Chip with simple program for Toy

[BobW]

"BobW" <nimby_NEEDSPAM@roadrunner.com> wrote in message
news:B7mdnSJSKb17P77VnZ2dnUVZ_obinZ2d@giganews.com...

"vic" <news@bidouille.org> wrote in message
news:48233453$0$20289$426a74cc@news.free.fr...
Hi,

I have two common anode 7-segments displays, and only one wire to drive
them. I need to achieve the following : when the control signal is +5V,
display1 is ON and display2 is OFF. When the signal is 0V, display1 is
OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.

I tried using a NPN transistor for display1 and a PNP for display2,
connecting their bases together. It works when the driving signal is
present, but when the signal is floating current flows from the base of
the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.

The circuit that didn't work :

VCC
+
|
|
___ |
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|
|
|
Display1
|
|
GND



Is there a way to achieve this ?

Thanks.

Try this:
5V
| 5V
R5 |
| |<---
|/-------------| Q3
----| Q1 |\--------- (to display 1 and then to GND)
| |>-- 5V
| | |
| | R3
| | |
---R1-----R2----------
| | |
| | R4
| |<--- |
|--| Q2 GND
|\
| |/--------- (to display 2 and then to 5V)
-------------| Q4
| |>----
R6 |
| GND
GND

Q1 and Q2 form a comparator so that when the input is floating they will
both be off. R3 and R4 set the input threshold.

When the input is high (5V) then Q1 will be on and that will turn on Q3.
When the input is low (GND) then Q2 will be on and that will turn on Q4.

You can figure out the resistor values. They shouldn't be too critical,
but R3 and R4 need to be small enough to ensure enough drive for the four
transistors.

I hope I got the ascii art right as I had to compose it in a separate word
processor.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

I see, now, that your displays are both common anode. You'll need to add
another PNP (common emitter mode) driven by Q4.

As Monica Lewinsky used to say, "Close, but no cigar."

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

 

[Dan Coby]

"petrus bitbyter" <pieterkraltlaatditweg@enditookhccnet.nl> wrote in message
news:48239730$0$3654$e4fe514c@dreader31.news.xs4all.nl...

"Dan Coby" <adcoby@earthlink.net> schreef in bericht
news:7vCdnV2xQMRN4b7VnZ2dnUVZ_oqhnZ2d@earthlink.com...
"ehsjr" <e.h.s.j.r.removethespampunctuation@bellatlantic.net> wrote in message
news:cxHUj.28009$zw.4986@trnddc04...
vic wrote:
Hi,

I have two common anode 7-segments displays, and only one wire to drive them. I need to achieve
the following : when the control signal is +5V, display1 is ON and display2 is OFF. When the
signal is 0V, display1 is OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.

I tried using a NPN transistor for display1 and a PNP for display2, connecting their bases
together. It works when the driving signal is present, but when the signal is floating current
flows from the base of the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.

The circuit that didn't work :

VCC
+
|
|
___ |
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|
|
|
Display1
|
|
GND



Is there a way to achieve this ?

Thanks.

Add diodes to isolate the bases, and resistors
to bias the transistors off when the desired
on signal is not present.

Ed

see below



VCC
+
|
+-------------+
| |
[R] |
| ___ |
o--------|<------+----|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o--->|-+-|___|--| GND
| |
[R] |
| |
| Display1
| |
+----------+
|
GND

The revised circuit does not solve the problem. Both diodes
will be conducting when the input is floating. As a result both
displays will still be on.

A simpler solution is to remove the diodes, change the
locations of the added resistors, and move Display1. I.e.:

VCC
+
|
+---------+
| |
[R1] |
| |
o-------------[R2]---+-------|
| |\
| VCC |
| + |
| | Display2
| Display1 |
| | |
| |/ |
Input---o-----[R2]-+--------| GND
| |
[R1] |
| |
+----------+
|
GND

The R1 / R2 resistor pairs need to be chosen so that
there is only about 0.5 volts across the transistors'
base-emitter junctions when the input is floating.
When the input is floating, the various resistors
will pull the input to VCC/2. The value of 0.5 volts was
chosen to be low enough to keep the transistors from
turning on when the input is floating but still allow the
transistors to be turned on when the input is being driven
to VCC or ground.


This may work, (so there are some ifs.)
Most important, the driving source must be able to provide the extra current (sink and source).

Actually there is not a very large increase in the driving source requirements..
The maximum voltages across the R1 resistors is 0.7 volts. If VCC is 5 volts,
then there is only about a 14% increase in the required drive currents. Some
diodes can be put in series with the transistor bases to decrease the extra
drive requirements and to add some more margin to the circuit.

The displays may need GND on one side. The OP did not said so, but his schematic suggests it.

Yes, this may be a concern.

 

[Paul E. Schoen]

"Dave" <dspear99ca@yahoo.com> wrote in message
news:u9IUj.1150$Yp.150@edtnps92...

"emilio_estevez" <poguen@gmail.com> wrote in message
news:34803565-2327-4a43-9b90-d7d7369139a1@e53g2000hsa.googlegroups.com...
My question came up when a friend of mine asked if I could change out
his Federal Pacific breaker box to a newer, safer one. My background
is in industrial controls and I don't normally deal with power at the
service level. I only have about two years experience so as I am now
trying to learn more on my own, I'm finding a lot of things that don't
add up.

Dude, don't go there. Have you ever heard of anyone changing out their
own breaker panel, much less doing it live? No? Huh, wonder why that
is?

The power that'll slam into you through 0 or 2 ga wire is HUGE and
UNFUSED... thousands of amps. The transformer on the pole has overload
protection, but it won't kick in until WAY after you're dead.

What you are considering is RIDICULOUSLY DANGEROUS. Trained
professionals don't do it.

You are right about that. The transformer is only fused at the primary to
protect it from overloads and malfunction, and many of these "pole pigs"
are 50 kVA or more. This means that a 120 VAC line to neutral or earth
current of 500 amperes would just start to make the fuse notice, and the
instantaneous current could be as high as 5000 amps (half a MegaWatt) for a
few cycles. Certainly enough to create a huge fireball and hurl globs of
molten copper and steel at the unfortunate amateur electrician who just
happened to let a scredriver slip across the mains.

A friend was working with another experienced test technician, doing
routine breaker testing in a large facility. Most of the switchgear was
disconnected, but the other test technician had to do some work in the main
fuse box, which I think was a 480 VAC feeder with heavy bus bar, probably
rated at 2000 amps or more. I think he had to tighten a bolt, and normally
he would have used a wrench that was mostly insulated with rubber tape, but
he was probably tired and in a hurry, and somehow the wrench slipped and
landed across the live bus. My friend saw it happen, and turned to the side
just as the fireball erupted, and it burned much of his face and body. The
technician who was responsible was badly burned, and soon died from his
injuries.

Here is a website with some images and movies of actual electrical arc
blasts so you can see the "potential" for danger:
http://205.243.100.155/frames/longarc.htm. Note that one of them is "only"
480 VAC.

Actually, I changed out the old breaker box in the house where I now live,
but I was completely remodeling it, and I had the utility company come out
and pull the meter. I had a separate feeder from my other house next door,
so I could provide temporary power. I actually located the new box on the
other side of the wall, next to the meter, inside an enclosed porch, and I
routed the 100 amp service cable into the new box while it was totally
dead. Even so, I treated it with respect, and I taped the exposed
conductors while I relocated them, and removed the tape only when I could
safely connect them to the main 100 amp breaker. It was several years later
that I was able to finish the bulk of the work, and I was able to have BGE
come out and replace the meter. I had tested the installation beforehand by
patching the other supply onto the mains, and I checked carefully for any
loose strands of the incoming service cable. Even so, I was a bit nervous
when the meter was reinstalled, but at least I had a main breaker directly
on the incoming line. The old breaker box did not have a main breaker.

More information and videos: http://www.lanl.gov/safety/electrical/

Paul

 

[Paul E. Schoen]

"Dave" <dspear99ca@yahoo.com> wrote in message
news:u9IUj.1150$Yp.150@edtnps92...

"emilio_estevez" <poguen@gmail.com> wrote in message
news:34803565-2327-4a43-9b90-d7d7369139a1@e53g2000hsa.googlegroups.com...
My question came up when a friend of mine asked if I could change out
his Federal Pacific breaker box to a newer, safer one. My background
is in industrial controls and I don't normally deal with power at the
service level. I only have about two years experience so as I am now
trying to learn more on my own, I'm finding a lot of things that don't
add up.

Dude, don't go there. Have you ever heard of anyone changing out their
own breaker panel, much less doing it live? No? Huh, wonder why that
is?

The power that'll slam into you through 0 or 2 ga wire is HUGE and
UNFUSED... thousands of amps. The transformer on the pole has overload
protection, but it won't kick in until WAY after you're dead.

What you are considering is RIDICULOUSLY DANGEROUS. Trained
professionals don't do it.

You are right about that. The transformer is only fused at the primary to
protect it from overloads and malfunction, and many of these "pole pigs"
are 50 kVA or more. This means that a 120 VAC line to neutral or earth
current of 500 amperes would just start to make the fuse notice, and the
instantaneous current could be as high as 5000 amps (half a MegaWatt) for a
few cycles. Certainly enough to create a huge fireball and hurl globs of
molten copper and steel at the unfortunate amateur electrician who just
happened to let a scredriver slip across the mains.

A friend was working with another experienced test technician, doing
routine breaker testing in a large facility. Most of the switchgear was
disconnected, but the other test technician had to do some work in the main
fuse box, which I think was a 480 VAC feeder with heavy bus bar, probably
rated at 2000 amps or more. I think he had to tighten a bolt, and normally
he would have used a wrench that was mostly insulated with rubber tape, but
he was probably tired and in a hurry, and somehow the wrench slipped and
landed across the live bus. My friend saw it happen, and turned to the side
just as the fireball erupted, and it burned much of his face and body. The
technician who was responsible was badly burned, and soon died from his
injuries.

Here is a website with some images and movies of actual electrical arc
blasts so you can see the "potential" for danger:
http://205.243.100.155/frames/longarc.htm. Note that one of them is "only"
480 VAC.

Actually, I changed out the old breaker box in the house where I now live,
but I was completely remodeling it, and I had the utility company come out
and pull the meter. I had a separate feeder from my other house next door,
so I could provide temporary power. I actually located the new box on the
other side of the wall, next to the meter, inside an enclosed porch, and I
routed the 100 amp service cable into the new box while it was totally
dead. Even so, I treated it with respect, and I taped the exposed
conductors while I relocated them, and removed the tape only when I could
safely connect them to the main 100 amp breaker. It was several years later
that I was able to finish the bulk of the work, and I was able to have BGE
come out and replace the meter. I had tested the installation beforehand by
patching the other supply onto the mains, and I checked carefully for any
loose strands of the incoming service cable. Even so, I was a bit nervous
when the meter was reinstalled, but at least I had a main breaker directly
on the incoming line. The old breaker box did not have a main breaker.

More information and videos: http://www.lanl.gov/safety/electrical/

Paul

 

[Tantalust]

"Mike Tomlinson" <mike@jasper.org.uk> wrote in message
news:jYz7iLNG69IIFwOj@jasper.org.uk...

In article <72ae0$48232b9a$4213eac2$18448@DIALUPUSA.NET>, bud--
remove.budnews@isp.com> writes

Not clamping phone wires to earth is a major surge suppression flaw.

It's simply not necessary in towns and cities in the UK. Occurrences of
damage caused by surges on phone lines are practically unheard of.
There are reports of damage caused by direct or nearly lightning
strikes, but of course nothing is going to protect against that.

Houses in villages and remote locations would probably benefit most from
additional protection. You can be sure that critical installations
(hospitals, data centres, etc.) will install additional protection.

British Telecom fit NTE (network termination equipment), also known as a
master socket, which does have surge arrestors built in, but they don't
clamp to earth, they're just across the line:

http://www.buzzhost.co.uk/nte5.php

has a circuit diagram of the NTE, and an interesting photo of damage
caused by a direct lightning strike further down the page (which, of
course, none of w_'s equipment would have prevented.)

An additional factor is that adding further surge protection devices can
affect the line characteristics, causing ADSL sync speeds to drop.

A service panel
suppressor doesn't help the voltage difference at all.

Obviously.

Surprising since the UK seems to be very good on electrical protection
in general.

As I said in an earlier post, a calm, intelligent assessment (not w_'s
level of hand-waving, gibbering hysteria) of each situation is needed
before deciding on the level of protection required.

It's clear that it's simply not needed for most UK domestic phone lines;
this will have been borne out by years and years of experience, looking
at the number of insurance claims, etc. I should think BT's attitude is
that if the customer wishes to install additional protection after the
demarc (NTE), that's up to them.

In the end, It's all about assessing risk and mitigating it.

I found this webpage rather amusing:

http://www.satcure.co.uk/tech/phonesurge.htm

but will leave it to others to comment

--
(\__/) Bunny says NO to Windows Vista!
(='.'=) http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.html
(")_(") http://www.cypherpunks.to/~peter/vista.pdf

Wow, it says phone lines there can have as much as 180 [ringing] volts on
them, interesting.

 

[Tom Biasi]

"Green Xenon [Radium]" <glucegen1@excite.com> wrote in message
news:4824d372$0$30183$4c368faf@roadrunner.com...

Hi:

Some video websites say that spatial aliasing causes jaggies, other say
it causes the Moire effect, still others say it causes pixelation. Who
is right? What does spatial aliasing look like?

Also, which video specs are measured in "Hz" -- regardless of whether the
video is analog or digital? I believe "frame rate" is one of them.

In digital video, what specs are measured in Hz?

What specs are measured in Hz in analog video?


Thanks,

Radium

Radium,
Why do you persist in these ventures?
It would be difficult for one to research the subject as you did and not
answer their own question.

I removed Xposts
Tom

 

[Paul E. Schoen]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:4824DBAD.AC2408EF@earthlink.net...

Sjouke Burry wrote:

Rich Grise wrote:
On Fri, 09 May 2008 09:15:55 -0700, John Larkin wrote:
On Fri, 09 May 2008 09:05:15 -0500, OverUnity

Umm, what names did you have in mind?
Well, in your case, I think John would fit nicely.


Those arguing over how many angels can dance on the head of a pin
know
who they are. They need to come to an understanding. I'm curious
about
I^2R losses and the impact distributive generation (private solar
etc.)
would have on those line losses.
Seems to me that a DC "local bus" should be standard within
solar/wind/whatever systems; I'd suggest 400 volts DC. Then a
standard, modular inverter topology could be used to connect to the
grid.

Just wondering, why 400? I've seen industrial-grade 90VDC motors, so
I'm guessing 90V is some kind of a standard; of course, 400VDC would
have less than 1/16 of the I^2*R losses, but why 400 and not, say,
480?
Or some other number? Should we start another religious war, this one
about "standard" local DC bus voltage selections? >:-


90 VDC motors are common, because that is the effective voltage you
get when you rectify the 120 VAC line voltage, and don't filter it.

Full wave rectified 120 VAC is still 120 V RMS (effective voltage), and 108
volts average, and 170 volts peak. But perhaps such motors are usually
driven with PWM or phase-controlled SCRs or TRIACs, and 90 volts is easily
produced. Motors can be driven at higher voltages for less than continuous
duty cycle, so this rating allows for that as well, and it also allows for
low line voltages.

DC bus voltages for use with V/F inverter drives are based on the peak
voltage of the sine wave, so it is about 180, 360, and 720 VDC for common
voltages of 120, 240, and 480 VAC. A DC local bus would be easy to use for
either adding or taking power. There is no phase angle or waveform to
contend with, and no power factor, so your watts in or out is just the
voltage times the current. And there is no ELF field to cause concerns, at
least on the DC bus.

Did I not hear somewhere that DC power is dangerous?
I.E. it kills you better than AC?


http://www.google.com/search?q=edison+elephants&rls=com.microsoft:en-us:IE-SearchBox&ie=UTF-8&oe=UTF-8&sourceid=ie7&rlz=1I7GWYA

This was his way to demonstrate the danger of AC, although the equivalent
amount of DC may have proven just as fatal. AC has the distinction of
causing fibrillation, while DC might cause your muscles to contract and
make you less able to release a "hot" conductor. Here it says (for dummies)
the main danger of DC is burns:

http://www.dummies.com/WileyCDA/DummiesArticle/The-Dangers-of-Electrical-Shock.id-2922.html

Try this powerpoint presentation (I'm still waiting on dial-up):
http://hightech.lbl.gov/presentations/dc-powering/DCP-overview-presentation.ppt

Here's a discussion a couple years ago:
http://www.control.com/thread/1026220674

This is pretty good:
http://www.ewh.ieee.org/cmte/ias-esw/pdfs/Hazards_of_Electricity.pdf

The bottom line for electricity: "Touch Me Not!"

Paul

 

[BobW]

"vic" <news@bidouille.org> wrote in message
news:48250145$0$5055$426a74cc@news.free.fr...

BobW wrote:
"BobW" <nimby_NEEDSPAM@roadrunner.com> wrote in message
news:B7mdnSJSKb17P77VnZ2dnUVZ_obinZ2d@giganews.com...
"vic" <news@bidouille.org> wrote in message
news:48233453$0$20289$426a74cc@news.free.fr...
Hi,

I have two common anode 7-segments displays, and only one wire to drive
them. I need to achieve the following : when the control signal is +5V,
display1 is ON and display2 is OFF. When the signal is 0V, display1 is
OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.

I tried using a NPN transistor for display1 and a PNP for display2,
connecting their bases together. It works when the driving signal is
present, but when the signal is floating current flows from the base of
the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.

The circuit that didn't work :

VCC
+
|
|
___ |
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|
|
|
Display1
|
|
GND



Is there a way to achieve this ?

Thanks.
Try this:
5V
| 5V
R5 |
| |<---
|/-------------| Q3
----| Q1 |\--------- (to display 1 and then to GND)
| |>-- 5V
| | |
| | R3
| | |
---R1-----R2----------
| | |
| | R4
| |<--- |
|--| Q2 GND
|\
| |/--------- (to display 2 and then to 5V)
-------------| Q4
| |>----
R6 |
| GND
GND

Q1 and Q2 form a comparator so that when the input is floating they will
both be off. R3 and R4 set the input threshold.

When the input is high (5V) then Q1 will be on and that will turn on Q3.
When the input is low (GND) then Q2 will be on and that will turn on Q4.

You can figure out the resistor values. They shouldn't be too critical,
but R3 and R4 need to be small enough to ensure enough drive for the
four transistors.

I hope I got the ascii art right as I had to compose it in a separate
word processor.

Bob
--
== NOTE: I automatically delete all Google Group posts due to
uncontrolled SPAM ==

I see, now, that your displays are both common anode. You'll need to add
another PNP (common emitter mode) driven by Q4.

As Monica Lewinsky used to say, "Close, but no cigar."

Bob

*gasp* 5 transistors needed to do what seemed simple at first glance ...

I don't quite understand what R5 and R6 are for, when Q1 and Q2 do not
conduct, the base current of Q3 and Q4 would be zero so the ressitors do
not seem necessary ?

Well I guess I could just try it and see if it works

There is always some leakage from collector to base, so R5 is there to
insure that Q3 doesn't conduct when Q1 is off. Same goes for R6/Q4.

Yeah, it's a lot of parts, but your requirements are kinda tough.

If you're only building one of these, and your supply is tightly regulated,
and your temperature range is limited, then the method (I forget who
proposed it) that keeps Vbe at about 0.4V when the drive is high impedance
may work okay.

If it were my project and the thing had to ALWAYS work (in production
quantities, over a varying range of Vcc, temperature, and Voh/Vol), then I'd
use this scheme.

If you have trouble with the resistor values then give a yell. R3 and R4
should be equal if your drive is from a CMOS output (i.e. the output swings
from supply to supply). If it's from an old TTL output, then the voltage at
the R3/R4 connection should be set to about ((2.4V-0.4V)/2)+0.4V = 1.4V
(i.e. the middle of the output swing range for worst-case TTL).

Have fun with it.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

 

[Alan]

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In article <bc45f$48248dbc$4213ea37$25354@DIALUPUSA.NET>,
remove.budnews@isp.com says...
krw wrote:
In article <OKadnRdUKpg42L7VnZ2dnUVZ_tPinZ2d@comcast.com>,
me@nomail.com says...
Mike Tomlinson wrote:

Phone wires were clamped to ground before the 1960s?
It was common to earth one leg of the incoming pair to either the house
ground or to its own rod. An earth connection also allowed "party
lines", where two houses could share one physical phone line pair, each
house with its own number. Disadvantage was that both lines could not
be used simultaneously.

http://en.wikipedia.org/wiki/Party_line_(telephony)

I think they used to ring between the red green for one party, yellow
green for the other party, black green, etc.

No, that would defeat the purpose of the party line. The ringers
either had "distinctive ring" (once for Mabel, twice for Maude) or
were frequency tuned.


They did frequency and distinctive rings. But for 2 parties you can ring
red-to-ground for one and green-to-ground for the other. It is in Mike's
Wikipedia link above. My recollection is black was ground and yellow was
sometimes used for a light in the phone (red and green are phone wires).

Princess phones used the yellow green pair for the dial light. A
transformer was hidden somewhere in teh house to supply the power
(IIRC, a standard 24VAC door bell transformer, but it's been a lot
of years).

6 volts as I recall. I had one of the transformers around ages ago,
it may still be stashed somewhere.

Not sure about the pair, though, since green/red is tip/ring of pair
one, black/yellow is tip/ring of pair two. Putting the transformer between
green and yellow would be putting the light current on the talk pair,
which would be inviting hum on the line.

More modern wiring uses:

colors:
main/stripe
-----------
white/blue green tip 1
blue/white red ring 1

white/orange black tip 2
orange/white yellow ring 2

white/green tip 3
green/white ring 3

white/brown tip 4
brown/white ring 4

( from http://www.ling.upenn.edu/~kurisuto/phone_wiring.html )


Similarly, I would question the reliability of ring on a single line referencing
ground, since party lines tended to be out longer distances -- the ground resistivity
would make it more difficult to get ring current to the phone(s).

I think the differing ring frequency would make more sense, since mechanical
resonance in the ringer provides a reasonable tuning mechanism.


Alan

 

[Andrew Holme]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:3eb67024-6caf-48c6-90b6-0a34a80e1319@y22g2000prd.googlegroups.com...

I made a low frequency crystal oscillator using a couple CMOS
inverters (CD4069) and a very old 31.5KHz crystal. It's a fairly large
crystal package that measures about 3/4 by 3/8 inch from the 1970s.

The first inverter feeds the second, and the output of the second
connects the crystal back to the input of the first inverter. The
first inverter has a 100K resistor connected from input to output and
there is a 470pF cap from the input to ground. Works well, but I'm
wondering what the best RC values should be? I was thinking the
reactance of the cap should be about equal to the resistor, but that
works out to about 50pF which doesn't work. Lots of different
combinations will work, such as 22K and 1000pF.

How does one determine the best values?

-Bill

The resistor controls loop gain. I would find the lowest value for which
oscillation starts reliably and multiply it by 10.

The loading capacitance will affect frequency. Since you are using a
crystal, presumably you care about frequency (if not, you could use a simple
RC oscillator) so choose the capacitor which gets you closest to 31.500 KHz.

I was recently playing with miniature 32.768 KHz tuning-fork watch crystals,
which require extremely low drive levels; but I presume your large 1970s
crystal is not of this type.

 

[Andrew Holme]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:3eb67024-6caf-48c6-90b6-0a34a80e1319@y22g2000prd.googlegroups.com...

I made a low frequency crystal oscillator using a couple CMOS
inverters (CD4069) and a very old 31.5KHz crystal. It's a fairly large
crystal package that measures about 3/4 by 3/8 inch from the 1970s.

The first inverter feeds the second, and the output of the second
connects the crystal back to the input of the first inverter. The
first inverter has a 100K resistor connected from input to output and
there is a 470pF cap from the input to ground. Works well, but I'm
wondering what the best RC values should be? I was thinking the
reactance of the cap should be about equal to the resistor, but that
works out to about 50pF which doesn't work. Lots of different
combinations will work, such as 22K and 1000pF.

How does one determine the best values?

-Bill

The resistor controls loop gain. I would find the lowest value for which
oscillation starts reliably and multiply it by 10.

The loading capacitance will affect frequency. Since you are using a
crystal, presumably you care about frequency (if not, you could use a simple
RC oscillator) so choose the capacitor which gets you closest to 31.500 KHz.

I was recently playing with miniature 32.768 KHz tuning-fork watch crystals,
which require extremely low drive levels; but I presume your large 1970s
crystal is not of this type.

 

[petrus bitbyter]

"vic" <news@bidouille.org> schreef in bericht
news:48250145$0$5055$426a74cc@news.free.fr...

BobW wrote:
"BobW" <nimby_NEEDSPAM@roadrunner.com> wrote in message
news:B7mdnSJSKb17P77VnZ2dnUVZ_obinZ2d@giganews.com...
"vic" <news@bidouille.org> wrote in message
news:48233453$0$20289$426a74cc@news.free.fr...
Hi,

I have two common anode 7-segments displays, and only one wire to drive
them. I need to achieve the following : when the control signal is +5V,
display1 is ON and display2 is OFF. When the signal is 0V, display1 is
OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.

I tried using a NPN transistor for display1 and a PNP for display2,
connecting their bases together. It works when the driving signal is
present, but when the signal is floating current flows from the base of
the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.

The circuit that didn't work :

VCC
+
|
|
___ |
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|
|
|
Display1
|
|
GND



Is there a way to achieve this ?

Thanks.
Try this:
5V
| 5V
R5 |
| |<---
|/-------------| Q3
----| Q1 |\--------- (to display 1 and then to GND)
| |>-- 5V
| | |
| | R3
| | |
---R1-----R2----------
| | |
| | R4
| |<--- |
|--| Q2 GND
|\
| |/--------- (to display 2 and then to 5V)
-------------| Q4
| |>----
R6 |
| GND
GND

Q1 and Q2 form a comparator so that when the input is floating they will
both be off. R3 and R4 set the input threshold.

When the input is high (5V) then Q1 will be on and that will turn on Q3.
When the input is low (GND) then Q2 will be on and that will turn on Q4.

You can figure out the resistor values. They shouldn't be too critical,
but R3 and R4 need to be small enough to ensure enough drive for the
four transistors.

I hope I got the ascii art right as I had to compose it in a separate
word processor.

Bob
--
== NOTE: I automatically delete all Google Group posts due to
uncontrolled SPAM ==

I see, now, that your displays are both common anode. You'll need to add
another PNP (common emitter mode) driven by Q4.

As Monica Lewinsky used to say, "Close, but no cigar."

Bob

*gasp* 5 transistors needed to do what seemed simple at first glance ...

I don't quite understand what R5 and R6 are for, when Q1 and Q2 do not
conduct, the base current of Q3 and Q4 would be zero so the ressitors do
not seem necessary ?

Well I guess I could just try it and see if it works

Didn't you see my four transistors, four resistors solution yesterday?

+--------+------------------+-------+--Vcc
| | | |
___ |< | ___ |/ |
in----+--|___|---| | +--|___|--| |
| |\ | | |> |
| | | | | |
| | |< | | |<
| +------| | +-----|
| | |\ | | |\
| .-. | | .-. |
| | | .---. | | | .---.
| | | | | | | | | |
| '-' | D | | '-' | D |
| | | | | | | |
| | '---' | | '---'
| | | | | |
------)------------+--------+------)-----------+-------+--GND
| |
+----------------------------+
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


petrus bitbyter

 

[petrus bitbyter]

"vic" <news@bidouille.org> schreef in bericht
news:48250145$0$5055$426a74cc@news.free.fr...

BobW wrote:
"BobW" <nimby_NEEDSPAM@roadrunner.com> wrote in message
news:B7mdnSJSKb17P77VnZ2dnUVZ_obinZ2d@giganews.com...
"vic" <news@bidouille.org> wrote in message
news:48233453$0$20289$426a74cc@news.free.fr...
Hi,

I have two common anode 7-segments displays, and only one wire to drive
them. I need to achieve the following : when the control signal is +5V,
display1 is ON and display2 is OFF. When the signal is 0V, display1 is
OFF and display2 is ON. When the signal is not connected (high
impedance), both displays are OFF.

I tried using a NPN transistor for display1 and a PNP for display2,
connecting their bases together. It works when the driving signal is
present, but when the signal is floating current flows from the base of
the PNP to the base of the NPN and both transistors turn each other on,
resulting in both displays being ON.

The circuit that didn't work :

VCC
+
|
|
___ |
o---------------------|___|--|
| |\
| VCC |
| + |
| | Display2
| | |
| ___ |/ |
Input---o---|___|--| GND
|
|
|
Display1
|
|
GND



Is there a way to achieve this ?

Thanks.
Try this:
5V
| 5V
R5 |
| |<---
|/-------------| Q3
----| Q1 |\--------- (to display 1 and then to GND)
| |>-- 5V
| | |
| | R3
| | |
---R1-----R2----------
| | |
| | R4
| |<--- |
|--| Q2 GND
|\
| |/--------- (to display 2 and then to 5V)
-------------| Q4
| |>----
R6 |
| GND
GND

Q1 and Q2 form a comparator so that when the input is floating they will
both be off. R3 and R4 set the input threshold.

When the input is high (5V) then Q1 will be on and that will turn on Q3.
When the input is low (GND) then Q2 will be on and that will turn on Q4.

You can figure out the resistor values. They shouldn't be too critical,
but R3 and R4 need to be small enough to ensure enough drive for the
four transistors.

I hope I got the ascii art right as I had to compose it in a separate
word processor.

Bob
--
== NOTE: I automatically delete all Google Group posts due to
uncontrolled SPAM ==

I see, now, that your displays are both common anode. You'll need to add
another PNP (common emitter mode) driven by Q4.

As Monica Lewinsky used to say, "Close, but no cigar."

Bob

*gasp* 5 transistors needed to do what seemed simple at first glance ...

I don't quite understand what R5 and R6 are for, when Q1 and Q2 do not
conduct, the base current of Q3 and Q4 would be zero so the ressitors do
not seem necessary ?

Well I guess I could just try it and see if it works

Didn't you see my four transistors, four resistors solution yesterday?

+--------+------------------+-------+--Vcc
| | | |
___ |< | ___ |/ |
in----+--|___|---| | +--|___|--| |
| |\ | | |> |
| | | | | |
| | |< | | |<
| +------| | +-----|
| | |\ | | |\
| .-. | | .-. |
| | | .---. | | | .---.
| | | | | | | | | |
| '-' | D | | '-' | D |
| | | | | | | |
| | '---' | | '---'
| | | | | |
------)------------+--------+------)-----------+-------+--GND
| |
+----------------------------+
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


petrus bitbyter

 

[Tom Biasi]

<sinebar@bellsouth.net> wrote in message
news:50dba210-dad9-466e-a0aa-db9fcc05f2ed@m3g2000hsc.googlegroups.com...

Please bear in mind that I am not an electrical engineer so my
knowlege is somewhat limited. Now my question.
I presume that storing high voltage at low current is more efficient
than storing low voltage at high current? I also presume that
capacitors can be charged much faster than a lead acid battery? So I
was thinking why not charge up a bank of high voltage capacitors with
high voltage low current and then step it down to low voltage high
current using a transformer which then could maintain a charge on a
conventional lead acid battery that is operating the cars electric
motor. My thinking is that the caps could be charged fast and the
high voltage which is useless to run an electric motor, could be
converted to high current which is usefull for running a motor or
maintaining a battery charge. Now, could such a system have the
advantage of extending the milage of an electric car significantly
with out having to wait hours to charge?

And, where are you going to get that DC transformer.

 

[Bob Eld]

<sinebar@bellsouth.net> wrote in message
news:50dba210-dad9-466e-a0aa-db9fcc05f2ed@m3g2000hsc.googlegroups.com...

Please bear in mind that I am not an electrical engineer so my
knowlege is somewhat limited. Now my question.
I presume that storing high voltage at low current is more efficient
than storing low voltage at high current? I also presume that
capacitors can be charged much faster than a lead acid battery? So I
was thinking why not charge up a bank of high voltage capacitors with
high voltage low current and then step it down to low voltage high
current using a transformer which then could maintain a charge on a
conventional lead acid battery that is operating the cars electric
motor. My thinking is that the caps could be charged fast and the
high voltage which is useless to run an electric motor, could be
converted to high current which is usefull for running a motor or
maintaining a battery charge. Now, could such a system have the
advantage of extending the milage of an electric car significantly
with out having to wait hours to charge?

Some people are trying to do that very thing, come up with a high voltage
capacitor that can store significant amounts of energy is a relatively small
volume and weight.

The reason for high voltage is that energy stored in a capacitor is
proportional to the voltage squared. To store lots of energy, its
advantageous to get the voltage as high as possible. One concept proposed a
voltage of 3,500 Volts on the caps.

Unfortunately there is no such thing with enough energy at the present time
and all efforts to get there have not panned out. It's a dream beyond our
present technology. Some still hope it can be done in some form, however.

 

[Bob Eld]

"Butter" <clannorm@yahoo.com> wrote in message
news:0228ddc7-48d4-42b4-8d97-611f0bac6bca@8g2000hse.googlegroups.com...

A guy I work with kept complaining that his breakers were tripping. I
went over and put an amp meter each of the two circuits providing
power to his electric heat in his house. I had first measured the OHms
in ea circuit. When first turned on the amp value in ea was where I
expected it to be about 3 amps. After a few minutes each had jumped to
about 20 and then a few minutes to 40 and finally to 65 which tripped
the breakers.
He didn't want to take some wall down to get at the heating unit then
so I'm wondering what is involved in a home electric heating system.
Breakers, wire, switch, thermostate and resistence wire. Is there any
more to it?
I rarely have internet acess and then only limited time and didn't
find anything useful on the internet.Any ideas?
Rosco

He needs to get a licensed electrician on it before he burns down his house.
Don't even try to fix it, you'll negate his insurance if there is a
fire....Stay Away!

There has to be some kind of physical short in the system that comes into
play as it heats up. Normal resistance wire increases in resistance (lower
current) when hot which is the opposite of what you describe.

 

[Tom Biasi]

"Boswell" <boswell6@sbcglobal.net> wrote in message
news:625dcd1b-0825-490e-947e-6f0e1cc1b282@k10g2000prm.googlegroups.com...

I have plans to assmble a SIMPLE Date-'clock' Basic parts being;
CLOCK CHIP,
DISPLAY DRVR,
LED,
LIGHT-CONVER. (power source).

Any reccommendations or Tips, would be appreciated.

ABI, La Mesa, CA

Follow the plans, when you need help come back.

Tom

 

[Tom Biasi]

"Boswell" <boswell6@sbcglobal.net> wrote in message
news:625dcd1b-0825-490e-947e-6f0e1cc1b282@k10g2000prm.googlegroups.com...

I have plans to assmble a SIMPLE Date-'clock' Basic parts being;
CLOCK CHIP,
DISPLAY DRVR,
LED,
LIGHT-CONVER. (power source).

Any reccommendations or Tips, would be appreciated.

ABI, La Mesa, CA

Follow the plans, when you need help come back.

Tom

 

[Bob Eld]

"BobG" <bobgardner@aol.com> wrote in message
news:df519ae7-5d11-4deb-8836-bcac68797535@l42g2000hsc.googlegroups.com...

Look at the Maxwell Boostcap web site. They make a cap the size of a D
cell that is 350 Farads, but only 2.7V. They put a bunch of em in a
box that will handle 125V and 100s of amps in and out for many
seconds. This is just the boost you need when your electric car starts
out. When cruiseing, 60V would be ok, but when accelerating, a big
relay could clack the cap bank in series with the battery and youd
have a big boost voltage to get up to speed. Also good for catching
all the juice the motor is putting out when slowing down. Better than
heating up the brakes.

Yep, but because of the low voltage, ultra-caps don't hold much energy. Do
forget, putting them in series divides the capacitance.

Fifty of them in a box for 125 Volts would be 7 Farads. The energy would be
54.7 kJ = 0.015kWh! That's the amount of energy in about 1 gram of
gasoline, less than a thimble full. Not very impressive. Consider that it is
less that 0.2 Cents worth at $4 a gallon. You'd have to charge these five
times to save a penny! Why do it?

 

[Bob Eld]

"BobG" <bobgardner@aol.com> wrote in message
news:df519ae7-5d11-4deb-8836-bcac68797535@l42g2000hsc.googlegroups.com...

Look at the Maxwell Boostcap web site. They make a cap the size of a D
cell that is 350 Farads, but only 2.7V. They put a bunch of em in a
box that will handle 125V and 100s of amps in and out for many
seconds. This is just the boost you need when your electric car starts
out. When cruiseing, 60V would be ok, but when accelerating, a big
relay could clack the cap bank in series with the battery and youd
have a big boost voltage to get up to speed. Also good for catching
all the juice the motor is putting out when slowing down. Better than
heating up the brakes.

Yep, but because of the low voltage, ultra-caps don't hold much energy. Do
forget, putting them in series divides the capacitance.

Fifty of them in a box for 125 Volts would be 7 Farads. The energy would be
54.7 kJ = 0.015kWh! That's the amount of energy in about 1 gram of
gasoline, less than a thimble full. Not very impressive. Consider that it is
less that 0.2 Cents worth at $4 a gallon. You'd have to charge these five
times to save a penny! Why do it?