Problems with the NTE 3083 (4N32) Optoisolator

[Jag Man]

I am using a pair of these as the output stage in my dual pulse generator
discussed
in several recent threads. When breadboarded, it worked. If I connected a
10k
resistor between +12 and the optoisolator collector, and grounded the
emitter,
I could see the optoisolator switch from open to conducting. When open
the collector was at +12 and when it conducted this voltage droped to 0
almost
instantaneously. When it opened again it went back up to +12, albeit a
little
slower than I had expected.

Now, after moving the circuit over to a PCB and soldering it up the
optoisolator
doesn't seem to be working. The diode section (between pins 1 and 2)
is getting switched, as I can see on a scope and by means of an LED I have
in
its circuit. However, the Darlington section doesn't switch. It seems to
be closed all the time. That is, when I power up the board the voltage at
the collector stays at nearly zero all the time.

I've checked and double checked my connections and joints and see no
problem there (famous last words, I know).

Only thing I can think of is maybe I cooked the Darlington while soldering.
I was careful to do most of the soldering before inserting the chips in the
sockets, but
I put them in to do some preliminary tests and did not remove before
soldering
the connections to pins 4 and 5. However, the data sheet says it should be
able to
withstand 260C for 10 seconds. I'm no pro put don't think I exceeded that.

Any other thoughts?

TIA

Ed

 

[John Fields]

On Fri, 19 Nov 2004 00:04:24 GMT, "Jag Man"
<Jag_Man653R-E-MOVE@hotmail.com> wrote:

However, the Darlington section doesn't switch. It seems to
be closed all the time. That is, when I power up the board the voltage at
the collector stays at nearly zero all the time.

---
Make sure that you have pullup resistors going from the collectors to
the positive supply, and if the pullups are there, make sure that
they're going to the positive supply voltage and not ground.

--
John Fields

 

[Robert Monsen]

Jag Man wrote:

I am using a pair of these as the output stage in my dual pulse generator
discussed
in several recent threads. When breadboarded, it worked. If I connected a
10k
resistor between +12 and the optoisolator collector, and grounded the
emitter,
I could see the optoisolator switch from open to conducting. When open
the collector was at +12 and when it conducted this voltage droped to 0
almost
instantaneously. When it opened again it went back up to +12, albeit a
little
slower than I had expected.

Now, after moving the circuit over to a PCB and soldering it up the
optoisolator
doesn't seem to be working. The diode section (between pins 1 and 2)
is getting switched, as I can see on a scope and by means of an LED I have
in
its circuit. However, the Darlington section doesn't switch. It seems to
be closed all the time. That is, when I power up the board the voltage at
the collector stays at nearly zero all the time.

I've checked and double checked my connections and joints and see no
problem there (famous last words, I know).

Only thing I can think of is maybe I cooked the Darlington while soldering.
I was careful to do most of the soldering before inserting the chips in the
sockets, but
I put them in to do some preliminary tests and did not remove before
soldering
the connections to pins 4 and 5. However, the data sheet says it should be
able to
withstand 260C for 10 seconds. I'm no pro put don't think I exceeded that.

Any other thoughts?

You could desolder the chip and then test it. If it's ok, you know that
it is the circuit.

If the collector of the darlington is always at 0V, then it is always
on. That could happen with a leakage of current through the diode, or if
its fried. Do you have a diode across the relay you are driving? If not,
then you can easily exceed the maximum voltage of the darlington, which
is 30V. Also, you need to ensure that the current through the thing is
below the max rating.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jag Man]

Looks like I've cooked the NTE 3083s. I pulled them out of the sockets and
set them up independently on a breadboard. Connected pin 1 to +12 through a
pushbutton,
and connected a 510 ohm resistor between 2 and ground. Thus the
LED section is activated by the pushbutton. Pin 5 is grounded,
and pin 4 goes to +12 through a 10k resistor. If it were working I would
expect
to see 12 volts at pin 4, droping to close to 0 when the button is pressed.
Instead, I see about 9 volts at pin 2 regardless of the button.

The LED section seems to be OK, as I get about 20 mA through the 510
resistor when the button is pressed.

Whatever I did wrong I did it consistanlty, as both of my 3083s now behave
that way.
The question is, is there a wiring error in my circuit, or were they heat
damaged while soldering? Guess I better go over the circuit again...

Ed

I am using a pair of these as the output stage in my dual pulse generator
discussed
in several recent threads. When breadboarded, it worked. If I connected a
10k
resistor between +12 and the optoisolator collector, and grounded the
emitter,
I could see the optoisolator switch from open to conducting. When open
the collector was at +12 and when it conducted this voltage droped to 0
almost
instantaneously. When it opened again it went back up to +12, albeit a
little
slower than I had expected.

Now, after moving the circuit over to a PCB and soldering it up the
optoisolator
doesn't seem to be working. The diode section (between pins 1 and 2)
is getting switched, as I can see on a scope and by means of an LED I have
in
its circuit. However, the Darlington section doesn't switch. It seems to
be closed all the time. That is, when I power up the board the voltage at
the collector stays at nearly zero all the time.

I've checked and double checked my connections and joints and see no
problem there (famous last words, I know).

Only thing I can think of is maybe I cooked the Darlington while
soldering.
I was careful to do most of the soldering before inserting the chips in
the
sockets, but
I put them in to do some preliminary tests and did not remove before
soldering
the connections to pins 4 and 5. However, the data sheet says it should be
able to
withstand 260C for 10 seconds. I'm no pro put don't think I exceeded that.

Any other thoughts?

TIA

Ed

 

[Jag Man]

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:Ybbnd.108598$HA.85604@attbi_s01...

You could desolder the chip and then test it. If it's ok, you know that
it is the circuit.

Fortunately, it is socketed so I just poped it out. Good idea though... see
my other posting today.

Do you have a diode across the relay you are driving? If not,
then you can easily exceed the maximum voltage of the darlington, which
is 30V. Also, you need to ensure that the current through the thing is
below the max rating.

So far, it's only been exposed to resistive load, 10k at 12 volts. But
thanks for
mentioning that. I'll put a diode in there so when I hook it to something
else it
will be protected.

Ed

 

[Robert Monsen]

Jag Man wrote:

Oops. My testing setup was in error. Pin 4, the emitter, should be grounded
rather than pin 5, the collector. When connected correctly the 3083s work.
They are OK. It cost me $6 to find out, though, as I bought new ones based
on my erroneous conclusions!

So, now it's back to the original question, i.e., why they don't work on my
circuit
board. Only difference is on the circuit board I have an LED,
a 400 ohm resistor, and a NTE123AP transistor in the cathode path to
ground.
Hmmm.

I'm unclear about whether it is always on or always off while in the
circuit. If you put a 10k resistor to 12V on the collector, and tie the
emitter to ground, the measured voltage at the collector is always at
ground, right? Also, you aren't connecting to the base pin, are you?

What kind of LED are you using? Does it light up when the opto is
supposed to trigger?

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Robert Monsen]

Jag Man wrote:

Robert,

The emitter of the 3083 (pin 4) is tied to ground in all cases except my
first
reported attempt to test the 3083, earlier today. The Collector
(pin 5) is fed from a 10k resistor that is tied to +12.

The LED is in the Cathode path to ground. I.e., pin 2 -> LED->380 ohm
resistor-
transistor->ground. It is a garden variety with about 1.8 volt offset. It
lights up
when the 3083's internal LED fires. That firing is caused by the transistor
in its circuit.
My circuit is shown in
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg

However, that diagram does not show the load that I put on the
3083.

thanks for your inputs.

Can you list the steps you have taken to determine if the optoisolator
is working? What were the results, in terms of voltages? What kind of
test equipment are you using?

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jag Man]

Robert,

Let me describe the two test setups I've used.

1. breadboard with just the 3083

+12 volts goes through a push-button to pin 1
pin 2 goes through 510 ohm resistor to ground
10k resistor between pin 5 and +12
pin 6 grounded.

Test: using digital VOM measure voltage at pin 5.
button not pushed, read 12 volts
Push the button, read 0 volts.

2. PCB with socketed ICs and all components shown in the diagram ( See
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg)
soldered in place.

Since the PCB does not have anything connected across pins 4 & 5 of the
3083,
I used a separate breadboard to simulate a load. On this breadboard I have
a 10k resistor connected
to +12, the other end of which is connected to pin 5 of the 3083 on the
PCB with alligator clips &
test leads. Pin 4 is also jumpered to ground.

Test: Using an Hitachi 212 oscilloscope monitor various voltages. For
example,
I measure the voltage at pin 3 of the 555 IC which feeds to the base of
the 3904 NPN transistor,
where I see a positive 12 volt pulse of width (2-10 ms) determined by the
pot going to pins 6 & 7 of the 555.
The frequency (6 - 43 Hz) is set by the pot on the 555 at the far left of
the diagram. What see here is what I would expect.
Then I measure the voltage at pin 5 of the 3083. What I see here is
anomalous and variable. It should be normally 12
volts, dropping down to 0 while the output of the 555 is high. What I see
instead is sometimes 6 volts, dropping
down to 0 when the 555 output is high. Yesterday, the normal was only
about 2 volts. Later today, it was coming back up to 12
volts, which would be normal. I don't know why it varies, perhaps
temperature. Also, I notice that the pulse width
doesn't seem to be stable at the 3083 ouptut, although it is at the 555
output.

My current interpretation of all this that the 3904 NPN transistors may
not be doing a very good job for me. For example,
perhaps they are letting current through sometimes when the 555 output is
low that would allow current to flow in
the 3083's LED section, thus turning on the output when it shouldn't.

That's about all I can say at this point.

Ed

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:VPKnd.540509$mD.371865@attbi_s02...

Can you list the steps you have taken to determine if the optoisolator
is working? What were the results, in terms of voltages? What kind of
test equipment are you using?

--
Regards,
Robert Monsen

 

[peterken]

"Robert Monsen" <rcsurname@comcast.net> wrote in message
news:uNRnd.542195$mD.430044@attbi_s02...

Jag Man wrote:
Robert,

Let me describe the two test setups I've used.

1. breadboard with just the 3083

+12 volts goes through a push-button to pin 1
pin 2 goes through 510 ohm resistor to ground
10k resistor between pin 5 and +12
pin 6 grounded.

Test: using digital VOM measure voltage at pin 5.
button not pushed, read 12 volts
Push the button, read 0 volts.

2. PCB with socketed ICs and all components shown in the diagram ( See
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg)
soldered in place.

Since the PCB does not have anything connected across pins 4 & 5 of
the
3083,
I used a separate breadboard to simulate a load. On this breadboard I
have
a 10k resistor connected
to +12, the other end of which is connected to pin 5 of the 3083 on
the
PCB with alligator clips &
test leads. Pin 4 is also jumpered to ground.

Test: Using an Hitachi 212 oscilloscope monitor various voltages. For
example,
I measure the voltage at pin 3 of the 555 IC which feeds to the base
of
the 3904 NPN transistor,
where I see a positive 12 volt pulse of width (2-10 ms) determined by
the
pot going to pins 6 & 7 of the 555.
The frequency (6 - 43 Hz) is set by the pot on the 555 at the far
left of
the diagram. What see here is what I would expect.
Then I measure the voltage at pin 5 of the 3083. What I see here is
anomalous and variable. It should be normally 12
volts, dropping down to 0 while the output of the 555 is high. What I
see
instead is sometimes 6 volts, dropping
down to 0 when the 555 output is high. Yesterday, the normal was only
about 2 volts. Later today, it was coming back up to 12
volts, which would be normal. I don't know why it varies, perhaps
temperature. Also, I notice that the pulse width
doesn't seem to be stable at the 3083 ouptut, although it is at the 555
output.

My current interpretation of all this that the 3904 NPN transistors
may
not be doing a very good job for me. For example,
perhaps they are letting current through sometimes when the 555 output
is
low that would allow current to flow in
the 3083's LED section, thus turning on the output when it shouldn't.

That's about all I can say at this point.

Ed


Ed:

There are a couple of possibilities I can think of.

1) The optoisolator isn't recovering fully. Try slowing down the
oscillator, and see if it gets better.

2) The 2N3904 isn't turning off all the way. Make sure that 10k resistor
at it's base is actually grounded, and not attached to Vcc.

3) You have some issue with the layout, maybe a small solder bridge that
is leaking current.

I'm sure the pros can think of some other possibilites.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God

There IS another possibility...
Some opto isolators have rather high leakage current on their output,
meaning from several hundreds of uA to even 1mA
using a 10k resistor (rather high value) "takes" all the voltage away due to
leakage, not due to signal....

 

[Robert Monsen]

peterken wrote:

There IS another possibility...
Some opto isolators have rather high leakage current on their output,
meaning from several hundreds of uA to even 1mA
using a 10k resistor (rather high value) "takes" all the voltage away due to
leakage, not due to signal....

You may have something here. The datasheet test resistance is about 180
ohms, giving an Ic of 50mA. I wonder why a darlington would leak so
much, though?

Maybe Jag could rerun the test with a 180 ohm resistor and see what happens.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jag Man]

Yes, I believe that's it! I put a 510 ohm resistor in there and it seems
to be behaving as it should. Guess I had the 10k in there simply because
I had one handy and figured it was big enough to be sure the thing wasn't
overloaded. Pure sloth on my part... even some mental artihmetic
should have told me that something in the 100s was more
appropriate.

I am still getting some jumping around of the pulse on the scope.
It's like the width snaps from one value to another, or the triggering point
changes from one sweep to the next. This thing operates at
very low frequencies, 6 to 40 Hz, and at the low end of that range
is where I see that behavior. But this may be a scope triggering issue.
My skills at using the thing (a Hitachi V-212) are minimal.

Thanks!

Ed

peterken wrote:

There IS another possibility...
Some opto isolators have rather high leakage current on their output,
meaning from several hundreds of uA to even 1mA
using a 10k resistor (rather high value) "takes" all the voltage away
due to
leakage, not due to signal....


You may have something here. The datasheet test resistance is about 180
ohms, giving an Ic of 50mA. I wonder why a darlington would leak so
much, though?

Maybe Jag could rerun the test with a 180 ohm resistor and see what
happens.

--
Regards,
..

 

[peterken]

check value of resistor (at least I hope a resistor is still connected to V+
?) in collector in output and verify if now a capacitive load is connected
to the pcb
if switching frequency is too high the transistor appears to conduct all the
time if capacitive load and resistor is too high (time constant)
lowering resistor value or buffering so no more capacitive load exists
solves things
and it's normal the voltage drops fast and rises slower : switching on
transistor givers a fast edge, but desaturation of a transistor (and
capacitive loads) slows down edges


"Jag Man" <Jag_Man653R-E-MOVE@hotmail.com> wrote in message
news:cMand.46145$QJ3.3360@newssvr21.news.prodigy.com...

I am using a pair of these as the output stage in my dual pulse generator
discussed
in several recent threads. When breadboarded, it worked. If I connected a
10k
resistor between +12 and the optoisolator collector, and grounded the
emitter,
I could see the optoisolator switch from open to conducting. When open
the collector was at +12 and when it conducted this voltage droped to 0
almost
instantaneously. When it opened again it went back up to +12, albeit a
little
slower than I had expected.

Now, after moving the circuit over to a PCB and soldering it up the
optoisolator
doesn't seem to be working. The diode section (between pins 1 and 2)
is getting switched, as I can see on a scope and by means of an LED I have
in
its circuit. However, the Darlington section doesn't switch. It seems to
be closed all the time. That is, when I power up the board the voltage at
the collector stays at nearly zero all the time.

I've checked and double checked my connections and joints and see no
problem there (famous last words, I know).

Only thing I can think of is maybe I cooked the Darlington while
soldering.
I was careful to do most of the soldering before inserting the chips in
the
sockets, but
I put them in to do some preliminary tests and did not remove before
soldering
the connections to pins 4 and 5. However, the data sheet says it should be
able to
withstand 260C for 10 seconds. I'm no pro put don't think I exceeded that.

Any other thoughts?

TIA

Ed

 

[Rich Grise]

On Sat, 20 Nov 2004 21:42:32 +0000, Jag Man wrote:

Robert,

Let me describe the two test setups I've used.

1. breadboard with just the 3083

+12 volts goes through a push-button to pin 1
pin 2 goes through 510 ohm resistor to ground
10k resistor between pin 5 and +12
pin 6 grounded.

Test: using digital VOM measure voltage at pin 5.
button not pushed, read 12 volts
Push the button, read 0 volts.

2. PCB with socketed ICs and all components shown in the diagram ( See
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg)
soldered in place.

Since the PCB does not have anything connected across pins 4 & 5 of the
3083,
I used a separate breadboard to simulate a load. On this breadboard I have
a 10k resistor connected
to +12, the other end of which is connected to pin 5 of the 3083 on the
PCB with alligator clips &
test leads. Pin 4 is also jumpered to ground.

For my money, you've got a short or open somewhere in your test setup. If
the LED lights, it's pretty safe to assume that the LED in the opto is
lighting. If so, the transistor should conduct. Now, did you say it's
acting short, or acting open? Where are pins 3 and 6 going to?

One thing you could do is measure the voltage between pins 1 and 2 of the
opto, just to make sure your LED is ledding, but since your schematic
doesn't show anything at all at the optos' transistors, it's impossible to
tell what your test setup is doing.

Could you please post the rest of the circuit?

Thanks,
Rich

 

[Jag Man]

Oops. My testing setup was in error. Pin 4, the emitter, should be grounded
rather than pin 5, the collector. When connected correctly the 3083s work.
They are OK. It cost me $6 to find out, though, as I bought new ones based
on my erroneous conclusions!

So, now it's back to the original question, i.e., why they don't work on my
circuit
board. Only difference is on the circuit board I have an LED,
a 400 ohm resistor, and a NTE123AP transistor in the cathode path to
ground.
Hmmm.

Ed

Looks like I've cooked the NTE 3083s. I pulled them out of the sockets and
set them up independently on a breadboard. Connected pin 1 to +12 through
a
pushbutton,
and connected a 510 ohm resistor between 2 and ground. Thus the
LED section is activated by the pushbutton. Pin 5 is grounded,
and pin 4 goes to +12 through a 10k resistor. If it were working I would
expect
to see 12 volts at pin 4, droping to close to 0 when the button is
pressed.
Instead, I see about 9 volts at pin 2 regardless of the button.

The LED section seems to be OK, as I get about 20 mA through the 510
resistor when the button is pressed.

Whatever I did wrong I did it consistanlty, as both of my 3083s now behave
that way.
The question is, is there a wiring error in my circuit, or were they heat
damaged while soldering? Guess I better go over the circuit again...

Ed

 

[Robert Monsen]

Jag Man wrote:

Robert,

Let me describe the two test setups I've used.

1. breadboard with just the 3083

+12 volts goes through a push-button to pin 1
pin 2 goes through 510 ohm resistor to ground
10k resistor between pin 5 and +12
pin 6 grounded.

Test: using digital VOM measure voltage at pin 5.
button not pushed, read 12 volts
Push the button, read 0 volts.

2. PCB with socketed ICs and all components shown in the diagram ( See
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg)
soldered in place.

Since the PCB does not have anything connected across pins 4 & 5 of the
3083,
I used a separate breadboard to simulate a load. On this breadboard I have
a 10k resistor connected
to +12, the other end of which is connected to pin 5 of the 3083 on the
PCB with alligator clips &
test leads. Pin 4 is also jumpered to ground.

Test: Using an Hitachi 212 oscilloscope monitor various voltages. For
example,
I measure the voltage at pin 3 of the 555 IC which feeds to the base of
the 3904 NPN transistor,
where I see a positive 12 volt pulse of width (2-10 ms) determined by the
pot going to pins 6 & 7 of the 555.
The frequency (6 - 43 Hz) is set by the pot on the 555 at the far left of
the diagram. What see here is what I would expect.
Then I measure the voltage at pin 5 of the 3083. What I see here is
anomalous and variable. It should be normally 12
volts, dropping down to 0 while the output of the 555 is high. What I see
instead is sometimes 6 volts, dropping
down to 0 when the 555 output is high. Yesterday, the normal was only
about 2 volts. Later today, it was coming back up to 12
volts, which would be normal. I don't know why it varies, perhaps
temperature. Also, I notice that the pulse width
doesn't seem to be stable at the 3083 ouptut, although it is at the 555
output.

My current interpretation of all this that the 3904 NPN transistors may
not be doing a very good job for me. For example,
perhaps they are letting current through sometimes when the 555 output is
low that would allow current to flow in
the 3083's LED section, thus turning on the output when it shouldn't.

That's about all I can say at this point.

Ed

Ed:

There are a couple of possibilities I can think of.

1) The optoisolator isn't recovering fully. Try slowing down the
oscillator, and see if it gets better.

2) The 2N3904 isn't turning off all the way. Make sure that 10k resistor
at it's base is actually grounded, and not attached to Vcc.

3) You have some issue with the layout, maybe a small solder bridge that
is leaking current.

I'm sure the pros can think of some other possibilites.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jag Man]

as far as i see your explanations i can only imagine the transistors
2N3904
don't get to the off-state completely

That's what it seems to me.

try measuring the collector voltage on them, see if it rises to V+

That's what I AM measuring, am I not? The scope is on the wire going from
the
pull-up resistor to the collector. It drops very close to zero, but does not
rise back up to +12... at least not all the time. I hedge here because When
I had a 10k pullup resistor it stayed so close to 0 that I though nothing
was happening at all. Then when I put in a 510 ohm it seemed to work...
I pronounced the problem solved. Then while finalizing things I ruined my
510
resistors so put 620 ohm ones in (happened to have ) and now it's not
working again. Comes
up to maybe 5 volts. When I parallel two 610 it comes up to perhaps 9 or 10
volts,
but not 12.

other thing is of course again leakage currents of the opto's, if it's say
250uA you get a voltage drop of 125mV on the output....

I have 4 NTE 3083 and I keep trading them around. It does make a difference,
but when it's not working I can't find one that does.

What I need explanined, however, is how are these things SUPPOSED to work?
IOW, what am I doing that is so different from what others do with the NTE
3083?
Can I do something to deal with the leakage?

TIA

Ed

 

[Jag Man]

Robert,

The emitter of the 3083 (pin 4) is tied to ground in all cases except my
first
reported attempt to test the 3083, earlier today. The Collector
(pin 5) is fed from a 10k resistor that is tied to +12.

The LED is in the Cathode path to ground. I.e., pin 2 -> LED->380 ohm
resistor->
transistor->ground. It is a garden variety with about 1.8 volt offset. It
lights up
when the 3083's internal LED fires. That firing is caused by the transistor
in its circuit.
My circuit is shown in
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg

However, that diagram does not show the load that I put on the
3083.

thanks for your inputs.


Ed

I'm unclear about whether it is always on or always off while in the
circuit. If you put a 10k resistor to 12V on the collector, and tie the
emitter to ground, the measured voltage at the collector is always at
ground, right? Also, you aren't connecting to the base pin, are you?

What kind of LED are you using? Does it light up when the opto is
supposed to trigger?

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Jag Man]

Hi Rich,

Here is the test configuration. All I do is connect the collector through a
pullup resistor
to +12 and ground the emitter.

http://sowell.ecs.fullerton.edu/jag/fuelingsys/testerTester.jpg

The resistor is nor 510 ohms, so I get better results. But it really does
seem that
there is a parallel path to ground somewhere, because it doesn't rise up to
12 volts
when the isolator switches off. I have inspected by board with a magnifiying
glass
and see no shorts. Only thing I can thing of is perhaps some solder ran
under the
socket somehow. Guess I'll try to unsolder them and look.

Any suggestions welcome.

Ed

"Rich Grise" <rich@example.net> wrote in message
newsan.2004.11.23.05.30.03.271724@example.net...

On Sat, 20 Nov 2004 21:42:32 +0000, Jag Man wrote:

Robert,

Let me describe the two test setups I've used.

1. breadboard with just the 3083

+12 volts goes through a push-button to pin 1
pin 2 goes through 510 ohm resistor to ground
10k resistor between pin 5 and +12
pin 6 grounded.

Test: using digital VOM measure voltage at pin 5.
button not pushed, read 12 volts
Push the button, read 0 volts.

2. PCB with socketed ICs and all components shown in the diagram ( See
http://sowell.ecs.fullerton.edu/jag/fuelingsys/tester111704.jpg)
soldered in place.

Since the PCB does not have anything connected across pins 4 & 5 of
the
3083,
I used a separate breadboard to simulate a load. On this breadboard I
have
a 10k resistor connected
to +12, the other end of which is connected to pin 5 of the 3083 on
the
PCB with alligator clips &
test leads. Pin 4 is also jumpered to ground.

For my money, you've got a short or open somewhere in your test setup. If
the LED lights, it's pretty safe to assume that the LED in the opto is
lighting. If so, the transistor should conduct. Now, did you say it's
acting short, or acting open? Where are pins 3 and 6 going to?

One thing you could do is measure the voltage between pins 1 and 2 of the
opto, just to make sure your LED is ledding, but since your schematic
doesn't show anything at all at the optos' transistors, it's impossible to
tell what your test setup is doing.

Could you please post the rest of the circuit?

Thanks,
Rich

 

[peterken]

"Jag Man" <Jag_Man653R-E-MOVE@hotmail.com> wrote in message
news:lwQpd.27517$6q2.9021@newssvr14.news.prodigy.com...

Hi Rich,

Here is the test configuration. All I do is connect the collector through
a
pullup resistor
to +12 and ground the emitter.

http://sowell.ecs.fullerton.edu/jag/fuelingsys/testerTester.jpg

The resistor is nor 510 ohms, so I get better results. But it really does
seem that
there is a parallel path to ground somewhere, because it doesn't rise up
to
12 volts
when the isolator switches off. I have inspected by board with a
magnifiying
glass
and see no shorts. Only thing I can thing of is perhaps some solder ran
under the
socket somehow. Guess I'll try to unsolder them and look.

Any suggestions welcome.

Ed

as far as i see your explanations i can only imagine the transistors 2N3904
don't get to the off-state completely
try measuring the collector voltage on them, see if it rises to V+

other thing is of course again leakage currents of the opto's, if it's say
250uA you get a voltage drop of 125mV on the output....

 

[peterken]

"Jag Man" <Jag_Man653R-E-MOVE@hotmail.com> wrote in message
news:zhSpd.49930$QJ3.3699@newssvr21.news.prodigy.com...

as far as i see your explanations i can only imagine the transistors
2N3904
don't get to the off-state completely

That's what it seems to me.

try measuring the collector voltage on them, see if it rises to V+

That's what I AM measuring, am I not? The scope is on the wire going from
the
pull-up resistor to the collector. It drops very close to zero, but does
not
rise back up to +12... at least not all the time. I hedge here because
When
I had a 10k pullup resistor it stayed so close to 0 that I though nothing
was happening at all. Then when I put in a 510 ohm it seemed to work...
I pronounced the problem solved. Then while finalizing things I ruined my
510
resistors so put 620 ohm ones in (happened to have ) and now it's not
working again. Comes
up to maybe 5 volts. When I parallel two 610 it comes up to perhaps 9 or
10
volts,
but not 12.


other thing is of course again leakage currents of the opto's, if it's
say
250uA you get a voltage drop of 125mV on the output....

I have 4 NTE 3083 and I keep trading them around. It does make a
difference,
but when it's not working I can't find one that does.

What I need explanined, however, is how are these things SUPPOSED to work?
IOW, what am I doing that is so different from what others do with the NTE
3083?
Can I do something to deal with the leakage?

TIA

Ed

How they are SUPPOSED to work is simple :
led on = output on, led off = output off
(some opto's have also a linear part, so might be used linear)

Dealing with leakage is simple :
Design as if ALWAYS the max leakage exists (datasheet, and add some to be
safe)
This means, calculate the output voltage of the opto in the off-state : Vo
= V+ - (load resistor * max leakage current)
thus meaning "using" the output may *not* rely on an exact V+, but rather on
a level calculated above to be very safe

One other thing I noticed in your schematic :
All your digitals are fed at 12V, yet the outputs are "only" divided by 2
(10k-10k) towards base of transistors....
Bad practice, I'd rather use say 10k towards base and 1k5 towards ground.
This gives a division of say 1:10 (or about 1.55V thereby ignoring the
junction of the transistor) as "control voltage" towards base of
transistors, which still is far over the junction voltage to "open" the
transistors.
This avoids the effect of say a "non-fully-zero-low" of the digitals causing
still opening the transistors a tiny bit and fouling the operation of the
circuit completely.
Tthis way the "low-level" of the digitals may be even upto say 3.5V *before*
any "opening" of the transistors arises.
In your setup a "low level" of say 1V already causes some opening the
transistors.

Also, the led of the opto's is directly connected (through the series
resistor) to the collector of the transistors.
ALSO the transistors have some leakage current, thereby *maybe* causing a
"continuous glow" of the leds of the opto, thus causing the output not to
behave correctly.
did you verify the leakage of the transistors ? Maybe a parallel resistor
to the leds can remedy this (say 4k7 to be safe to "catch" any leakage)
(tip : short or disconnect the led input of the opto's, this verifies if the
leds in the opto do glow continuously caused by leakage currents)

One other thing :
(must say, I didn't go check it so sorry if I'm wrong here)
Doesn't the 555 need a pullup at it's output ? (assuming it's an open
collector output here)

Tip learned in a 24 years carreer in R&D :
As a designer *never* assume anything works perfectly, *always* ask yourself
"what if..."
This is called "worst case design", and as a result the circuits are more
reliable.