Chip with simple program for Toy

[Andrew Holme]

Jack// ani wrote:

Hi there,

In 3-phase AC wiring, if phase to neutral voltage is 110V, then why is
phase to phase voltage 220? I know phase difference between any two
phases differ by 120 degree, so they should add up to give something
less than 220V! It should sum up to give 220V if the phase difference
were 0 degree or 360degree!

Thanks

They would *add* to give 110V, but I suspect you're talking about the
potential difference between phases. Using the cosine rule, the PD between
phases is:

sqrt(110*110 + 110*110 - 2*110*110*cos(120)) = 190.5

 

[Roger Johansson]

BobG wrote:

Can someone sum up the top couple of rules of algebra for him? How
about something like: 'An equation has an expression on each side of
the equal sign. To solve the equation for any of the variables, you
need to get that variable over to the left side of the equal sign. To
eliminate a variable on one side, multiply both sides of the equation
by the inverse of that variable. This doesnt change the equality,
because you are multiplying both sides by the same number.' Is this
the necessary and sufficient information needed to solve ohms law for
3 variables?

I can add some to your text above.

You can do anything to an equation as long as you do it to both sides
equally, the equation is still valid.
(an exception is dividing by zero, which gives meaningless results)

The methods you can use to isolate one variable on one side are
addition, subtraction, multiplication, division, inverting, squaring,
square root, substitution, etc..

Somebody who does not know these methods should take some time to learn
basic algebra, especially equation solving.

--
Roger J.

 

[Larry Brasfield]

"Jack// ani" <nospam4u_jack@yahoo.com> wrote in message
news:1112901176.778444.270260@f14g2000cwb.googlegroups.com...

I think you got something wrong, or I didn't expressed it correclty!

I'll go with that set of alternatives.

Say you have two AC sources of 110V, now if I put them in series they
should add up to give 220V if their instantaneous phases are same(0 or
360) or if they are 180 phase out they should sum up to zero. I think
these two AC sources are just like two phases of 3-phase AC supply
which are 120degree phase apart. And they should give a voltage less
than 220V when summed up.

Your above statements are consistent with phasor
arithmetic as I understand it, as long as "just like",
"less than", and "summed up" are interpreted in a
way most favorable to your understanding.

Any Help...Thanks

You'll need to describe your issue more specifically
to get any help with it, I believe.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Roger Johansson]

Peter Bennett wrote:

I've been working in electronics for some 40 years, and have no idea
what the "P.I.R.E. wheel" is - I just remember E = IR and P=EI, and
shuffle things around as needed. The same "shuffling" rules apply to
any simple equation.

Here is such a wheel, if you are curious.

http://www.the12volt.com/ohm/ohmslaw.asp

Here is a lesson in simple equation solving algebra

http://www.algebrahelp.com/lessons/equationbasics/index.htm

An online equation solver, for really lazy people

http://www.algebrahelp.com/calculators/equation/calc.jsp


It is incredible how much stuff you can find on the web today, you only
need to put together the right search words.

More advanced lessons in algebra. Ask dr Math!

http://mathforum.org/library/drmath/view/57620.html

http://www.ifigure.com/math/algebra/algebra.htm


--
Roger J.

 

[Peter Bennett]

On 7 Apr 2005 10:58:33 -0700, "Jack// ani" <nospam4u_jack@yahoo.com>
wrote:

Thaks all. A little doubt left..... is there any relation between
maximum charging current and ampere-hour rating of the battery?

Like maximum charging current above was 40% of the AH rating!

Thanks again

Depends on the battery. For most flooded lead-acid batteries, a
maximum charge rate of 1/5 of the AH rating is generally suggested.
Some gel and AGM batteries can be charged at higher rates.



--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[John Fields]

On 7 Apr 2005 14:43:43 -0700, george_holland222@hotmail.com (george)
wrote:

Hi,

I would like to know the Kva for the following appliances.

4 freezers at 5 amps each

3 chiller units at 3 amps each

2 chillers at 35 amps each.

I do not know if they are resistive or conductive(if that is necessary
for the answer)

Would the answer be 99 * 240 /1000 = 23.76??

Hope you can help.

---
It's not "conductive", it's "reactive", but if you're taking the
ratings from the nameplates and you're running the appliances on 240V
mains then, yes, it's 23.76 kVA,

--
John Fields
Professional Circuit Designer

 

[John Popelish]

js5895 wrote:

Thanks, I know basic high school algebra, but I just never understood
how to apply it to real world problems. I keep reading my electrical
book on that it says "Current is directly proportional to voltage" and
"Current is inversely proportional to resistance" and then I look at
the

Another definitions of "ohms" is volts per ampere. So, for any fixed
resistance, the ratio of volts divided by amperes (volts per ampere)
equals the value of the resistance. So resistance is the constant of
proportionality that relates volts to amperes. 100 ohms means that
the voltage is always 100 times the amperes.

P.I.R.E. wheel, trying to remember the whole wheel just by remembering
those statements and some algebra. I'm looking at it like a puzzle and
noticing some patterns like, that the power formulas you have to square
or square root to find an answer, so I can see that proportional and
inversely proportional part. I'm trying to figure out how they got
something like this "I = E/R" from that statement, looking at that
formula, thinking "I" is proportional to "E" and "I" is inversely
proportional "R", and I'm thinking why did they divide?. I'm racking my
mind and I know this is a simple basic DC formula compared to other
electrical formulas like, the AC ones.

The basic definition of resistance R=E/I (ohms equals volts per
ampere) can be rearranged to I=E/R or E=I*R.

The second basic formula on those wheels is P=E*I. But you can
substitute I*R for E (from the above rearrangement of R=E/I) to get
P=I*I*R or substitute E/R for I to get P=E*E/R

That is all there is on that wheel.

 

[Bob Monsen]

On Thu, 07 Apr 2005 19:40:44 -0700, dave.shattuck wrote:

I am hoping someone in this group can explain why, the below, is ok to
do. I'm a little cautious in implementing it because I already did
something similar (without pulling up the 4-lines high with resisters)
and it appears to have shorted the ports on my development board. I
don't want to make another costly mistake! If you can respond to
this...I'll be very indebted!

I'm attaching a schematic of what Zilog recommends in terms of how to
connect a keyboard to the GPIO ports. The problem I have is that I
don't understand it. Specifically:

Looking at the schematic - If you imagine that the GPIO pins PB4 - PB7
are output and set to high...and that the pins PB0 - PB3 are set to
input. And that by pressing a key you make a direct connection from
PB7 out through line 1 of the keypad then it would see that you would
have a problem. with the current going through PB0. Because it is
being sent through PB7 through the keypad and out of line 1 and into
PB0 directly (there is no resister on that end) you would have a
problem, wouldn't you?

If PB0 is an input, it won't act like it's an output. It acts like a 1MEG
resistor, unless you have pullups enabled, in which case it'll act like a
resistor of whatever value the pullups are. I don't know which processor
you are using (your schematic didn't make it) so I can't help there.

Initially I think I burnt out my development board
because I had half of
the GPIO pins tied to the input of the other half of the pins with no
current limiting resister. It appears that this schematic does the same
thing, doesn't it?

It may, or it may not. Again, no schematic.

Some further information about the pins when they are set to output and
input modes is in the other attachment, GPIO modes.jpg.

The problem I'm having is that - based on the table you see in that
attachment it appears that there should not be any problem connecting
the output of one pin to the input of another. Because it is 'high
impedance' it limits the current and, therefore, I don't need a limiting
resister.

This is true

If that's the case - how did I destroy my original
development board?

I dont' know. However, it's possible that the processor comes up with all
outputs. I know the PICs don't do this, for this very reason; they all
start out as inputs.

If you have a software bug where you've made them outputs by mistake,
they'll pass as much current as they can.

Another possibility is that the input was whacked by SCR latchup. If you
put too much current into an input, and it is higher or lower than the
power pins, it can get stuck into a state where the pin is sinking large
amounts of current. If you don't catch it and stop it, it'll destroy the
input port.

One simple thing you can do is read the datasheet, and find out what the
maximum current is into a pin, and put a resistor there that will keep it
from getting to that maximum current. So, if the max current is, say,
20mA, and your power rail is 5V, you put a 5/.02 = 250 ohm resistor in
front of the port. That way, no matter what you do, you won't be able to
hurt it.

I didn't use resisters but I did tie the output to the input of
another
pin.

Links to the two examples I'm talking about can be found:
ftp://www.wirelessjugah.net/Examples/

Thanks very much for your help!

 

[Ban]

Lessie wrote:

The schematic concerned is at the end of the microchip
document on microchip serial programming
http://ww1.microchip.com/downloads/en/AppNotes/91013b.pdf

The circuit generates VPP for the programming phase. D1 is
the reference of 12.7V and Q2 forms a negative feedback
with the opamp. This part I do understand. However, Q2 is
strange, what does it do? It seems that it will be in a
cut out state because base will be higher than emitter by
0.6V due to Q2 ?

Q1 is there to discharge the 12.7V line capacitance when Vpp_in goes low. In
programming operation it is isolated as you have rightly understood. But
when Vpp_out stays high, when the next instruction is send, it will be
misinterpreted as a programming bit, so the line has to be forced low within
one clock cycle and Q1 guarantees that.
--
ciao Ban
Bordighera, Italy

 

[Lord Garth]

<dave.shattuck@gmail.com> wrote in message
news:1112928044.093482.22720@o13g2000cwo.googlegroups.com...

I am hoping someone in this group can explain why, the below, is ok to
do. I'm a little cautious in implementing it because I already did
something similar (without pulling up the 4-lines high with resisters)
and it appears to have shorted the ports on my development board. I
don't want to make another costly mistake! If you can respond to
this...I'll be very indebted!

I'm attaching a schematic of what Zilog recommends in terms of how to
connect a keyboard to the GPIO ports. The problem I have is that I
don't understand it. Specifically:

Looking at the schematic - If you imagine that the GPIO pins PB4 - PB7
are output and set to high...and that the pins PB0 - PB3 are set to
input. And that by pressing a key you make a direct connection from
PB7 out through line 1 of the keypad then it would see that you would
have a problem. with the current going through PB0. Because it is
being sent through PB7 through the keypad and out of line 1 and into
PB0 directly (there is no resister on that end) you would have a
problem, wouldn't you?

Initially I think I burnt out my development board because I had half
of the GPIO pins tied to the input of the other half of the pins with
no current limiting resister. It appears that this schematic does the
same thing, doesn't it?

Some further information about the pins when they are set to output
and input modes is in the other attachment, GPIO modes.jpg.

The problem I'm having is that - based on the table you see in that
attachment it appears that there should not be any problem connecting
the output of one pin to the input of another. Because it is 'high
impedance' it limits the current and, therefore, I don't need a
limiting resister. If that's the case - how did I destroy my original
development board? I didn't use resisters but I did tie the output to
the input of another pin.

Links to the two examples I'm talking about can be found:
ftp://www.wirelessjugah.net/Examples/

Thanks very much for your help!

Forgetting all else, if PB4 - PB7 are output and set high, you can't get
anything useful like this. You must ripple a low (0) from PB4 through
PB7 with your keyboard routine. 0111, 1011, 1101, 1110 rapidly.
Between each increment, you read PB0 through PB3 for a low input.
Pull-up resistors wouldn't hurt anything and is recommended.

Knowing which output bit is low and which input bit read that low
allows you to know which key was pressed. Any low that is read should
stop scanning and re-read the input to eliminate bounce. Scanning should
resume when the inputs are again all high.

The tricky part comes when more than one key is simultaneously pressed.

 

[Lord Garth]

"Ace" <acefrehley@btopenworld.com> wrote in message
news:1112909778.572011.180360@o13g2000cwo.googlegroups.com...

Hi

I have a slightly unusual project I wonder if someone can help with. I
need to customise a guitar to include 20 4.8v 3A torch batteries. They
will be powered by 2 4.5V Alkaline batteries, the tricky bit is that i
need the lights to flash on and off in groups of 4 or 5. Basically I
think i am looking at having 4 or 5 loops of lights presumambly in
serial, connected to 4 or 5 channels. What I am lacking is a simple
controller to switch channels on a timer (once per second would be
about right)

I think the controller itself should be relatively simple but I don't
know how to design it. Can anyone email me a basic diagram and list of
components i would require? The 2 principle considerations are that it
would need to be basic as I am not an electrician, and secondly that it
would be fairly compact as it needs to fit in a guitar.

Any help would be greatly appreciated!

Trevor

Link us to these 20) 4.8V 3A batteries! That sounds real strange. I don't
think you're communicating well enough. 4 or 5 channels running 4 or 5
lights...
simultaneously or sequentially? What does each channel do? Does each
channel
respond to a different frequency range or do these groups of lamps simply
turn on
one channel after another at a 1Hz rate?


>

 

[Bob Monsen]

On Thu, 07 Apr 2005 17:00:48 -0700, Lessie wrote:

The schematic concerned is at the end of the microchip
document on microchip serial programming
http://ww1.microchip.com/downloads/en/AppNotes/91013b.pdf

The circuit generates VPP for the programming phase. D1 is
the reference of 12.7V and Q2 forms a negative feedback
with the opamp. This part I do understand. However, Q2 is
strange, what does it do? It seems that it will be in a
cut out state because base will be higher than emitter by
0.6V due to Q2 ?

It's a bizzare way to draw the circuit. However, if you notice, the
emitters are both connected to the same node, which is also the feedback.
Thus, it's just a follower. Why they feel they need low
impedance for Vpp is beyond me, though. Usually, Vpp draws almost no
current, so the opamp by itself should work fine. Whatever.

----
Regards,
Bob Monsen

 

[Lord Garth]

"Lessie" <lessie@btinternet.com> wrote in message
news:7c3b95d1.0504071600.53d89316@posting.google.com...

The schematic concerned is at the end of the microchip
document on microchip serial programming
http://ww1.microchip.com/downloads/en/AppNotes/91013b.pdf

The circuit generates VPP for the programming phase. D1 is
the reference of 12.7V and Q2 forms a negative feedback
with the opamp. This part I do understand. However, Q2 is
strange, what does it do? It seems that it will be in a
cut out state because base will be higher than emitter by
0.6V due to Q2 ?

Q1 & Q2 rapidly, cleanly and alternately switch Vpp between Vcc and Gnd.
Q2 is an NPN transistor. When its base is (very) positive WRT its emitter,
Vcc
on the collector less the (insignificant) drop across the transistor,
appears
at the R13-emitter junction and 1mA flows through R13.
Q1 now has 5 volts on its emitter and gnd at its collector. When U1A
switches to low, the PNP transistor Q1 turns on and Q2 turns off which pulls
Vpp to gnd.

 

[Lord Garth]

"js5895" <JoshTmp@nycap.rr.com> wrote in message
news:1112915344.658435.134080@g14g2000cwa.googlegroups.com...

Thanks, I know basic high school algebra, but I just never understood
how to apply it to real world problems. I keep reading my electrical
book on that it says "Current is directly proportional to voltage" and
"Current is inversely proportional to resistance" and then I look at
the
P.I.R.E. wheel, trying to remember the whole wheel just by remembering
those statements and some algebra. I'm looking at it like a puzzle and
noticing some patterns like, that the power formulas you have to square
or square root to find an answer, so I can see that proportional and
inversely proportional part. I'm trying to figure out how they got
something like this "I = E/R" from that statement, looking at that
formula, thinking "I" is proportional to "E" and "I" is inversely
proportional "R", and I'm thinking why did they divide?. I'm racking my
mind and I know this is a simple basic DC formula compared to other
electrical formulas like, the AC ones.

The term "inverse" means 1/whatever just as the term "per" means "for
every".
When someone says "percent" they mean "for every 100". One cent is 1 of
100.
As some smart man has said, "Words have meanings". Now if he could only
pronounce "nuclear" properly.

 

[Bob Monsen]

On Thu, 07 Apr 2005 14:38:06 -0700, SklettTheNewb wrote:

I have created a simple light sensing circuit using 2 voltage dividers
and a comparator. Basically, high resistance and the output from
comparator is high.

I need to explore other options due to the fact the comparator needs
more power than I have available. I would like to make this circuit
with 1.5 volts and I've found a couple LED flash circuits that will
allow me to handle that end w/ 1.5v
(http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm#15flash.gif)

but the comparator.... I dunno.

I thought I could use a voltage divider with a transistor, but when I
sat down to actually make this circuit, I realized I couldn't figure it
out.

I've researched op-amps as a comparator but that doesn't seem like a
good fit either.

I would be interested in any ideas that you guys might have.

Sorry to post so many questions here, I've been really active on this
project lately and I'm learning a lot here so it's hard to not come
back

-Steve

You can use a simple two-transistor oscillator to pump up the voltage
sufficiently to light up a white LED from a 1.5V source. Using another NPN
transistor to kill off oscillations. Here is a try:

1.5V
.----------------o--------------------.
| | o-----.
| | | |
| ' | |
| .-. 100u C| |
.-. | |1k C| -
| | | | C| ^ ->
| | Variable '-' 10k 330p | |
'-' Resistor | | |
| | ___ || | |
| .---------o---|___|--o---||----o-----'
| | | | || |
| | o----------)-. |
| ' | | | |
| |/ \| | | |/
'----| |--------' '-----|
|> <| |>
| | |
| | |
| | |
'---------o--------------------'
GND

NPN = 2n4401
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Your variable resistor may not be right for this. As the resistance
increases above about 470k, The thing will start oscillating, causing the
LED to fire up. If you want it to go when there is light, you'll need to
use a slightly different scheme. As it gets lighter, your LDR probably
has less resistance. Thus, replace 'variable resistor' with a resistor
that approximately equals the resistance of the LDR at the point you want
it to turn on at. Then, put the LDR from the base of the leftmost
transistor to ground. When it gets lighter, it'll turn of the leftmost
transistor, allowing the oscillator to go.

---
Regards,
Bob Monsen

 

[Lord Garth]

"jason" <cheanglong@gmail.com> wrote in message
news:1112931342.655967.172020@l41g2000cwc.googlegroups.com...

Hello All
Say I have a resistor R, inductor L and capacitor C connected together
in series

The easiest way to find the total impedance is just plug the formula
a+ bj

Z= square root (a^2 + b^2 )

without drawing any phasor diagram right?

<snip>

yep, the resistance is the horizontal axis,
the reactance is the vertical axis
and
the impedance is the hypotenuse.

Sounds just like the Pythagorean theorem, doesn't it....

 

[mike rothe]

It really depends on what type of alarm clock you use(AC/DC) and what you want to drive with the relay(AC/DC/Load/etc...)

If you just want to set up a relay to go off at a certain time of day, I would use a cheap wrist watch, connect the piezo beeper wires to an opto coupler which will drive a tranistor and in turn a relay, plus it's nice an small so it will fit between the sticks of dynamite!!!

(Thats just a joke BTW!, If you would like to explain what you are going to use this for I would be happy to email you a schematic+parts list, but not if it's for a bomb)

Mike ROthe


--------------= Posted using GrabIt =----------------
------= Binary Usenet downloading made easy =---------
-= Get GrabIt for free from http://www.shemes.com/ =-

 

[Lord Garth]

"jason" <cheanglong@gmail.com> wrote in message
news:1112944014.015990.204210@l41g2000cwc.googlegroups.com...

Yes that's right Garth ,.. Thank you

Sometimes in some cases, we compare the reactance of a capacitor with
resistance of a resistance like the below case

1/wC << Rs

So my question is , what does it mean by this comparison?
reactance < resistance

What does it tell us?

Should not we compare resistance with resistance
while comparing reactance with reactance
Both of them have the unit ohm?

What is the implication of using such a comparison , what does it tell
us?

Thank you

I was trying to recall the original question.....

You had a series RCL I think that all you are seeing is that
the inductive reactance and the capacitive reactance are nearly
equal but opposite therefore they cancel. The circuit appears to
be very much like a resistance only.

If 1/wC << Rs the resultant phasor is very close to Rs so it seems
to not be very significant in a first approximation.

I guess I'm not seeing why wL couldn't also be << Rs resulting in
a phasor in quadrant 1 rather than Q4

In either case, the circuit would be dominated by Rs, the power factor
would be close to 1 therefore the current and the voltage would be
(pretty much) in phase. The crest factor would be close to minimum.
Rs would have to handle near maximum power. It sounds like reasonable
matching for maximum power transfer.

 

[Lord Garth]

"Ace" <acefrehley@btopenworld.com> wrote in message
news:1112973304.069143.44730@z14g2000cwz.googlegroups.com...

Sorry my mistake. the bulbs are 4.8v 3A. There are 20 lights I want to
install on the guitar, my idea is to have a 4 or 5 channel controller,
each channel operating a group of 4 or 5 lights so that a different
group of lights flashes on every 1 second. The lights would only need
to operate for 5-10 minutes so battery life shouldn't be a major
concern.

I hope this calarifies

Much better, thanks...

My first suggestion would be to substitute high brightness white LEDs for
the (I think) incandescent bulbs. First, LEDs are more reliable and use
far
less current which makes them easier to use with electronic switching.

What sort of power source do you have available? It would help to have
the type of battery, its chemistry and its approximate mAH rating. If you
use an array of batteries, the final voltage needs to be known as well.

The basic circuit would be a 555 oscillator outputting a 1Hz pulse followed
by a divide by 4 or 5 counter and then a decoder. This basic circuit has
several places where it can be made better / different. For instance, the
counter / decoder could be replaced with a cheap micro which has been
programmed
to walk an output exclusively among 5 outputs or the counter / decoder
could
be replaced with a simple shift register where the final output is looped
back to
its input. You just have to power up setting one stage while resetting the
others.
All are easy to build but I suspect wiring the guitar will be more
difficult.

What all of these three possibilities have in common is the need for a final
driver to actually fire the lamps. If you decide to lose the incandescent
lamps
and go with the LEDs, so much the better.

The lamps should be connected such that the wiring is minimized. 5 lamps in
5
groups means 10 wires so I suppose it's good that you want each groups lamps
all on at the same time.

 

[Lord Garth]

"hartly" <hartlyuk@yahoo.com> wrote in message
news:1f73b571.0504080742.19670c8d@posting.google.com...

Why does it say '4 Amp-hrs.(20 hour rate)' on a battery?.
I know it means it can give out 1 amp. for 4 hours.
But does 20 hr.rate mean that in reality it can only give out 0.2
amps.for 20 hours(making 0.2x20=4Ah.)?.
Thanks, Hartly

You are correct ... 4AH/20H=.2A

A 12 volt lead / acid battery is dead when the potential falls to 10.5
volts.