Chip with simple program for Toy

[Gene the Skeptic]

There are 2 major differences between the European and US electric systems:
(I use here layman language for comprehensibility)
1. The 240V US voltage is obtained by using two 120V voltages, 180 degree
out of phase from each other and connecting the load between them, thus
summing the voltages. In Europe the residential 220V is a single phase; one
terminal is equivalent to the US white wire, the other to the black wire,
only that the voltage is 83% higher.
2. The frequency in US is 60Hz and in Europe is 50Hz. When the same voltage
is applied, this causes AC-motors to overheat, because the an iron core is
designed for 60Hz and will saturate at 50Hz . To avoid the iron core
saturation at 50Hz one has to decrease the voltage to 50/60 x 240= 200V .
Now to your problem. This solution is only HYPOTHETICAL! One could connect
the European 220V to the BLACK and RED and create an "artificial" WHITE
using an insulation-transformer, with the primary connected to the 220V
(BLACK and RED) and the secondary, at 120V, connected one side to BLACK and
the other to WHITE (it must be an insulation transformer, not an
autotransformer for safety reasons). Because I do not know if the
electronics (supplied by the 120V) in the dryer is sensitive to phase
differences between BLACK and RED, I cannot tell you if it will work. What
is left is now the issue with the motor overheating, which perhaps will not
be excessive with just 10% voltage (200V to 220V). If you are willing to
experiment assisted by an electrician that knows the local code...that is
OK, but of course all the cautions and warnings do apply.
Gene
P.S. I appreciate comments from fellow EE's particularly regarding the
safety and the overheating issues.

 

[Gene the Skeptic]

OOPS, I made a mistake. An isolation tranformer connected in the way I
suggested, "becomes" equivalent to a autotransformer :-(

 

[Eric R Snow]

On Sun, 31 Oct 2004 07:22:09 -0500, "news" <news@abc.com> wrote:

Interesting idea. Yes I know that I will have to still use some wire. It
will just be so much easier with smaller gage. I can run it into the house
put it along baseboards etc. I would still like to use a small electronic
switch with low voltage wire. Can anyone point me to where I can find this
kind of thing.

Thanks.

You can use X10 devices for this. Google X10. There are lots of

dealers. X10 sends a signal over the house wiring. So, you would mount
a reciever inside the lamp housing and the transmitter in a convenient
switch box inside the house.
ERS

 

[BobGardner]

Does anyone know how to measure the Q value of an inductor,
============================

Its just wL/R isnt it? Or resonate it against a known C, measure hi and lo 3dB
down freq points, then compute f0/bw

 

[BobGardner]

Many of these
transistors have no markings on them at all. Others say things like
"NTCMP S3706" or "NTC P N4916".
=============================

You can tell if its pnp or npn with an ohmmeter.... if you know the pinout....
red on base, black on emitter showing some lo ohms... its npn

For the ones with numbers, type em into google... mps3706 sounds like a
motorola (or On semiconductor) number, 2N4916 might be a vaild number...

 

[john jardine]

"BobGardner" <bobgardner@aol.comma> wrote in message
news:20041031163200.28581.00002282@mb-m20.aol.com...

Does anyone know how to measure the Q value of an inductor,
============================
Its just wL/R isnt it? Or resonate it against a known C, measure hi and lo
3dB
down freq points, then compute f0/bw

Yes it is wL/R but the damned R will refuse to be easily pinned down. A
simple way is to series resonate the inductor and capacitor. That way there
is fewer complications from shunt loading effects dropping the measured Q
value.
Use a signal generator with a 50ohm output. Feed the output voltage to a
47ohm 1ohm potential divider. Feed the circuit from across the 1ohm
resistor. Measure the voltage feeding the circuit (across the 1ohm) and the
magnified resonant voltage across the L or C. Voltage ratio is the Q value
at that test frequency. It's not perfect but it'll get you up and running.
regrds
john

 

[John Fields]

On 31 Oct 2004 06:40:28 -0800, tek1940@hotmail.com (George) wrote:

Thanks John,

John Fields <jfields@austininstruments.com> wrote in message news:<bk97o0tvmm7ra6497kuak2526fnil3t54s@4ax.com>...
snip
---
Since you haven't mentioned anything to the contrary, it seems you're
_assuming_ that the battery voltage will remain constant at 8.2V
regardless of the load on it. It won't.


This is the method I've been using:
1. Measure all the components separately (no load).
2. Using Ohm's Law, calculate the expected result.
3. Connect the circuit
4. Measure with a MM.
I did the first and last tests again and measured the battery under
load.
In the first test (highest R), the battery under load showed same 8.2
V.
In the last test (lowest R), it showed 6.15 V.
Question: Does the Law refer to unloaded or loaded voltage source?

---
It refers to the current which will be forced to flow through a
resistance with a known voltage across it. _Measure_ the voltage
across the resistance, then divide that voltage by the resistance of
the resistor and you'll have the current flowing through the resistor.

If you don't, and you make the assumption that a 9V battery will stay
at 9V regardless of whether 1 milliamp or half an amp is being taken
from it you'll be wrong.
---

You also seem to be having trouble with the concept of the internal
resistance of the battery. Consider it a variable resistance in
series with the battery, located in the same housing as the battery,
but which you can't physically get to.

But, you _can_ measure it.
snip

I understand there is internal resistance in the battery. But I
assumed it is internal and that whatever is produced by the battery is
*after* the fact.

---
You still don't understand. It's not after the fact, it's part of the
deal.

The internal resistance is the same as a resistor that you can't get
rid of that's in series with the battery, and since the current in
series resistances is the same in all the resistances, the internal
resistance will drop some voltage when the load is drawing current,
and that voltage won't be available for the load to get the current it
needs.


Let's say you have a 9 volt battery with a 1 ohm internal resistance
hooked up to some load. It'll look like this:


+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| [LOAD]
| |
+--------------------------------+


Now let's say the load is 9 ohms. That'll make the circuit look like
this:

+------------------------+
+-|-[9V BATTERY+]--[1 OHM]-|-----+
| +------------------------+ |
| [9R]
| |
+--------------------------------+

Since resistors in series add, and the current in series resistances
is the same in all the resistances, we have 9 volts and 10 ohms, so
the current flowing in the circuit will be

E 9V
I = --- = ---- = 0.9A
R 10R


Now, if your 9 ohm load is a 9 watt light bulb and you expect to get a
certain amount of light from it when you connect it across the 9 volt
battery, you're in for a surprise, since you'll only be able to get
0.9A out of the battery because of its internal resistance.
---

I did a search on "battery internal resistance", found this quote: "A
battery's internal impedance increases with decreasing capacity due to
various conditions such as age, ambient temperature, discharge history
etc."

Note it does not say it changes according to the load.

---
So you don't know what the hell you're talking about, but I'm wrong
because you looked up something that didn't mention it?
Get a goddam clue.
---

Instruments that measure battery internal resistance have an upper
range of 40 ohms. Other web pages had graphs showing an upper limit of
420 milli ohms, I don't know what size battery they were talking
about, possibly those used in a cell phone.

Also note that my MM is an inexpensive model. In the first test I
reported that it showed .06 mA. But in my second test I noticed it
also showed .05 and .07. My point is that some of my data may not be
accurate enough.

---
The problem is that you didn't take _enough_ data. At every point
where you measured the current you should also have measured the
voltage across the resistor.
---

This is a great learning experience.

---
You're welcome.

--
John Fields

 

[Peter Bennett]

On 31 Oct 2004 18:54:13 -0800, tek1940@hotmail.com (George) wrote:

snip

Question: Does the Law refer to unloaded or loaded voltage source?

When you use Ohm's Law to calculate the current through a resistor,
you must use the voltage across the resistor at the time the current
is flowing - that is the only voltage the resistor knows about.


If I measure the battery at 8.2 V, can you not assume that is what is
realized in the circuit?

No - the actual battery voltage once you connect a load will depend on
the current drawn by the load, and the age of the battery.

If you're taking an exam and you calculate an answer based on the
unloaded battery (before connecting to the circuit), will the answer
be correct?

Exams are not real life - on exams you can probably assume that you
are given an ideal voltage source, which will provide the specified
voltage, regardless of how much current you try to draw.

A lead-acid battery (car battery) will have a very low internal
resistance. A small battery, like the tiny cells in a 9 volt battery,
will have a much greater internal resistance. Both the cell size and
chemistry affect the internal resistance.

Do you know how much internal resistance there is on a new 9 V
battery?
Why would it be higher than a 12 V car battery?

It would definitely be much higher than a car battery, but I don't
know by how much - and I don't want to risk damaging my meter (or
killing a battery) trying to find out.

A car battery has very large plates, so that it can deliver the 200
amps or so that are required to operate the starter without the
voltage dropping too much.

I've sent an inquiry to a 9 V manufacturer; it's not stated on their
data sheets.
I'm not convinced a 9 V battery could have 18167 ohms.

That sounds very much too high for a fresh battery.

I'd guess you might be able to get 250 mA or so through a short
circuit - that would be about 36 ohms. However, if you attempt to
draw 250 mA from a 9 volt battery, the chemicals near the plates will
be rapidly depleted, so the internal resistance will rise, and the
current and voltage will drop. If you let the battery rest for a
while, it will recover somewhat, so you will be able to draw a fairly
high (but probably not so high) current for another short time.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Robert Monsen]

Erik Durling wrote:

We're a couple of students doing an electronics project. We are to
construct a battery indicator circuit. We had a finished circuit on
paper and we had simulated it with Multisim 7. But when we tested the
circuit in real life, it didn't work at all like we had planned (or
according to simulation). The circuit was based upon the idea of
zenerdiodes having a constant voltage drop, and not conducting if the
circuit isn't able to supply the needed voltage across it. But the
zener-diodes doesn't stop conducting at all, and the voltage drop is
far from constant.

This is the main circuit: http://ersker.com/circuit.gif

The zener diodes used in real life are: BZX55 and not BZC55, but that
ought not change anything (right?).

We did a DC Transfer Analysis in the simulator and got three nice
curves showing the voltage at the three collectors (compared to
ground) as the voltage from the source goes from 0-4.5 volts, and
everything seems right there: http://ersker.com/dctrans.gif

So, the problem we're having with our real circuit is that the LEDs
never turn off (the transistors are always saturated). The zener
diodes doesn't stop conducting when the voltage across them goes under
their specified value.

What's our mistake/misconception?

Do the zeners only operate correctly at a certain current?

As I've pointed out elsewhere, Zener models appear to suck.

Here is a simple circuit that will do what you need:

Vin
|
|
o------.
| |
| |
e |
.---b [1k]
| c |
| | |
| | |
{10k] [1k] |
| | e
| o-----b
| | c
zener | |
| [10k] LED
| | |
'----o------'
GND

When Vin is above the zener voltage + a bit, the left transistor will
conduct. If its conducting, then the base of the right transistor will
be higher than the emitter + 0.7, that transistor will be off, turning
the LED off.

When the voltage drops below the zener voltage, the left PNP transitor
will turn off, causing the right PNP transistor to turn 'sharply' on. As
the voltage continues to decay, the LED will remain on, but will get dimmer.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

+--------------------------+
| |_
| +-[9V SOURCE+]--[Rint]--->_|+9V
| | |
| | |
| | |_
| +------------------------>_|-9V
| |
+--------------------------+

You guys actually have the patience to do these great "schematic" sketches in
your mail reader, or is there a secret I don't know about that makes it easier
than it appears?

They are handy, and given the right font do a great job. Just seems rather
labor intensive to me.

Hey, if its by hand, and takes as long as I think, kudos to all who do it as
its a big help.

John

 

[Clarence]

"dB" <dmb06851@yahoo.com> wrote in message
news:1757808.0411010252.608b8c53@posting.google.com...

Read this.

http://laenga.com/extras/politics.htm

Had nothing to say.
Political BS!

 

[Tom Grayson]

The major player in answering this question is "volume"
Because the manufacturer can expect to sell millions of thses things,( DVD
rom Drives) The development cost can be split, a few dollars over each item
with no major impact in the overall price. After the development costs have
been covered the manufacturer can then turn a reasonable profit before
market saturation.

Not so with larger drives. The thing is there are a lot more "Players" in
the motor drive market and in reality, if a maunfacturer goes to market with
say a 2KW Drive there are probably more then 30 manufacturers out there each
trying for their share of a rather limited market, so the prospective sales
will be maybe only in the 1,000's and not the millions that the DVD rom
makers have at their disposal.

another factor is PC users tend to have a rather "Geeky" approach to their
equipment, with a fair market wanting the Latest and greatest, So they would
be perfectily willing to throw away a perfectly good "Widgett" to replace it
with another widgett of similar cost but more functions or better
specifications. Usually in the industrial drive market, once a device is up
and running, There it stays until it breaks or the whole machine is ready
for an upgrade.

The other thing is the DVD Rom Drives are a dedicated application, Known
Supply Voltage, Known load inertia, One job description. as we all know
the application of Variable speed AC Drives has about as many diferent jobs
and supply voltages as you immagination can put together. any reputable
drive maker must design his drive to be as adaptible as possible so that it
can be applied to any of these aapplications. This all adds Cost to the
development.

Having said that, there, is no reason why the cost of drives will not fall
as time progresses, as we learn how to do things more cost effectively and
new technology either takes away some of the obsticales or makes them easier
to contend with.

Well there's my 2cents worth.

"~Dude17~" <dude17@sacbeemail.com> wrote in message
news:b959931f.0409102228.5a14eff2@posting.google.com...

X-No-Archive: Yes

I took apart a DVD-ROM drive for the heck of it and the control LSI is
amazing stuff.

The driver chip has everything to directly drive all the mechanical
parts in a CD-ROM drive from loading tray, focus, tracking, sled and
the spindle motor. I find the spindle motor control the most
fascinating.

http://www.rohm.com/products/databook/optdisc/pdf/bd7902cfs.pdf

The chip controls the spindle motor using three phase PWM and reading
the controller documentation leads me to believe the thing can be
controlled somehow with pin 24.

The range is rather wide 230RPM while playing back audio CD at outer
diameter and about 10,000RPM at 48x CAV mode. The chip can also apply
reverse torque to quickly bring the disc to stop.

Is it difficult to make a variable speed drive using the spindle motor
and the LSI pulled from a DVD-ROm drive to let me run the motor
anywhere from 280 to 10,000RPM outside of the original drive? It
would surely make a cool project part.


If this sophisticated control can be built into a $20 DVD-ROM drive,
how expensive would it be to integrate a similar controller with
beefier drive circuit to drive a motor in few hundred watt to a few
kilowatt range?

 

[John Fields]

On 28 Oct 2004 06:03:21 -0700, egd.bydalen@swipnet.se (Erik Durling)
wrote:

We're a couple of students doing an electronics project. We are to
construct a battery indicator circuit. We had a finished circuit on
paper and we had simulated it with Multisim 7. But when we tested the
circuit in real life, it didn't work at all like we had planned (or
according to simulation). The circuit was based upon the idea of
zenerdiodes having a constant voltage drop, and not conducting if the
circuit isn't able to supply the needed voltage across it. But the
zener-diodes doesn't stop conducting at all, and the voltage drop is
far from constant.

This is the main circuit: http://ersker.com/circuit.gif

The zener diodes used in real life are: BZX55 and not BZC55, but that
ought not change anything (right?).

We did a DC Transfer Analysis in the simulator and got three nice
curves showing the voltage at the three collectors (compared to
ground) as the voltage from the source goes from 0-4.5 volts, and
everything seems right there: http://ersker.com/dctrans.gif

So, the problem we're having with our real circuit is that the LEDs
never turn off (the transistors are always saturated). The zener
diodes doesn't stop conducting when the voltage across them goes under
their specified value.

What's our mistake/misconception?

Do the zeners only operate correctly at a certain current?

---
I've read some of the replies to your original post, above, but your
last post seems to show that you're still pretty much in the dark, so
here goes:

If you hook up this very simple circuit:

+VIN>---+
|
|K
[Z4.3V]
|
+---->VOUT>---+
| |
[R1] [VOLTMETER]
| |
GND>----+-------------+

You'll notice that if you start with +VIN at 0V and then make it more
and more positive, the voltage across R1 will start to increase long
before +VIN gets to 4.3V. The reason for that is because the Zener
isn't a perfect switch and it will start allowing current to flow
through itself before its Zener voltage is reached. Zeners are
designed to be shunt regulators, and if you look at a data sheet for a
Zener you'll find that the Zener voltage is only guaranteed to be
within a certain range of voltages if the current through the Zener is
the "test current", usually 20mA for 1/2 watt diodes with a Zener
voltage of 12V or less.


Now, if we take a look at your circuit:

VIN>-----+-------------+
| |
| [LED]
[1K] |
| [150R]
| |
| C
+--[<ZENER]--B Q1
| E
[10K] |
| |
GND>-----+-------------+

we can see that as VIN starts to go more and more positive, more and
more current will start to flow through the base-to-emitter junction
of the transistor, just because the Zener isn't a switch and will
start to conduct well below its Zener voltage. If you have a
transistor with a reasonably high beta (100 to 300) then the current
which is flowing through the Zener (and also through the b-e
junction)will cause a collector current to flow which will be 100 to
300 times higher than that. So, even if you're way below the Zener's
knee and you have, say, 10ľA of reverse current flowing, the collector
current will be somewhere between 1mA and 3mA, which will be enough to
light the LED through that 150 ohm resistor.

If what you're trying to do is build something with LEDs which light
sequentially as the supply voltage increases, then you need to use
three comparators, a voltage reference, three LEDs and a handful of
resistors.

Want a schematic?













--
John Fields

 

[Squidster]

On Sun, 31 Oct 2004 09:11:34 -0800, James Howe wrote:

"NTCMP S3706" or "NTC P N4916". Is there a reasonable way to learn

NTC is a Taiwanese semiconductor manufacturer; Nanya Technology
(http://www.nanya.com), and probably OEMs for Motorola/ONSemi for the
MPS3706, which is Motorola's original implementation of the generic
2N3706. Datasheet is here:
http://www.datasheetarchive.com/search.php?search=mps3706&mfg=ALL

PN4916 is originally from Fairchild, in your case is probably OEM-sourced
implementation from NTC too.
Datasheet is here:
http://www.datasheetarchive.com/search.php?search=pn4916&mfg=ALL

It all boils down to exposure and experience. Just like everybody knows
what brands of PC exist in the market. You'll soon learn what are the
dominant seminconductor manufacturers are there...

 

[CBarn24050]

Subject: Re: zener trouble
From: Robert Monsen rcsurname@comcast.net
Date: 01/11/2004 07:28 GMT Standard Time
Message-id: <cAlhd.282869$wV.19497@attbi_s54



Erik Durling wrote:
We're a couple of students doing an electronics project. We are to
construct a battery indicator circuit. We had a finished circuit on
paper and we had simulated it with Multisim 7. But when we tested the
circuit in real life, it didn't work at all like we had planned (or
according to simulation). The circuit was based upon the idea of
zenerdiodes having a constant voltage drop, and not conducting if the
circuit isn't able to supply the needed voltage across it. But the
zener-diodes doesn't stop conducting at all, and the voltage drop is
far from constant.

This is the main circuit: http://ersker.com/circuit.gif

The zener diodes used in real life are: BZX55 and not BZC55, but that
ought not change anything (right?).

We did a DC Transfer Analysis in the simulator and got three nice
curves showing the voltage at the three collectors (compared to
ground) as the voltage from the source goes from 0-4.5 volts, and
everything seems right there: http://ersker.com/dctrans.gif

So, the problem we're having with our real circuit is that the LEDs
never turn off (the transistors are always saturated). The zener
diodes doesn't stop conducting when the voltage across them goes under
their specified value.

What's our mistake/misconception?

Do the zeners only operate correctly at a certain current?

Hi, if you look at the datasheet you will see why it doesn't work. Why you
would even consider simulating such a circiut is beyond me.

 

[karlgabel]

Ran into this looking for replacements for a dying (due to parts shortages)
board at work and had to reply .

The first reply concerning "A) the file to put into the chip and B) the
means to get the file into the chip" is in error. The programming
information for the chip is located off the chip in either a serial EPROM
(xc17XX - where xx is the size of the EPROM), parallel EPROM (27256 or
some variant) or could be loaded by a processor.

Still looking for a xc3030-70? I have some XC303-50PC84C that may work if
the package is correct. The part I have is faster than the -70 so it
should not be a problem. Digikey (http://www.digikey.com) also carries
the XC3030 in various packages.

Let me know if you are interested. I have six parts laying around
gathering dust.

 

[karlgabel]

Ran into this looking for replacements for a dying (due to parts shortages)
board at work and had to reply .

The first reply concerning "A) the file to put into the chip and B) the
means to get the file into the chip" is in error. The programming
information for the chip is located off the chip in either a serial EPROM
(xc17XX - where xx is the size of the EPROM), parallel EPROM (27256 or
some variant) or could be loaded by a processor.

Still looking for a xc3030-70? I have some XC303-50PC84C that may work if
the package is correct. The part I have is faster than the -70 so it
should not be a problem. Digikey (http://www.digikey.com) also carries
the XC3030 in various packages.

Let me know if you are interested. I have six parts laying around
gathering dust.

 

IF I do without the "sidetone"(mic voice reproduced in earphone of
same handset) what do I have to do AT A MINIMUM to make 2 handsets
into "intercom" ?
On Tue, 2 Nov 2004 15:32:18 -0800, "Joel Kolstad"
<JKolstad71HatesSpam@Yahoo.Com> wrote:

"Roger Johansson" <no-email@home.se> wrote in message
news:Xns9595BEEF61A2886336@130.133.1.4...
But the sound from each mic needs to be fed to both earphones,
because you need to hear yourself as well as the other person in your
earphone. The sound which reaches your ear becomes very confusing if you
do not hear your own voice as well as the other person.

^^^^
As far as I can tell, there are a lot of cell phones out there that don't
provide this 'sidetone,' and people don't seem to get too confused using
them.

---Joel

 

[Active8]

On 2 Nov 2004 18:33:12 -0800, JeffM wrote:

You guys actually have the patience
to do these great "schematic" sketches in your mail reader,
or is there a secret I don't know about that makes it easier ?
John (uvcceet)

:One thing that's interesting about it is that
:this way schematics can get archived as text files
:by outfits that don't archive binary files, like Google,
:and yet the data's there, and worth a thousand words.
: John Fields

And that thousand word savings in bandwidth leaves us plenty of
room for off-topic stuff.

Yup. What he said.

The name of the program Fields mentioned is:
Andy's ASCII Circuit by Andreas Weber
http://groups.google.com/groups?&q=andy%27s-ascii-circuit&meta=group%3Dsci.electronics.*
As you hang around, you will see this appended to some drawings.

No he won't, it's:

|
||-+
||<-
-||-+
|
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

BTW, are you really using OS/2?

--
Best Regards,
Mike

 

[Active8]

On 31 Oct 2004 09:11:34 -0800, James Howe wrote:

I recently purchased a 'grab bag' of components which included a bunch
of transistors. Most of the transistors are what I think of as a
typical transistor, black case, flat on one side. Many of these
transistors have no markings on them at all. Others say things like
"NTCMP S3706" or "NTC P N4916".

In a male or female voice? That's a big clue.

Is there a reasonable way to learn
more information about the specifications for these transistors?

Besides those other good suggestions, electronics parts stores (real
ones) have cross-reference guides. They gave me one (the NTE book) a
long time ago. Note, the specs of the original and replacement might
not be an exact match, but hey, it's a grab bag

Maybe Bainsville Electronics in b'more (or someone like that) has a
cross-ref search on their web-site.
--
Best Regards,
Mike