my only problem is I dont get how?to get the
chips to divide by 10 then by 6 to get the 1hz
|
Shift-register and synchronous counters
Good day to all. I am new in this group and would like to know if
this is the right place to ask for help on the mentioned subjects? If
not can you please refer me to the correct group or person.
I really would appreciate it because a need some information urgently.
Thakx
The Crack
|
ok first of all I would like to thank you for your
help how you could tell it had to go from low to high I have no clue Iv been
trigering it with a micro button but it seems to skip some binary number like 1
is hard to get it to go to do you have a simple ocilator I could use to triger
it at about 1hz or less or a way to make the button less sensitive so it doesent
skip.
this chip was bought with 5 others being told I
could make a clock with this chip by dividing a 60hz input by 10 and then 6 to
make the second and so on is that posible with this chip or do I need somthing
different?
Nick Graber
?
|
Dearest PcBuilders,
First of all, i like to say sorry for interrupting
the recent topics. Because of the urgent matter ie.: Bad sector was?stated
on the bargraph during scandisk?by start up floppy disk. Last week,
when?i did maintained the harddisk with?Scandisk(mode: thorough), the
missery begin. At start the bargraph showed a B sign on one of the
clusters.??When the scandisk ended, it's reported that no bad cluster
at all...!!!.? How come..???
For reference, the harddisk was divided by two
partition ie: C: and D:.
This problem occurred if i do scandisk in DOS mode
( Start up disk ). I am verry appreciated to whom that very concern and reply
promptly for this matter.
Best regards,
sie liong gin?
|
Nick,
??? There are probably a thousand or more different ways of making either a
1Hz oscillator or a debounce switch, with a thousand or so different IC's, one
of the simplest methods (to kill two birds with one stone so to speak) would be
to use the LM/NE555 timer IC. And perhaps (although I'm not so sure it would be
accurate enough to make say - a clock) it would even be able to function as a
timebase for your clock. Configure it in monostable (one-shot) mode for a switch
debouncer or astable (multivibrator) mode for a timebase.
?
Come to think of it, I may have been initially
wrong with my "how you are going to make a clock with those I don't know"
statement. Come to think of it, it probably wouldn't be all that difficult. If
you do it in the way you describe, it is entirely possible, although perhaps not
all that accurate, using the public power supply (which isn't known to be
absolutely spot on, but I guess you can always add another timebase later if
need be.
?
Yes, you just need to feed low voltage un-rectified
AC (i.e. 60Hz) to the clock pin of the counters and feed the corresponding
outputs into perhaps some AND gates (if the binary value needed uses more than
one bit) which feed into the clock pin of some more counters, and feed the
output of the digit-counters into some binary-to-BCD converters and then into
some LED 7-seg display units. It can be done ... perhaps with a lot of chips,
but it can be done.
?
There is loads of info on 555 timers about the
place, but if you come up?blank ... post another message and I will post up
some suggestions for circuits.
?
?
Let me know how you get on,
?
?
JOn
?
?
?
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----- Original Message -----
Sent: Wednesday, August 08, 2001 6:05
PM
Subject: Re: [Electronics_101] Re:
counting ic's
ok first of all I would like to thank you for
your help how you could tell it had to go from low to high I have no clue Iv
been trigering it with a micro button but it seems to skip some binary number
like 1 is hard to get it to go to do you have a simple ocilator I could use to
triger it at about 1hz or less or a way to make the button less sensitive so
it doesent skip.
this chip was bought with 5 others being told I
could make a clock with this chip by dividing a 60hz input by 10 and then 6 to
make the second and so on is that posible with this chip or do I need somthing
different?
Nick Graber
?
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Re: Shift-register and synchronous counters
Crack-man,
Ask away, we'll see what we can come up with.
Jon
|
The PCB is the circuit board, a shortened term for Printed Circuit Board. As
for the jack, it should be a separate part that is attached to the PCB. it
may or may not be enclosed in plastic. Under where the jack is located, you
sill see two or three silver colored pads that correspond to the legs of the
jack, check these for secure attachment, also check to ensure the traces (if
its a green coated board, the lighter of the green colors) are not broken.
if anything is amiss, you can 1) repair the item yourself, or 2) take it to
a friend or shop who knows how to repair electronic equipment. in the case
of 1 above, you would need a light duty pencil type soldering iron of about
20-30 watts. if any traces are broken, you can scrape the coating off with a
sharp knife until you expose the copper. Then bridge the break with small
gauge wire, soldered on both sides of the break. You problem is more likely
to be at the pads for the jack though, and simply reheating them with a
soldering iron so that the connection is repaired should solve your problem.
Shawn
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-----Original Message----- From: akirarpg@... [mailto:akirarpg@...] Sent: Monday, August 06, 2001 10:50 AM To: Electronics_101@... Subject: [Electronics_101] Re: AC input shaky --- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote: I have to agree with Troy here as one of two possibilities. I
recently found this exact problem on one of my DC adapter plugs. The
solder connections at the PCB had broken and needed touch up. The
other failure point is the plug on the end of the cable. The easiest way to determine the problem is inside or outside before you open it
up is try another DC adapter. Be sure to use one of the same
voltage, and polarity connection, and something that meets the
minimum current capabilities too.
Larry Hendry Sorry for the late response...I don't know what a PCB is. I opened up the drum machine and it appears that the little jack for the DC input (maybe that is the PCB?) is actually a part of the circuit board. I have tried other adapters with this one, and all the adapters that are finicky with my 606 work great on other equipment, so I don't think that's the problem. My theory is that the problem is in the actual hole where the end of the adapter cable goes, but I don't know what's wrong with it or how to fix it. I could take a picture of it if that would help you to figure out what's wrong. To unsubscribe from this group, send an email to: Electronics_101-unsubscribe@... Your use of Yahoo! Groups is subject to
|
--- In Electronics_101@y..., Larry Hendry <hendrysr@y...> wrote: I have to agree with Troy here as one of two possibilities. I
recently found this exact problem on one of my DC adapter plugs. The
solder connections at the PCB had broken and needed touch up. The
other failure point is the plug on the end of the cable. The easiest way to determine the problem is inside or outside before you open it
up is try another DC adapter. Be sure to use one of the same
voltage, and polarity connection, and something that meets the
minimum current capabilities too.
Larry Hendry Sorry for the late response...I don't know what a PCB is. I opened up the drum machine and it appears that the little jack for the DC input (maybe that is the PCB?) is actually a part of the circuit board. I have tried other adapters with this one, and all the adapters that are finicky with my 606 work great on other equipment, so I don't think that's the problem. My theory is that the problem is in the actual hole where the end of the adapter cable goes, but I don't know what's wrong with it or how to fix it. I could take a picture of it if that would help you to figure out what's wrong.
|
Nick,
Ok, that's cool ... now what you need to do is tie the reset line low
(via say a 5k resistor), and tie the clock pin to +Vcc via another say 5k
resistor, then pull the clock pin down to low, it should then count per
pulse received. It needs a transitional voltage (i.e. needs to go from high
to low) to trigger properly. I would also suggest that in your final circuit
you tie all output lines to ground via 10k (or thereabouts) resistors. When
dealing with logic it is always a good idea (i.e. safe practice) to tie any
lines, used or unused, to a particular level.
If you want to trigger it, apply 0V to the clock pin without disconnecting
the resistor tied to +Vcc, to reset, apply +Vcc to MR1/2, again without
disconnecting the resistor tied to ground.
Hope this helps!
Regards,
JOn
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----- Original Message ----- From: "Nick" <Nickgraber@...> To: <Electronics_101@...> Sent: Sunday, August 05, 2001 12:13 AM Subject: [Electronics_101] Re: counting ic's ok well I did that I used a new chip just to make shure everyting is
good and I still cant get it to count. I was wondering are all of
the outpst low to start because it is counted 0 so it would have an
output of LLLL = 0000 = 0 but when I apply the negitve to the triger
for the counter I still get 0000. I am now using 2n3904 transistor to
run the led.
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ok well I did that I used a new chip just to make shure everyting is
good and I still cant get it to count. I was wondering are all of
the outpst low to start because it is counted 0 so it would have an
output of LLLL = 0000 = 0 but when I apply the negitve to the triger
for the counter I still get 0000. I am now using 2n3904 transistor to
run the led.
|
Nick,
Without spending a lot of time studying the chip, it looks like a simple
2-stage binary counter ... not sure how you're going to a make a clock out
of it but I will leave that up to you.
Pin 1 (clock inverted) needs a negative trigger, so hook up your switch to
ground and you then should get some action happening on your outputs. I also
would suggest to you that you use a transistor drive on the output's if you
are driving any more than logic level signals (i.e. driving a LED etc.) the
device has a very low current rating on the output and you will potentially
destroy something if you try to drive anything directly.
Good luck!
JOn
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----- Original Message ----- From: "Nick" <Nickgraber@...> To: <Electronics_101@...> Sent: Saturday, August 04, 2001 2:45 PM Subject: [Electronics_101] counting ic's hello I am new at this eletronics stuff so talk slow.
I am working tords building a digital clock simple enuff well atleast
teh concept was. well my question is I bought some counting ic that
I was told would work for a clock and well the story goes I cant get
it to count so I am asking for some help. First off the ic's were
ordered from digikey the link to the parts pdf info file is ...
I took and put
+ to the vcc at 4.9v and grounded pin 7 so now the ic should have
power next I took and put a led in series with 1q0 I think it was and
the ground because the output should be high and took and hooked a
normaly open button from the + to the 1cp and pushed it which sould
of made 1q0 high and turned on the led in all of my theory.
Nick Graber
NickGraber@...
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hello I am new at this eletronics stuff so talk slow.
I am working tords building a digital clock simple enuff well atleast
teh concept was. well my question is I bought some counting ic that
I was told would work for a clock and well the story goes I cant get
it to count so I am asking for some help. First off the ic's were
ordered from digikey the link to the parts pdf info file is ...
I took and put
+ to the vcc at 4.9v and grounded pin 7 so now the ic should have
power next I took and put a led in series with 1q0 I think it was and
the ground because the output should be high and took and hooked a
normaly open button from the + to the 1cp and pushed it which sould
of made 1q0 high and turned on the led in all of my theory.
Nick Graber
NickGraber@...
|
Pedro,
I think you will find that this is the Maximum Reverse Voltage on the
leds (also known as the reverse breakdown voltage), which means that if
there is any more than 5 volts applied across the leads in reverse, the
diode will breakdown and probably start smoking and sputtering and do all
sorts of nasty stuff ...
What you need to look at is the Forward Voltage (denoted as Vf), typical
standard LED's have a Vf of about 2V (between 1.9 & 2.1V). The actual
formula for calculating the resistance based on forward voltage drop, supply
voltage, and maximum continuous forward current is as follows:
R=(E-Vf) x 1000 / I
Where:
R = Resistance needed (Ohms)
E = DC Supply Current (Volts)
I = Total LED current draw (mA)
I've attached a circuit of what I'm pretty sure you are trying to do;
calculated on the values I could work out ... based on a Vf of 2V,
Continuous Forward Current of 30mA, with a supply voltage of 12VDC, which
should be pretty safe ... you can recalculate if need be to accomodate a
higher forward voltage.
Regards,
Jonathan
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----- Original Message ----- From: "Pedro de Oliveira" <olive_@...> To: <Electronics_101@...> Sent: Friday, August 03, 2001 12:43 AM Subject: [Electronics_101] Re: Newbie LED question Thanks to you all for your excellent suggestions.
The reverse voltage on these LED's is 5V. I take this to mean that
they "want" 5V supply voltage to them.
I have therefore decided (due to your help) to run eight parallel
arrangements of two LED's in series from the 12V supply. My LED's are
30mA and so I would need a 0.5A power supply. I am using a 30A supply
so this should not be a problem (even after distributing power to the
other components).
Since these LED's are 5V, I assume I need a resistor to drop the rest
of the 12V supply (i.e. 2V) so about 66 Ohm (or closest available). I
will connect these to each series array so I will need eight. I am
thinking of connecting the resistors to the negative side of LED.
Does this sound convincing or have I got it all in a mess????
Cheers for your help again.
Pedro de Oliveira
|
Thanks to you all for your excellent suggestions.
The reverse voltage on these LED's is 5V. I take this to mean that
they "want" 5V supply voltage to them.
I have therefore decided (due to your help) to run eight parallel
arrangements of two LED's in series from the 12V supply. My LED's are
30mA and so I would need a 0.5A power supply. I am using a 30A supply
so this should not be a problem (even after distributing power to the
other components).
Since these LED's are 5V, I assume I need a resistor to drop the rest
of the 12V supply (i.e. 2V) so about 66 Ohm (or closest available). I
will connect these to each series array so I will need eight. I am
thinking of connecting the resistors to the negative side of LED.
Does this sound convincing or have I got it all in a mess????
Cheers for your help again.
Pedro de Oliveira
|
Jon.....
?
Honestly, I don't even know what type of LED I have
and bought these because of their ultra-bright characteristics from a surplus
electronics component shop at US$ 1.20 for 30 pieces. ?Overdriving
the? LED is done?on purpose to give the high visibility and intensity
that I am looking for.? How long these LED will last is another story. I
would expect the LED on the brake light will last longer as these are only
switch on intermitently during braking while the back light is switch on all the
time. As the system is of an experimental project, I might have to make changes
to ensure more reliability over time.
?
My scooter is running with 2x12volt/12ah SLA
battery connected in series for 24volt operation. Perhaps, an additional?12
volt voltage regulator circuit stepping down from the 24 voltage system would
help to stabilise voltage.
?
Thank you for your comments and
regards.
?
Bob Wong - Singapore
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----- Original Message -----
Sent: Thursday, August 02, 2001 8:56
AM
Subject: Re: [Electronics_101] Newbie LED
question
Robert,
?
Depending on the type of LED you are using, let's
assume that Vf is 2V - so when you put 5 of
them in series, you have an actual voltage drop of 10V, leaving you 2V
in?excess ... so in actual fact you are over-driving your LED's by
0.4V?(say 20% per each) for short periods of time - no big deal, it will
reduce the life of your LED's somewhat, but probably not enough to notice. I
will agree that it is a slightly more efficient usage of current, in that you
are not dissapating 16 seperate lots of heat - certainly a
potential?issue when you are running on limited supply
source.
?
?
If you are using it on a scooter another thing to
point out would be that if you are running it on the main drive batteries,
when in use the voltage at the terminals will actually be a lot less that 12V,
if you are running it on a seperate?battery, as the battery discharges
you will be running closer to the 10V mark too, and so getting closer to the
LED's capabilities.
?
l8r
?
Jon
----- Original Message -----
Sent: Thursday, August 02, 2001 8:17
AM
Subject: Re: [Electronics_101] Newbie
LED question
I have been experimenting using leds
for?lighting system on my electric scooter and have?built a
combined back/brake light with 3 rows of 5 red leds in series operating on
12 volts SLA. The brake light is connected parallel 2 rows of 5 leds in
series and the back light is1 row of 5 leds in series. Current draw is
approx. 90ma for the brake light and 45ma for the back light (total current
draw=135ma). No current limiting resistors were? used and the leds are
operating at maximum light intensity. I have used the light for more than 6
months and did not encounter any problem, the circuit diagram is posted
below should you be interested.
?
?
Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08
AM
Subject: Re: [Electronics_101] Newbie
LED question
16 x 30mA = 480mA.?? Therefore, the 12V source must be able
to put out half an amp for the LEDs.
?
half an amp!? That's as
much as a television draws!? Wow...
?
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Robert,
?
Depending on the type of LED you are using, let's
assume that Vf is 2V - so when you put 5 of them
in series, you have an actual voltage drop of 10V, leaving you 2V in?excess
... so in actual fact you are over-driving your LED's by 0.4V?(say 20% per
each) for short periods of time - no big deal, it will reduce the life of your
LED's somewhat, but probably not enough to notice. I will agree that it is a
slightly more efficient usage of current, in that you are not dissapating 16
seperate lots of heat - certainly a potential?issue when you are running on
limited supply source.
?
If you are using it on a scooter another thing to
point out would be that if you are running it on the main drive batteries, when
in use the voltage at the terminals will actually be a lot less that 12V, if you
are running it on a seperate?battery, as the battery discharges you will be
running closer to the 10V mark too, and so getting closer to the LED's
capabilities.
?
?
l8r
?
?
Jon
toggle quoted message
Show quoted text
----- Original Message -----
Sent: Thursday, August 02, 2001 8:17
AM
Subject: Re: [Electronics_101] Newbie LED
question
I have been experimenting using leds
for?lighting system on my electric scooter and have?built a combined
back/brake light with 3 rows of 5 red leds in series operating on 12 volts
SLA. The brake light is connected parallel 2 rows of 5 leds in series and the
back light is1 row of 5 leds in series. Current draw is approx. 90ma for the
brake light and 45ma for the back light (total current draw=135ma). No current
limiting resistors were? used and the leds are operating at maximum light
intensity. I have used the light for more than 6 months and did not encounter
any problem, the circuit diagram is posted below should you be
interested.
?
?
Bob Wong - Singapore
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08
AM
Subject: Re: [Electronics_101] Newbie
LED question
16 x 30mA = 480mA.?? Therefore, the 12V source must be able
to put out half an amp for the LEDs.
?
half an amp!? That's as much
as a television draws!? Wow...
?
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to:
Electronics_101-unsubscribe@...
Your
use of Yahoo! Groups is subject to the .
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|
I have to agree with Troy here as one of two possibilities. I recently found this exact problem on one of my DC adapter plugs. The solder connections at the PCB had broken and needed touch up. The other failure point is the plug on the end of the cable. The easiest way to determine the problem is inside or outside before you open it up is try another DC adapter. Be sure to use one of the same voltage, and polarity connection, and something that meets the minimum current capabilities too. Larry Hendry --- Troy C <troy1@...> wrote: The first thing that I would do is open
it up and see where the jack sits on the
board and wiggle it around a little to
see if the solder joints are loose. If
so, just hit them with a soldering iron
and you should be good. __________________________________________________ Do You Yahoo!? Make international calls for as low as $.04/minute with Yahoo! Messenger
|
Not Neccesarily .... a Television draws 0.5A @
110/240V (depending on where you are in the world) so your?television is
sucking 55/120W (P=VA; 110 x 0.5 = 55)? whereas the LED's running @ 12V are
drawing a little under 6W.
?
?
16 x 30mA = 480mA.?? Therefore, the 12V source must be able to
put out half an amp for the LEDs.
?
?
half an amp!? That's as much as
a television draws!? Wow...
|
I have been experimenting using leds
for?lighting system on my electric scooter and have?built a combined
back/brake light with 3 rows of 5 red leds in series operating on 12 volts SLA.
The brake light is connected parallel 2 rows of 5 leds in series and the back
light is1 row of 5 leds in series. Current draw is approx. 90ma for the brake
light and 45ma for the back light (total current draw=135ma). No current
limiting resistors were? used and the leds are operating at maximum light
intensity. I have used the light for more than 6 months and did not encounter
any problem, the circuit diagram is posted below should you be
interested.
?
?
Bob Wong - Singapore
toggle quoted message
Show quoted text
----- Original Message -----
Sent: Thursday, August 02, 2001 4:08
AM
Subject: Re: [Electronics_101] Newbie LED
question
16 x 30mA = 480mA.?? Therefore, the 12V source must be able to
put out half an amp for the LEDs.
?
half an amp!? That's as much
as a television draws!? Wow...
?
To unsubscribe from this group, send an email
to:
Electronics_101-unsubscribe@...
Your
use of Yahoo! Groups is subject to the .
|
16 x 30mA = 480mA.?? Therefore, the 12V source must be able to
put out half an amp for the LEDs.
?
half an amp!? That's as much as
a television draws!? Wow...
?
|
