Maker Pro
Maker Pro

Problems with the NTE 3083 (4N32) Optoisolator

J

John Fields

Jan 1, 1970
0
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?

---
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:

1. The transistor driving the LED may not turn off completely.
You can test for this by measuring the current through the LED with
the [LED driver] transistor in its off state. Simply put a DMM set
to milliamps or microamps in series with the LED, turn the
transistor off and read the meter.

2. Bad opto. Test for this by grounding the LED anode and cathode and
the base of the transistor and then measuring the current which
flows out of the power supply with V+ connected to the collector
and V- connected to the emitter.

3. Non-isolated base. For this one you'll need to have a leakage path
between pin 6 and a positive voltage somewhere. With a gain of
50000, all you need to get a couple of mA into the output is 40nA,
so that's what I'd be looking for. Check for contamination
(moisture, solder flux residue, dirt, etc.) around the opto's base
lead and if you find any, clean it up.

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.
 
J

Jag Man

Jan 1, 1970
0
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.

Yes, I now see that. I have now replaced the base-to-ground resistor with
1.5K.
Circuit works very well now! Thanks!




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)

Don't know...
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.

While I have reasonable training (engineering/Comp sci) and good basic
understanding
of circuits I have little practical experience. I'm afraid I've been
accepting proposed
designs here without full understanding. By and large, it's been very
helpful. The baisc
disign of my circuit is due to Chris Foley. I think it's very clever.
Mistakes in implementations
are often my own!

Thanks much

Ed
 
J

Jag Man

Jan 1, 1970
0
1. The transistor driving the LED may not turn off completely.
You can test for this by measuring the current through the LED with
the [LED driver] transistor in its off state. Simply put a DMM set
to milliamps or microamps in series with the LED, turn the
transistor off and read the meter.


I think this was the problem. See my reply to peterken.

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.

I've wondered about that. Chris Foley proposed them and I thought why not.
Still have not tried to drive actual circuit (the ECU on my Jag), so
don't even know if the optos will provide the needed current. Since the ECU
is normally driven by a pair of tiny Hall effect transistors on a circuit
board in the distributor, I expect the 125 mA max of the optos will
suffice.
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.

Thanks. That's very kind. I'll let you know if I need them.

Ed
 
Top