Chip with simple program for Toy

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.*
created by Andy s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

Thanks for those links. I will be checking them out.

BTW, are you really using OS/2?

Yep. VIrus and spyware free since 1992

John

 

[Roger Johansson]

colklink@newsguy.com wrote:

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" ?

If you use one op-amp you can have sidetone for no extra cost.

Mix the two mic signals on the input of the op-amp, send the output to
the earphones.


Separating the signals so you get no sidetone will cost more components.
You will need two amplifiers (op-amps) then.

The mics you have there are probably electret mics, which need a power
supply to work, but you can hopefully fix that from the same battery
which drives the opamp. Use a battery instead of a power supply, it will
reduce noise and hum problems and make the circuit portable.

Look up a standard op-amp circuit called audio mixer.

So the minimum is one op-amp and a few resistors and capacitors, and a
battery.

But a quad op-amp in one capsule is just as cheap as a single op-amp, so
I would recommend to use one op-amp for each mic as mic amplifier, one
op-amp as audio mixer, and the last op-amp as loudspeaker amp, in case
you will need it in the future.

Look up electret mics power circuits, study the mics you have and find
out how they need to be supplied with power.

Or describe the equipment you have in detail, and hope for somebody to
give you a readymade circuit schematic.

Don't you have any specifications on these headsets, what does it say,
technical details like impedances, mic power supply, connectors details,
etc..


--
Roger J.

 

[Nick Hull]

In article <20041025192821.17053.00001820@mb-m26.aol.com>,
bobgardner@aol.comma (BobGardner) wrote:

I need to determing how much electricity a solar water heater is
replacing.

The water heater is either on or off. If you could get a 220V clock, you could
gator clip that onto the heating element and see how many hrs a dat its
running. Quick... how many BTUs in a KWHR? ans: 3412

Since 220 volts uses 2 110 volt legs a 110 volt clock can be made to
work. Use analog not digital.

--
Free men own guns, slaves don't
www.geocities.com/CapitolHill/5357/

 

[John Fields]

On 2 Nov 2004 20:58:15 -0800, tek1940@hotmail.com (George) wrote:

John,
R1 = battery internal resistance
R2 = meter internal resistance


snip

Now to see how the internal resistance of the battery affects the
battery output voltage, let's connect a 20000 ohms-per-volt analog
multimeter across the battery and see what happens. Let's say the
meter has a 10 VDC range. That would make its internal resistance
200000 ohms if we selected the 10VDC range and, with a 1 ohm internal
resistance for the battery we wind up with a reading of:


E1R2 9V * 200000
E2 = ------- = --------------- = 8.999955V
R1+R2 1R + 2000000R

See the voltage falling? That's because more voltage is being dropped
across the battery's internal resistance as it's required to supply
more current.


Isn't the current increase due to the lower value of R2 (meter internal resistance)?

---
Well, in a manner of speaking. Before the meter was connected to the
battery there was _no_ current flowing through R1, so battery + was
sitting at 9.0V. Then, when the meter was connected, the infinite
resistance across the + and - terminals of the battery went to 200k
ohms, so current stared to flow out of the battery, through R1,
through the meter, and then back to the battery. That current flowing
through R1 caused a voltage drop across it, so the meter could only
measure the voltage from the end of R1 not connected to the battery to
battery minus, and that's why it didn't read 9.000000V.
---

So far, with the voltmeter loads, the drop has been small, but it
starts to matter when you start drawing significant current from the
battery. for example, assume you have a 20 ohm load and that because
you've figured out that

E 9V
I = --- = ----- = 0.45A
R 20R

you expect the 9V battery to deliver 450mA into the load.


Well, if you look at

E1R2 9V * 20R
E2 = ------- = ------------ = 8.57V
R1+R2 1R + 20R

you'll find that, because of the battery's internal resistance you can
only get 8.57V across the 20 ohm resistor, which is only going to
allow about 429mA to flow through the load.

That's also borne out if you do:

E 9V
I = ------- = ----- = 0.42857...A
R1+R2 21R


Why is R2 (meter internal resistance) not utilized here?

---
Because it's insignificant compared to the internal resistance of the
battery and the load it's in parallel with, but that's an excellent
observation and you bring up a valid point which we'll get to in a
second.

The meter is still connected.

Yes, it is, and it's in parallel with R2, so the circuit now looks
like this:

+-----+---->E1
| |
| [R1]
+| |
9V +------+--->E2
| | |
| [R2] [R3]
| | |
+-----+------+--->0V

where R3 is the internal resistance of the voltmeter.

R2 and R3 are in parallel, and to determine the total resistance of
the pair we can write:



R2 * R3
Rt = ---------
R2 + R3


Where Rt is the total resistance of the pair.


For your digital multimeter, with a 10M input resistance we'll get:



R2 * R3 20 * 10E6
Rt = --------- = ---------- = 19 999 960 ohms
R2 + R3 20 + 10e6


and for the 20,000 ohms-per-volt analog multimeter we'll get


R2 * R3 20 * 200000
Rt = --------- = ------------- = 19 998 000 ohms
R2 + R3 20 + 200000

The Load is a new type of resistance.
Shouldn't it be represented as R3?

---
It could be, since once we've figured out the total resistance we can
call that value anything we want. But, for the moment, just to keep
us from forgetting that there are really two resistors there let's
keep on calling it Rt.

OK, now if there are no meters in there we'll have:


+-----+---->E1
| |
| [R1]
+| |
9V +---->E2
| |
| [R2]
| |
+-----+---->0V

and if we start off with E1 = 9V, R = 1 ohm and R2 = 20 ohms,

we know from last time that we can write:


E1 * R2 9 * 20
E2 = --------- = -------- = 8.571429V
R1 + R2 1 * 20



Now, if we substitute Rt for R2 we'll have:

+-----+---->E1
| |
| [R1]
+| |
9V +---->E2
| |
| [Rt]
| |
+-----+---->0V


and, if we plug in the value we got for Rt with a 10Mohm DMM in there
we'll have:


E1 * Rt 9 * 19.999960
E2 = --------- = ------------- = 8.571428V
R1 + Rt 1 + 19.999960


If we plug in Rt for the analog voltmeter we'll have


E1 * Rt 9 * 19.998
E2 = --------- = ------------- = 8.57138
R1 + Rt 1 + 19.998



So you can see that things only start to get significant about four or
five places to the right of the decimal because the resistances of the
voltmeters are so large compared to the internal resistance of the
battery.

--
John Fields

 

[john jardine]

"sk" <yshashi@gmail.com> wrote in message
news:9b0842f2.0411031108.7ce0b410@posting.google.com...

Hey
I'm actually designing a comparator circuit


|--------feedback resistor------|
| |
| |---------------------| |
RC CIRCUIT ----|------|+ | |
| COMPARATOR CIRCUIT | |
INPUT----------|- |--|---------OUT
|---------------------|
The task is to design this cirucit in such a way that when an input
pulse (0.5,-2.5) with a pulse width of 8micro seconds is given it
should produce me an output which remains high for atleast like
64micro seconds..i tried using R=2248.5k and C=8500pF and the feedback
resistor i used was 90k..i was able to get an o/p which remains high
for 64microseconds but when looked at it carefuly i was able to figure
out that the outout remained high just beacuse of the large difference
between the resistors(the storage capacitance is not much observed in
the o/p graph)...it should basically work like a monostable
multivibrator (without using ic 555 timer though)...it would be of
great help to me if u could give me some suggestions as how could i do
this..

thanks
sk

Homework or not, I needed a rest from programming a damned PIC and as a
starter, suggest this ...
Should give a -repeatable- timed pulse of about 65uS.

+5V +5V +5V +5V +5V +5V
| | | | | |
.-. | --- .-. | |
| | | --- | | | |
| |1k | 1n| | | | .-.
'-' | | '-'10k | | |
|| | ___ |< | | | | |1k
o--||--o-|___|--| BC556 | | | '-'
IN || 10k |\ | | |\| |
o 1n | | o----|-\ |
| | | | | >--o---o
| '------o----------|+/ OUT
| | | |/|LM393 o
| | | | |
| .-. .-. | |
| | | | | | |
| | | | | | |
| 56k'-' '-'6k8 | |
=== | | | |
GND === === === ===
GND GND GND GND

(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

 

[john jardine]

"john jardine" <john@jjdesigns.fsnet.co.uk> wrote in message
news:cmbt70$ht0$1@news8.svr.pol.co.uk...

"sk" <yshashi@gmail.com> wrote in message
news:9b0842f2.0411031108.7ce0b410@posting.google.com...
Hey
[clip]

No good. Belay that. I Didn't read the question. The 8uS input pulse is the
critical feature.

 

[Peter Bennett]

On Wed, 03 Nov 2004 17:29:41 -0600, John Fields
<jfields@austininstruments.com> wrote:

R2 and R3 are in parallel, and to determine the total resistance of
the pair we can write:



R2 * R3
Rt = ---------
R2 + R3


Where Rt is the total resistance of the pair.


For your digital multimeter, with a 10M input resistance we'll get:



R2 * R3 20 * 10E6
Rt = --------- = ---------- = 19 999 960 ohms
R2 + R3 20 + 10e6

Watch those pesky periods! That should be 19.999 960 ohms

and for the 20,000 ohms-per-volt analog multimeter we'll get


R2 * R3 20 * 200000
Rt = --------- = ------------- = 19 998 000 ohms
R2 + R3 20 + 200000

And that is 19.998 000 ohms - both such a tiny bit under 20 ohms, that
we may as well call it 20 ohms.


--
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

 

[Rich Webb]

On 31 Oct 2004 09:11:34 -0800, googlegroup@wingspread.imap-mail.com
(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". Is there a reasonable way to learn
more information about the specifications for these transistors?

Radio Shack has a handly little (and inexpensive) tester that will
classify common small signal transisters by type (NPN, PNP) and flag the
base and collector leads. Part # 22-330. Not sure that this is in their
current catalog but you might be able to find a store with one.

Also, many general-purpose multimeters have a socket that accepts a
transistor and that then reads out hfe (more or less). It helps to know
the type and CBE order, though, so even with one of these the little RS
gadget is helpful.

--
Rich Webb Norfolk, VA

 

[John Fields]

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

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

---
Only if you measure the battery voltage when the battery is under
load.
---

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?

---
No, but if the the purpose of the exam is just to see whether or not
you know how to use Ohm's law, it won't matter. However, if the
purpose of the exam is to see whether you know what internal
resistance is about, you'll flunk!
---

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. (Example, say my company makes a product. We buy a
component widget. The widget company may have various "internal
resistance" to make it, but when it arrives at our company, all we
care is that meets our specs.)

The internal resistance of the battery is indeed inside the battery,
but it is still in series with the battery terminals, so it will cause
the battery terminal voltage to drop when current is drawn from the
battery.

Not understanding this yet. If you measure the V at the battery's
external posts, does this not include the internal resistance?

---
AHA!!! There it is...

Yes. it does, but you don't see it because the resistance of the meter
is so much higher than the internal resistance of the battery.

Taking a look at this circuit,

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

the 9v source and Rint represent the battery and its internal
resistance, and the box around them represents the case; just like a
regular 9V battery is put together.

Now, let's say that Rint is one ohm and that you've got your voltmeter
hooked from +9V to -9V, like this:

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

Let's also say that the internal resistance of your voltmeter is 10
megohms.

Then, for convenience, we can redraw the circuit to look like this:


+-----+---->E1
| |
| [R1]
+| |
9V +---->E2
| |
| [R2]
| |
+-----+---->0V

Where R1 is the internal resistance of the battery, R2 is the internal
resistance of the meter, E1 is the voltage from the 9V source, and E2
is the voltage across the meter.


Now, if we calculate the current being drawn from the battery, we can
say:

E 9V
I = ------- = ------------- = 0.00000089999991A = 899.9991nA
R1+R2 10,000,001R

and if R2 is equal to 10 megohms, the voltage the meter will see will
not be 9.0V, it'll be 8.9999991V, because the other 0.0000009V will
have been dropped across R1, the battery's internal resistance.

Considering that R1R2 is a voltage divider, there's a much more
convenient way to find out what the voltage across the meter will be,
and that's to use


E1R2
E2 = -------
R1+R2


For the 10 Megohm case you'll get:


E1R2 9V * 10000000R
E2 = ------- = ----------------- = 8.9999991V
R1+R2 1R + 10000000R

which is just what we got before.


Now to see how the internal resistance of the battery affects the
battery output voltage, let's connect a 20000 ohms-per-volt analog
multimeter across the battery and see what happens. Let's say the
meter has a 10 VDC range. That would make its internal resistance
200000 ohms if we selected the 10VDC range and, with a 1 ohm internal
resistance for the battery we wind up with a reading of:


E1R2 9V * 200000
E2 = ------- = --------------- = 8.999955V
R1+R2 1R + 2000000R

See the voltage falling? That's because more voltage is being dropped
across the battery's internal resistance as it's required to supply
more current.

If we use a voltmeter with a 1000 ohm iron vane movement to measure
the battery voltage, it'll draw 9mA from the battery when it's
connected and it'll read:


E1R2 9V * 1000R
E2 = ------- = ------------ = 8.991001V
R1+R2 1R + 1000R


The reading dropped again because of the higher current flowing
through the battery's internal resistance, and the higher the load
current goes, the higher the drop across the internal resistance will
be.

So far, with the voltmeter loads, the drop has been small, but it
starts to matter when you start drawing significant current from the
battery. for example, assume you have a 20 ohm load and that because
you've figured out that

E 9V
I = --- = ----- = 0.45A
R 20R

you expect the 9V battery to deliver 450mA into the load.


Well, if you look at

E1R2 9V * 20R
E2 = ------- = ------------ = 8.57V
R1+R2 1R + 20R

you'll find that, because of the battery's internal resistance you can
only get 8.57V across the 20 ohm resistor, which is only going to
allow about 429mA to flow through the load.

That's also borne out if you do:

E 9V
I = ------- = ----- = 0.42857...A
R1+R2 21R

Does the internal resistance only get activated when a load is connected?

---
No, it's always there but it doesn't matter until you start taking
current out of the battery. Kind of like even though the electricity
from the power company is always there, on the other side of the
switch, you don't get charged for it until you turn on the switch and
start using it.
---

I'd like to see a circuit diagram of the innards, showing the path
from one post to the other, thru the resistance.
Evidently I missed "Batteries 101".

---
Read what follows "AHA!!! There it is...", previous.
---

Several web sites I've read indicate the internal resistance only
increases with age; a higher discharge history would increase the
resistance faster.
I didn't see indications that it's a variable resistance changing with
the load, as John stated.

---
Never mind what you read, find out for yourself what the deal is.
Make your measurements again measuring the battery voltage and the
current through the load for each measurement you make, then calculate
the various internal resistances you get for the battery with
different loads and post what you find.
---

Do you know how much internal resistance there is on a new 9 V
battery?

---
Measure it
---

Why would it be higher than a 12 V car battery?

---
because it's smaller and the chemistry is different.
---

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.

---
Measure it.

--
John Fields

 

[John Larkin]

On Mon, 01 Nov 2004 07:28:08 GMT, Robert Monsen
<rcsurname@comcast.net> wrote:

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.

That's even worse than the original circuit. More parts, too.

John

 

THANKS TO ALL

 

[John Popelish]

ShadowTek wrote:

OK thanks.
I thought 3 Amps for a AA battery sounded strange but I was confused
by the markings on the battery.
They were "LRG AM3 1.5V".
I guess LRG just means large and AM3 means short curcuit current.

I doubt that the designation LRG AM3 has anything to do with size or
amperage.
Here is a summary of some Duracell specs:
http://www.web-ee.com/Component%20Pages/battery_info.htm
Note that the MN1500 AA cell is rated for 2850 milliampere hours and
its maximum rated load is 43 ohms for a load current of 28
milliamperes. Now, .028*43=1.2 volts, so this implies that a typical
internal resistance must be dropping the rest of 1.5 volts. So the
internal resistance is about .3volts/.028 amperes or 10 ohms. So the
short circuit current should be about 1.5 volts / 10 ohms = .15
amperes. I am quite sure that you will get more than this with a
fresh cell, but this is typical for the mid life point.
Here is another site that summarizes battery characteristics:
http://www.zbattery.com/zbattery/batteryinfo.html

Note the resistance of the alkaline AA cells.

I was further thrown off by trying to read the current with my
multimeter from terminal to terminal.
I guess that I was createing a short curcuit which was trying to empty
the battery.

Yes. Current meters have very low resistance.

(BTW, my multimeter only reads up to 250mA so do you think reading a
3A short curcuit for no more than 2 seconds would have damaged the
MM?)

If it reads more than zero then you haven't blown the fuse in series
with the current meter function.

The exact specs for the LEDs were 2.8V-Min. 3.2V Typ. 3.8V-Max..

Yes, the required voltage varies with temperature and lot to lot.

So if intend to run the LED at 2.9V then I should still be able to run
1 LED off 2 1.5V cells right?

The two cells with their 20 ohms of total internal resistance will
push some current through the LED, but the exact amount is hard to
predict, except that it will be low.

.1 divided by .02 = 5 Ohm reistor?
So 2 AA batteries and a 5 Ohm resistor will run 1 of these LEDs for
150 hours?

The resistor is in addition ot the 20 ohms of internal resistance, and
at this low voltage the current will be much lower than 20
milliamperes, so the battery life will be very long.

Or I could wire several LEDs in parallel so long as a 5 Ohm resistor
prefixed every led?

Yes. But if the LEDs vary all over the specified voltage range, you
can expect the one with the lowest drop to be drawing much more
current than the one with the highest drop.

This is why I recommended that you use 3 cells and a much higher
resistance in series with each. It makes the current matching much
better and more independent of LED variations and sag in battery
voltage.

Do LEDs suffer any damage from being underpowerd?

Just the opposite. Their light output is roughly proportional to
their current, but their lifetime goes up dramatically as the current
goes down.

Since we are talking about batteries then the constant drain would be
a problem.
I have a 3 white LED headlamp and I have run the batteries in it all
the way down several times.
It uses 2 flat 3V lithium batteries.
It still seems to work ok.

BTW, radio shack sells their white LEDs for 6$ a peice.
If you are used to payin that much then you should check out the link
I posted earlier.

--
John Popelish

 

[Dmitri(Cabling-Design.com]

news wrote:

Im an amateur/novice at electronics. Mostly I'm interested in
electronics to
fix things and make life a little easier around the house.

My problem is that I have a light that is on a switch outside the
house. I
want to turn it on and off from inside the house. It's in an awkward
place
and makes it hard and unsightly to run 120v Wires and switch box etc.
into
the house. I would like to build an electronic switch that would only
require me to run a small wire to an electronic switch from the light
into
the house. Can anyone point me to a schematic for this type of thing
that I
can build. I want the switch to provide its own power to the low
voltage
switch in the house. I don't want to be fussing with some sort of AC/DC
adapter to provide the switch power.

Can anyone help me out.

Thanks.

That can be easily handled by a pair of X-10 devices: a lamp module for
that light (screw-in or wall switch replacement) and a mini-controller,
which you install in a convenient place. Once you try it, you'll want all
your house on X-10 switches ;-). Very easy to install, everything is
pretty much plug-and-play. Do eBay search for X10 (X 10, X-10, you got the
idea), you'll normally find dozens of sweet deals there.

--
Dmitri Abaimov, RCDD
http://www.cabling-design.com
Cabling Forum, color codes, pinouts and other useful resources for
premises cabling users and pros
http://www.cabling-design.com/homecabling
Residential Cabling Guide
-------------------------------------






##-----------------------------------------------##
Article posted with Cabling-Design.com Newsgroup Archive
http://www.cabling-design.com/forums
no-spam read and post WWW interface to your favorite newsgroup -
sci.electronics.basics - 5366 messages and counting!
##-----------------------------------------------##

 

[BobGardner]

I tried to read up on electrical stuff but there is a lot I dont
understand.
=======================

Most of it is derived from Ohms Law.... E=IR (voltage=current x ohms). The gist
of it is, voltages in series add, but the current in a series circuit is the
same in all series elements of that circuit... makes sense if you think about
it. The trend is to use special led driving voltage regulators that drive
strings of several (6 or 7) leds in a string, 20ma each string. The regulator
keeps the right voltage and current even as the battery voltage starts
dropping. You could use a 12v battery and drive 3 leds in series with a
resistor... 12V from batt - 9v across leds leave 3v/20ma for the resistor to
soak up... 150 ohms. You could drive as many of these 3 diode and a 150 ohm
resistor strings as you can afford.

 

[BobGardner]

Transformers are supposed to be 90+ percent efficient, unless they are
saturating... A 1st cut is the VA 'apparent power'.... stick a multimeter in
series with the primary and see how much current its drawing.... 10
watts/120V=83ma

 

[John G]

"Frederic C" <cecile.frederic@telefonica.net> wrote in message
news:db3f0fdc.0411051408.4f8ec054@posting.google.com...

Thank you Gene,
First of course the 240V is AC, not DC.
I am willing to try your proposal. I do have a power transformer 240V
=> 110V, but how do I know if it is an insulation-transformer ?
If I understand well, I connect the 240V power to black and red, and
the 110V power (coming out of the transformer) to black and white ...
or should it be the red and white ?

Thanks again for your help.
Frederic.


I can only repeat what I said in an earlier reply:-

"You need help from a knowledgeable local electrician."

If you have to ask about an insulation transformer then you need help
It is an ISOLATion transformer you are looking for.

And to rely on advice over the net about wire coloUrs could be
suicidal.
--
John G

Wot's Your Real Problem?

 

[Rusty Wright]

My favorite "help me learn electronics" site on the web is

http://www.ibiblio.org/obp/electricCircuits/

When you click on each volume, scroll down and there's a pdf file you
can download and read offline. Or print (watch out, Volume 1 - DC is
538 pages). The graphics and diagrams are clearer and better in the
pdf files, compared to the online html version.

The author, Tony Kuphaldt, doesn't assume that you know anything and
explains things very clearly, which is what I need. I've found them
immensely helpful.

Any time someone asks a question demonstrating that they don't know or
understand electronics, I would point them there.

"Graham Knott" <g.knott@ntlworld.com> writes:

Have a look at
http://homepage.ntlworld.com/g.knott/elect75.htm

 

[John Popelish]

John Popelish wrote:
(snip)

Here is a summary of some Duracell specs:
http://www.web-ee.com/Component%20Pages/battery_info.htm
Note that the MN1500 AA cell is rated for 2850 milliampere hours and
its maximum rated load is 43 ohms for a load current of 28
milliamperes. Now, .028*43=1.2 volts, so this implies that a typical
internal resistance must be dropping the rest of 1.5 volts. So the
internal resistance is about .3volts/.028 amperes or 10 ohms. So the
short circuit current should be about 1.5 volts / 10 ohms = .15
amperes. I am quite sure that you will get more than this with a
fresh cell, but this is typical for the mid life point.
Here is another site that summarizes battery characteristics:
http://www.zbattery.com/zbattery/batteryinfo.html

Note the resistance of the alkaline AA cells.

Correction. I mistook the ohms listed as the cell internal resistance
(which seemed high to me). Then I read the note above the list that
says the listed resistance is the load used for the test data (I
assume this means the ampere hour test data that took the batteries to
the cut off voltage).

--
John Popelish

 

[The Phantom]

On Fri, 5 Nov 2004 14:48:47 -0800, "Joel Kolstad"
<JKolstad71HatesSpam@Yahoo.Com> wrote:

Would anyone have a pointer to where I might found information about the
'allowable' S-plane locations for poles and zeroes in two port networks
parameters for passive structures, specifically Y, Z, and S parameters?

These are my thoughts:

-- Whatever applies to Y probably applies as Z as well.
-- Since Z11, Z22, Y11, and Y22 can all be seen as simple input
impedances/admittances, none of them should have right half-plane (RHP)
poles OR ZEROES.
-- Z12, Z21, Y12, and Y21 probably can't have RHP poles but probably can
have RHP zeroes.
-- S parameters are a complex enough transformation that it's unclear to me
whether or not there are any particular restrictions.

The reason I ask... I generate equivalent circuit models from measurement
data using curve fitting; the results are in the form of a rational
polynomial that can be readily factored to its pole/zero formulation. It'd
be nice to have an 'eyeball' method of ensuring the resultant ECM is stable.

Pointers to network theory books discussing this would also be appreciated.

Thanks,
---Joel Kolstad

This topic seems a little advanced for this group, but no matter.

You will find what you are looking for in most any textbook on
network synthesis, such as Guillemin or Van Valkenburg for example.

For passive networks, the facts are these:

The zeroes and poles of a driving point immittance must be in the LHP
or on the jw-axis (in the latter case, they must be simple).

The poles of the transfer immittance must be in the LHP or on the
jw-axis (and simple if on the axis). The zeroes may be anywhere in
the complex plane (for a ladder they may not be in the RHP) and for
grounded networks (which would be most typical networks), they may not
be on the positive real axis. Poles may not be at the origin or at
infinity. The degree of the numerator may equal but not exceed the
degree of the denominator.

This is straight out of Van Valkenburg.

 

[Terry]

"John Popelish" <jpopelish@rica.net> wrote in message
news:418975A2.2E11B2BE@rica.net...

ShadowTek wrote:

I just bought 50 3.2V 20mA white LEDs for about $30US total off of
EBay.

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=66952&item=3847009435&rd=1&ssPageName=WDVW
I tried to read up on electrical stuff but there is a lot I dont
understand.
For instance, if I use 2 1.5V AA betteries in a series that are 3 Amps
each,

The OP does not appear understand the relationship between volts and amps
(basic Ohm's Law).
So there's little chance will understand the niceties of powering LEDs,
variations to specification and current limiting.