-- 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.
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
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.
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
soldering.
the
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
soldering.
the
Fortunately, it is socketed so I just poped it out. Good idea though... see my other posting today.
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
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
apressed.
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,
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
However, that diagram does not show the load that I put on the
3083.thanks for your inputs.
Ed
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.
Robert,
Let me describe the two test setups I've used.
Test: using digital VOM measure voltage at pin 5. button not pushed, read 12 volts Push the button, read 0 volts.
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.
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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.
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
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happens.
.Hi Rich,
Here is the test configuration. All I do is connect the collector through a pullup resistor to +12 and ground the emitter.
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
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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....2N3904
That's what it seems to me.
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.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
--- When you force current through the LED it generates infrared photons which fall on the sensitive area of the phototransistor and allow current to flow through the collector-to-emitter junctions of the Darlington connected transistors.
The Darlington has very high current gain; on the order of 50000, so even a very small quiescent leakage current flowing into the base of the first transistor will cause a much larger quiescent current to flow through the output, which seems to be what you're experiencing.
The leakage can come from three places:
There _is_ a fourth choice, getting rid of that damned Darlington in the first place and sidestepping the problem. You don't need the gain since you've got plenty of drive available for the IRLED, so why buy trouble? If it was me I'd go for a vanilla 4N35, 4N36, or 4N37.
Matter of fact, I've got some 4N35's left over from a project, so if you want a few of them, gratis, email me with a physical address where I can send them and I will.
-- John Fields
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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.
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