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
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--- 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
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----- 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@...
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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
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----- 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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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...
?
|
Hi Pedro,
If your LED doesn't produce light, simply reverse the leads and it
should be in the forward current direction. Each LED has an anode and
cathode end and is polarity sensitive.
John
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--- In Electronics_101@y..., "Pedro de Oliveira" <olive_@h...> wrote: Hi and thanks for your help
I understood all that your message said and will now get 12x330 Ohm
resistors to wire before each of the LED's. There is just a bit I
don't understand in your message and this is cut and pasted below:
<SNIP>
Insure that the polarity is such, that the LED's are forward
biased, negative through the series resistors to the cathodes, and
positive voltage to the anodes. Hope this helps. </SNIP>
What does this mean, do they all have to lie the same way or +ve to
+ve
Thanks again
Pedro
|
Hi and thanks for your help I understood all that your message said and will now get 12x330 Ohm resistors to wire before each of the LED's. There is just a bit I don't understand in your message and this is cut and pasted below: <SNIP> Insure that the polarity is such, that the LED's are forward
biased, negative through the series resistors to the cathodes, and
positive voltage to the anodes. Hope this helps. </SNIP> What does this mean, do they all have to lie the same way or +ve to +ve Thanks again Pedro
|
It should also be noted that it is preferable to use one (330 ohm) resistor
per LED rather than one (e.g. 20 ohm) resistor for the whole lot, because of
internal resistance deviations in each LED, as one device may end up drawing
more current than the rest and cause some damage.
Just my 2 cents ...
Jon
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----- Original Message ----- From: <jperrynew@...> To: <Electronics_101@...> Sent: Wednesday, August 01, 2001 4:57 PM Subject: [Electronics_101] Re: Newbie LED question Hi Pedro, If you want to run the 16 LED's at 30 mA each, your 12V power supply must have at least 500 mA (16 X 30 mA=480 mA) or better capability. A power supply with 1A capability would run much cooler. Each of the LED's, if they don't already have a built-in resistor, will need a series limiting resistor to insure your 30 mA of current is not exceeded. The forward voltage (VF)drop is not mentioned in your spec, but the LED's are probably dropping about 2 V, that would mean that you have about 10 V dropped across your limiting resistors. Using Ohms Law R = E/I you can calculate the resistor value to be R = 10V/.03A = 333 ohms. Resistors come in standard values, the closest value for the above application is a 330 ohm. The power rating of the resistor can be found by using the formula P = E X I, in this example P = 10V X .03A = 300 milliwatts. Therefore the 330 ohm resistors should have a power rating of at least .5 watts (500 milliwatts). If you are desiring to have all 16 LEDs on at the same time, connect each of the LEDs with their respective resistors in series and then all of them across the 12 V source. Insure that the polarity is such, that the LED's are forward biased, negative through the series resistors to the cathodes, and positive voltage to the anodes. Hope this helps. John --- In Electronics_101@y..., "Pedro de-Oliveira" <olive_@h...> wrote: Hi all
I am a newbie to the art of electronics so please be gentle with me :-)
I am trying to run 16 LED's from a single 12v power source. The
LED specs are as follows:
Parameter Value Units
DC Forward Current[1] 30 mA
Peak Forward Current 100 mA
Average Forward Current 30 mA
Power Dissipation 120 mW
Reverse Voltage (IR = 100 mA) 5 V
LED Junction Temperature 100 ???C
Operating Temperature Range -40 to +80 ???C
Storage Temperature Range -40 to +100 ???C
What I would like to know is what would be the best way to hook these things up and what value resistors I would need to do it.
Cheers
Pedro
_________________________________________________________________
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Hi Pedro, If you want to run the 16 LED's at 30 mA each, your 12V power supply must have at least 500 mA (16 X 30 mA=480 mA) or better capability. A power supply with 1A capability would run much cooler. Each of the LED's, if they don't already have a built-in resistor, will need a series limiting resistor to insure your 30 mA of current is not exceeded. The forward voltage (VF)drop is not mentioned in your spec, but the LED's are probably dropping about 2 V, that would mean that you have about 10 V dropped across your limiting resistors. Using Ohms Law R = E/I you can calculate the resistor value to be R = 10V/.03A = 333 ohms. Resistors come in standard values, the closest value for the above application is a 330 ohm. The power rating of the resistor can be found by using the formula P = E X I, in this example P = 10V X .03A = 300 milliwatts. Therefore the 330 ohm resistors should have a power rating of at least .5 watts (500 milliwatts). If you are desiring to have all 16 LEDs on at the same time, connect each of the LEDs with their respective resistors in series and then all of them across the 12 V source. Insure that the polarity is such, that the LED's are forward biased, negative through the series resistors to the cathodes, and positive voltage to the anodes. Hope this helps. John --- In Electronics_101@y..., "Pedro de-Oliveira" <olive_@h...> wrote: Hi all
I am a newbie to the art of electronics so please be gentle with me :-)
I am trying to run 16 LED's from a single 12v power source. The
LED specs are as follows:
Parameter Value Units
DC Forward Current[1] 30 mA
Peak Forward Current 100 mA
Average Forward Current 30 mA
Power Dissipation 120 mW
Reverse Voltage (IR = 100 mA) 5 V
LED Junction Temperature 100 ¡ãC
Operating Temperature Range ¨C40 to +80 ¡ãC
Storage Temperature Range ¨C40 to +100 ¡ãC
What I would like to know is what would be the best way to hook these things up and what value resistors I would need to do it.
Cheers
Pedro
_________________________________________________________________
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|
>I am trying to run 16 LED's from a single 12v power source. The LED specs
>are as follows:
>DC Forward Current[1] 30 mA
>Peak Forward Current 100 mA
>Average Forward Current 30 mA
>What I would like to know is what would be the best way to hook these things
>up and what value resistors I would need to do it.
?
?
Use a resistor that will limit the current to about 30mA so that not too much current is drawn and the LEDs will be bright enough.?? An LED will drop about 2V so, that would leave 10V to drop across the resistor.?? 10V divided by 30mA is 333.3 ohms, so a 330 ohm resistor should work fine provided that the 12V is regulated and it will put out enough current to drive all 16 LEDs.?? If each LED and resistor draws 30mA, then the 16 will draw a total of 16 x 30mA = 480mA.?? Therefore, the 12V source must be able to put out half an amp for the LEDs.
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No...sorry. I didn't mean to make you type out that long
explanation; I hope it helps someone else. The actual *input* for
the AC cable is messed up. Like on the cheap older keyboards - I
have to position the jack that goes into it just so to get it to
work. Is there a way to fix this? 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. Although you really can't be sure what needs to be done until you open it up and have a look.
|
