4N33 - Opto Isolator - What to do with unused base ?
Daniel Quinz
isolated output for a photo flash unit and I am not sure
what to do with the 'base' input since I do not use it.
I drive the LED with a 470 ohm resistor to ground to turn the
photo-transistor on and this works just fine.
I am not 100% clear on how the light from the LED and externally applied
current to the base interact. Do they both, independantly,
influence the collector current (by superposition)?
I have tried reading about the opto-coupler in "The Art of Electronics"
by Horowith & Hill. They mention that the transistor can be
used as a photo-diode or phototransistor, but they do not go into detail
as to how the base relates to the LED effect.
Any suggestions would be welcome.
Regards,
Daniel Quinz, P. Eng.
Mironetics CES
e-mail: da...@micronetics-ces.com
Jack Schidt
you need it. Also improves very low current transfer characteristics. In
on/off applications, with adequate LED current, it is usually unconnected.
If the circuit is working, apply Jack's Law:
'If the damn thing works, leave it alone'.
Best wishes
JS.
Daniel Quinz wrote in message <36523B...@micronetics-ces.com>...
OptoEngr
resistor from base to emitter. This is essential if you want to switch the
transistor on and off at high rates, as in data transmission. This
resistor especially effects turn-off time by depleting stored charge that
keeps the transistor on for a while after you turn off the LED. However,
your application does not require fast turn off time at all, so you probably
don't need the resistor.
BTW, I don't subscribe to 'Jack's Law'. I believe that you should try to
understand how much margin your circuit has before deciding to leave it
alone.
Paul Mathews
Aengineering Co.
non-contact sensor design expertise
see the LED FAQ at: http://www2.whidbey.net/optoinfo/LED_FAQ.html
Jack Schidt
Thanks for your reply.
I believe I just might understand this.
He gets a usable output with a 470 Ohm in series with the LED, and is
working at a low frequency, and/or driving a high load impedance.
This was your conclusion as well.
Other than reducing the drive current by increasing the LED resistance
value, which you did not suggest, how do you 'understand how much margin
your circuit has'?
He and you still do not 'understand how much margin your circuit has', and
likely do not need to.
Just trying to inject a bit of humor.
Loosen up.
Richard Kilsby
proxy.airnews.net>, Jack Schidt <ja...@wintel.net> writes
>Paul,
>
>Thanks for your reply.
>
>I believe I just might understand this.
>
>He gets a usable output with a 470 Ohm in series with the LED, and is
>working at a low frequency, and/or driving a high load impedance.
>
>This was your conclusion as well.
>
>Other than reducing the drive current by increasing the LED resistance
>value, which you did not suggest, how do you 'understand how much margin
>your circuit has'?
>
>He and you still do not 'understand how much margin your circuit has', and
>likely do not need to.
>
>Just trying to inject a bit of humor.
>
>Loosen up.
>
Problem is that optos also " loosen up " ;-)
OK not in this application but when optos are "normally on" then the
current transfer ratio will degrade over time. ( years )
Fitting a base emitter resistor effectively pre degrades the device so
that it will continue to work over the years. It will also reduce the
CTR spread amongst otherwise identical parts.
I hope I have got a good sense of humour but I don't subscribe to Jack's
Law either. I read the spec sheets instead.
I also have a good chuckle when I come across one of Jack's circuits
that worked at one time.
Rick Kilsby
Jack Schidt
Agreed. I have several optocoupler data books as well. I saw no problem.
This is also an extremely common LED current for an extremely common
optocoupler.
Point was, the prior poster brayed about dangers, then offered no creative
suggestions.
And you followed suit. Exactly.
I do not believe this three component circuit was my design.
Was that your impression?
JS.
OptoEngr
It seems to me that if you can spout off about 'Jack's Laws', I should be
able to make generalities in response. 'Creative' is not what I would call
your suggestion to, in effect, leave well enough alone.
I also have a hard time imagining anyone making any use of your comments
about 'biasing in the linear mode'.
However, for details about how to design with optocouplers and analyze the
margins, there are many available app notes on the web, including several
you can find here:
http://www.sci.siemens.com/opto/
Paul Mathews
Aengineering Co.
non-contact sensor design expertise
see the LED FAQ at: http://www2.whidbey.net/optoinfo/LED_FAQ.html
Jack Schidt wrote in message ...
Richard Kilsby
>
>Agreed. I have several optocoupler data books as well. I saw no problem.
>This is also an extremely common LED current for an extremely common
>optocoupler.
>
>Point was, the prior poster brayed about dangers, then offered no creative
>suggestions.
>
>And you followed suit. Exactly.
>
>I do not believe this three component circuit was my design.
>
>Was that your impression?
Whoa there Jack.
All I put in was what I believe to be the most common use for the base
connection. ie to normalize the CTR. This was missing from the thread.
The opto design was not yours.
I certainly did not wish to impugn any of your designs or circuits.
My reference to "Jack's circuits" was meant to be a poke at all those
designs (and I have done them too) where they work straight out of the
box only to come back and bite you later.
It is almost as if it is best for the circuit not to work properly first
time and then you have to go right into the finest detail to get it to
go. That way you know the darn thing will work for ever.
Hope I didnt dent your sense of humor.
Rick Kilsby
Jack Schidt
Agreed and acknowledged on the base connection details. I saw no need to
drag every possible nuance into forum concerning an occasional on/off
through an optocoupler.
This is not a 1000 KB/S NSC link, and does not need that level of treatment.
Even now, the original poster's head is swimming- and that is who's question
I was responding to. If he is still awake - 'Hey- it will work without the
base resistor!'.
I have in fact been haunted by recreating designs that myself, and worst
yet, others have rigged into functionality. Physics is a tough customer.
Call me any time on a design flaw, And i will learn from it.
Like all, I make mistakes (especially in typing ASCII schematics). I do not
take the time to drag out every reference I own, and enumerate every detail
on every aspect of every application of every question on every UseNet post.
I charge for that sort of thing.
No offense taken. Thanks again for responding.
Nothing wrong with a bit of exchange.
Have a good day.
JS.
Daniel Quinz
>
> Rick,
>
<snip>
> Even now, the original poster's head is swimming- and that is who's question
> I was responding to. If he is still awake - 'Hey- it will work without the
> base resistor!'.
Yup! Definitely awake and certainly enjoying the swim :)
As for it working without the base resistor, I knew it would.
What I was really concerned about was what effect ambient electrical
noise would have on that "floating" base input?
I understand well how conventional transistors are used (I studied them
a while ago in university) but I have never really had to work
up close with a phototransistor before and am not very clear on how the
external base line is supposed to be used.
I want to make sure that the product does not suffer from "ghost"
behaviour once it goes out the door because of a floating base input.
Up until now, I still don't have a clear vision of how the base input
interacts with the LED effect.
I really don't like building a circuit, let alone shipping one, that I
don't understand 100%.
Regards,
Daniel Quinz, P. Eng.
Micronetics CES
e-mail: da...@micronetics-ces.com
Bruce Nepple
>interacts with the LED effect.
What!! Are you saying that after all these great responses you still don't
know the answer to your fine question? Well, that's probably because
everyone is so busy dispensing advice and being top dog that they don't
bother to explain. Happens a lot here in s.e.d.
I'm not an optical guy (I do 1's and 0's, in verilog if I can get away with
it), but my understanding is that the LED "shoots" (that should push
someone's button :>)) photons into the base region, causing current flow in
the base the same as if you injected a current through a pullup resistor.
(Or, you can think of it it as a photo diode from the base to the collector)
Another path that might add current is Collector Base leakage (which, in
some applications is very significant). Either of these will create a
voltage at the base terminal, which , as you probably know, will be around
0.6-0.7 volts when the transistor is on. Adding a resistor from the base to
ground will divert some of that current making it less sensitive to light,
less sensitive to leakage, and faster since the effective base capacitance
has a discharge path (as was stated earlier)
That's how they interact. The question now is: How to determine how much
current is coming from the LED, what resistor makes sense for your
application, and what margins you have. I have to go home now, but if no
one gives you an answer I'll take a shot at it tomorrow. (besides, I
couldn't find a single optoisolator datasheet so I could see how they spec
it.)
As an experiment, you might measure the beta of the transistor with a pullup
resistor and no light. Then remove the pullup an see what amount of light
it takes to turn the transistor on the same amount. Then add a pulldown and
see how much more light it takes. Then you could equate all this with the
spec sheet and make a design decision.
Bruce
Daniel Quinz wrote in message <365378...@micronetics-ces.com>...
>Jack Schidt wrote:
>>
>> Rick,
>>
>
><snip>
>
>> Even now, the original poster's head is swimming- and that is who's
question
>> I was responding to. If he is still awake - 'Hey- it will work without
the
>> base resistor!'.
>
>Yup! Definitely awake and certainly enjoying the swim :)
>
>As for it working without the base resistor, I knew it would.
>
>What I was really concerned about was what effect ambient electrical
>noise would have on that "floating" base input?
>
>I understand well how conventional transistors are used (I studied them
>a while ago in university) but I have never really had to work
>up close with a phototransistor before and am not very clear on how the
>external base line is supposed to be used.
>
>I want to make sure that the product does not suffer from "ghost"
>behaviour once it goes out the door because of a floating base input.
>
>
Jack Schidt
consider a resistor across the junction. Don't forget the long-term effects
of having this LED turned on ( a whopping 8 mA ) continuously.
Scientifically, I reiterate:
"Use a larger LED limiting resistor", and make sure (50%- this is new, a
good figure) margin exists.
If you still have questions, you should always check with the component
manufacturer first, and believe UseNet last.
JS.
Jack Schidt
Charles Mosher
: I recently wired a test circuit using a 4N33 opto isolator to provide an
: isolated output for a photo flash unit and I am not sure
: what to do with the 'base' input since I do not use it.
: I drive the LED with a 470 ohm resistor to ground to turn the
: photo-transistor on and this works just fine.
: I am not 100% clear on how the light from the LED and externally applied
: current to the base interact. Do they both, independantly,
: influence the collector current (by superposition)?
: I have tried reading about the opto-coupler in "The Art of Electronics"
: by Horowith & Hill. They mention that the transistor can be
: used as a photo-diode or phototransistor, but they do not go into detail
: as to how the base relates to the LED effect.
: Any suggestions would be welcome.
: Regards,
: Daniel Quinz, P. Eng.
: Mironetics CES
: e-mail: da...@micronetics-ces.com
Most opto-isolators have rather high transistor gain in the
phototransistor, perhaps around 1000, so if the overall ctr were to be 100
%, ie, 1:1, then there is only about 10^-3 ctr from LED to photo current.
The function of the base to emitter resistor is to steal some of this
photo current, so as to make the curve more NON-linear, and to introduce
some tolerance of common mode capacitive coupling (most have perhaps 1/4
pF of such coupling, even when mounted on a circuit board with guard
traces, and this coupling is present no matter how you choose to amplify
the photocurrent), so it is not prudent to ignore this. This resistor may
however introduce larger apparent gain variations from unit to unit, so if
the device is not characterized for use with a particular value, you may
have to investigate this.
Let's see if I can draw some ASCII art illustrating this (I will not use
tabs for this, so it should come through right):
Ic out
| /
| /
| /
| /
| /
+---------------------------------- I led in
^
|
corresponds to stolen photocurrent
This stolen base current greatly speeds transistor turnoff, and for
digital applications, is highly recommended. Sometimes one can choose a
value that permits stuffing a wider range of components than leaving the
base open.
If you need more bandwidth than you can get this way, I recommend shorting
the base to the emitter, and driving the photodiode (at zero bias to make
bypassing easy -- namely, none required) straight into an op amp input,
with perhaps a 1 M feedback resistor, and an adjustable feedback capacitor
of several pF, used as a peaking control. These photodiodes have
capacities of the order of tens of pF, and hence this capacity can not be
neglected in the behavior of the circuit. I typically get rise times of a
half microsecond in this way using a LF347.
Hope this is not too much detail for you.
--
Charles Mosher
ratr...@svpal.org
Jack Schidt
I say NO. You said no earlier, as well. You are welcome to disagree now.
Do we need a new law, Paul's law:
"Where everything irrelevant has a place, and nothing is decided."? (SAFE)
You tell me, and those reading!
OptoEngr
precision required in the timing, and the temperature range of operation, I
don't really know for sure. However, I suspect that the answer is NO. As I
wrote in my first posting:
....However,
your application does not require fast turn off time at all, so you probably
don't need the resistor....
This seems clear enough to me. I'll try to look for the humor in Jack's
future postings.
Paul Mathews
Aengineering Co.
non-contact sensor design expertise
see the LED FAQ at: http://www2.whidbey.net/optoinfo/LED_FAQ.html
Jack Schidt wrote in message
<5FF318F058143335.9CB02AB2...@library-proxy.airnews.ne
t>...
Daniel Quinz
I understand things much clearer now.
I went to the Seimens website and donwloaded an application note on
opto-couplers. The note described pretty much everyhing
you summarized in your post, but with lots of schematics.
I appreciate your (as well as everones elses) input.
I now understand how this device works and am confident that my circuit
design is valid.
Regards,
Daniel Quinz, P. Eng.
Micronetics CES
e-mail: da...@micronetics-ces.com
Daniel Quinz
>
> Hope this is not too much detail for you.
>
> --
> Charles Mosher
> ratr...@svpal.org
Not at all! It's very helpful!
Thank you kindly for the feedback!
John Woodgate
>take the time to drag out every reference I own, and enumerate every detail
>on every aspect of every application of every question on every UseNet post.
The trouble is, some people demand no less, and usually even more. See
my .sig!
--
Regards, John Woodgate, Phone +44 (0)1268 747839 Fax +44 (0)1268 777124.
OOO - Own Opinions Only. You can fool all of the people some of the time, but
you can't please some of the people any of the time.
Jack Schidt
>....However,
>your application does not require fast turn off time at all, so you
probably
>don't need the resistor....
>
That would be the case in a Photoflash circuit. The whopping 8 mA of LED
excitation current (aperiodic, and sub-audio frequency) does not cause me to
lose sleep either.
Again, perhaps you could loosen up. Perhaps a little.
LASCR's were used for years, and they were Gawdaful slow.
Thanks again for your input.
JS.