the highest resistance
winding(s), connect it to a variac or variable transformer. My guess is that
terminal one connects to terminal four and terminal three connects to
terminal six and the line connects between terminals one and three, but that
is just a guess for 120 VAC. It doesn't matter what the power rating is
because you are not going to draw much current for testing purposes. The
purpose of the variac is to keep from blowing breakers if you have the
connections wrong. Advance the variac and while monitoring the primary
current with a clamp on ammeter if you have one, otherwise listen carefully
if it sounds like it is drawing high current with excessive hum or even
blowing a breaker or a fuse. If that happens, you obviously have the wrong
primary connections. When operating properly, the variac should get to the
line voltage without excessive current draw, noise or hum or popping
breakers.

With line voltage now on the transformer, measure the secondary terminals
with an AC voltmeter and map out all of the measurements. I suspect that
there are four secondary windings that can be series'd or paralleled for a
number of different voltages.
Bob