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Driving a line from an opto-isolator

Hello,

I have a board with a 74HCT573 with 10K pull-ups on the inputs. This is
connected to a sensor with an opto-isolator output (4N25) via 2m of
cable. The resistor for the opto led is 10K on a 5V supply, so
according to datasheet the CTR can be as low as 20% at that drive
level. So the transistor output could be pulling as little as 80uA. I
can see them asking me to increase my 10K pull-ups by a factor of >
6.25.

What is the limiting factor here? Surely you cannot keep reducing the
current and expect the cct to work. Even the 500uA seems low to me.

regards

IanM
 
J

John Popelish

Jan 1, 1970
0
Hello,

I have a board with a 74HCT573 with 10K pull-ups on the inputs. This is
connected to a sensor with an opto-isolator output (4N25) via 2m of
cable. The resistor for the opto led is 10K on a 5V supply, so
according to datasheet the CTR can be as low as 20% at that drive
level. So the transistor output could be pulling as little as 80uA. I
can see them asking me to increase my 10K pull-ups by a factor of >
6.25.

What is the limiting factor here? Surely you cannot keep reducing the
current and expect the cct to work. Even the 500uA seems low to me.

The minimum current that can drive the CMOS as a pull up
depends on the leakage current into the COS, and the off
state leakage current of the opto. But it is possible that
you cannot get very close to this limit, because there is a
speed penalty to pay when the currents are small. The node
capacitances still need to be charged up by the current.
And that takes time.
 
John said:
The minimum current that can drive the CMOS as a pull up
depends on the leakage current into the COS, and the off
state leakage current of the opto. But it is possible that
you cannot get very close to this limit, because there is a
speed penalty to pay when the currents are small. The node
capacitances still need to be charged up by the current.
And that takes time.

Thanks John,

that makes sense

Ian
 
C

Chris

Jan 1, 1970
0
Hello,

I have a board with a 74HCT573 with 10K pull-ups on the inputs. This is
connected to a sensor with an opto-isolator output (4N25) via 2m of
cable. The resistor for the opto led is 10K on a 5V supply, so
according to datasheet the CTR can be as low as 20% at that drive
level. So the transistor output could be pulling as little as 80uA. I
can see them asking me to increase my 10K pull-ups by a factor of >
6.25.

What is the limiting factor here? Surely you cannot keep reducing the
current and expect the cct to work. Even the 500uA seems low to me.

regards

IanM

Hi, Ian. You're right -- 500uA of LED current seems really small for
an optoisolator. And the current transfer ratio for the optoisolator
is specified at one level of drive current (typically If = 10mA).
You'll also notice when looking at the datasheet:

http://www.fairchildsemi.com/ds/4N/4N25-M.pdf

that current transfer ration (CTR) goes way down when If goes below a
couple of mA (see Fig. 3 and Fig. 4, p.6 of the datasheet).

If I were looking at this, I'd suggest that the HCT573 can easily drive
more than 1/2mA, and I'd look at bumping up that current by reducing
the series resistor from the HCT573 to the opto input. The data sheet
says it can drive 15 ea. LSTTL loads, so it can sink (or source) 6mA.
Assuming you're sinking current to turn on the opto, have a 5V supply,
and the opto LED takes up 1.7V when on, you should be able to use a 560
ohm series resistor quite easily.

http://www.onsemi.com/pub/Collateral/MC74HCT573A-D.PDF

You're asking a really intelligent question. The answer is in the data
sheets.

Good luck
Chris
 
Hi Chris,

I wasnt clear in my original post. There are 2 boards involved here
with 2m of cable inbetween. The 74HCT573 with 10k pull-ups is on my
board (it is the one I have control over - but not my design) and the
opto-isolator is on the other board (I have no control over this). The
500uA referred to is the current required to pull down my boards
inputs, and I mentioned I thought this was on the low side. As the
opto-isolator is being under-driven it cannot sink this current, so I
can see them asking me to increase the pull-up value. This seems wrong
to me, hence my question about limits.


regards

Ian
 
J

Jonathan Kirwan

Jan 1, 1970
0
I wasnt clear in my original post. There are 2 boards involved here
with 2m of cable inbetween. The 74HCT573 with 10k pull-ups is on my
board (it is the one I have control over - but not my design) and the
opto-isolator is on the other board (I have no control over this). The
500uA referred to is the current required to pull down my boards
inputs, and I mentioned I thought this was on the low side. As the
opto-isolator is being under-driven it cannot sink this current, so I
can see them asking me to increase the pull-up value. This seems wrong
to me, hence my question about limits.

If the opto cannot be expected to sink more than 80uA (you haven't
disclosed the rates required and that is important to know, along with
the details, I suppose, about what composes the 2m of cabling
impedance), you need to be sure that your end can work correctly with
that much capability. Your inputs require at least 500uA, I gather,
mostly because of the existing 10k pull-ups on your end (5v/10k =
500uA, and ignoring any HCT input requirements.) That simply needs to
change.

You could increase the pull-ups on your end and yes that may very well
impact the supportable rates. If so, you need to find another way
than just a simple modification of a pull-up to get the job done, I'd
imagine. Modifying a simple passive isn't necessarily the right
solution, if you are trapped by other requirements, and you should be
prepared to find a route that is correct for the application and not
simply limited to finding a resistor value you aren't comfortable
with. In other words, perhaps consider using an active front end.

John's point about the node capacitance is important and this will
include everything from the 4N25 transistor itself, though the cable,
and onto your input circuit, whatever that is. The 4N25 will have to
drive it all with what it has in place at rates that work for you. You
may need to think closely about this.

I gather that the 10k pull-up on your side is the only explicit
pull-up that the 4N25 output transistor sinks against and that you
don't have access to the transistor base at the 4N25? My guess is
that you aren't doing rates much in excess of 1kHz (and I'm wondering
about meeting your HCT input thresholds as it is.) Increasing that
resistor will probably demolish your rate, my guess. If you can only
expect about 80uA, then you might consider using a PNP on your end
before the HCT input, with appropriate resistors. I think you should
be able to acheive the same rates on that current. Also, perhaps,
consider a current steering arrangement using two BJTs.

Jon
 
J

Jonathan Kirwan

Jan 1, 1970
0
<snip>
Also, perhaps,
consider a current steering arrangement using two BJTs.

Actually, now that I think about it, you probably don't need a BJT to
develop the steered current source. Just your larger valued pull-up
resistor (say, 56k?) So one cheap NPN BJT, your increased value
pull-up, and a collector resistor for the NPN (6.8k-10k?) would be
enough, I'd imagine.

Jon
 
Jonathan said:
Actually, now that I think about it, you probably don't need a BJT to
develop the steered current source. Just your larger valued pull-up
resistor (say, 56k?) So one cheap NPN BJT, your increased value
pull-up, and a collector resistor for the NPN (6.8k-10k?) would be
enough, I'd imagine.

Jon

Hi Jon,

You have lost me here, I understood the PNP method, but not this.
Rates are not an issue by the way (apart from 500ns max rise/fall time
into 74HCT573), you can consider it DC.

I am going to try to get the manufacturer of the sensor to put a
reasonable drive current into the emitter of the opto-coupler!!
Otherwise it is a mod to our board similar to your description - nice
mod, all 8 inputs are used!

regards,

Ian
 
J

Jonathan Kirwan

Jan 1, 1970
0
Hi Jon,

You have lost me here, I understood the PNP method, but not this.

Probably my fault. I'm just a hobbyist and see things in my own way,
using words that may not be the best to use.

What I meant is simply this:
: +5
: +5 |
: | |
: | \
: | / R1
: \ \ 10k
: / R2 /
: \ 56k |
: / +-----> HCT input
: | |
: | |
: | |/c Q1
:from OPTO-+----| NPN
: |>e
: |
: |
: gnd

In the above case, the pull-up is now 56k, which is about the current
you are talking about, I think. Since the NPN base and opto collector
node itself stays within 6/10ths of a volt of ground, the current
through R2 above stays roughly the same. But the current either is
sunk by the opto's collector or else by the NPN base-emitter. In
other words, the current doesn't change much but is 'steered' one way
or another.
Rates are not an issue by the way (apart from 500ns max rise/fall time
into 74HCT573), you can consider it DC.

Slow. That's good. That 56k or 68k pull-up on your side, if the
rates are truly close to DC (under 100Hz or so) and you don't have
much voltage pickup on the cable, might be okay too.

Or either the PNP being pulled active or else the NPN steering
arrangement will be fine, too, I think. But I can see that you want
the least mod possible. And under the circumstances, just increasing
your pull-up to 56k seems fine.
I am going to try to get the manufacturer of the sensor to put a
reasonable drive current into the emitter of the opto-coupler!!
Otherwise it is a mod to our board similar to your description - nice
mod, all 8 inputs are used!

Ah... design push-back!

Well, I'm not so sure you need to mod it that much, given your comment
about the signal being close to DC. Probably, you _can_ get by with
just a resistor change, though I've no idea what your 2m of cable
winds though and may pick up along the way.

Jon
 
C

Chris

Jan 1, 1970
0
Hi Chris,

I wasnt clear in my original post. There are 2 boards involved here
with 2m of cable inbetween. The 74HCT573 with 10k pull-ups is on my
board (it is the one I have control over - but not my design) and the
opto-isolator is on the other board (I have no control over this). The
500uA referred to is the current required to pull down my boards
inputs, and I mentioned I thought this was on the low side. As the
opto-isolator is being under-driven it cannot sink this current, so I
can see them asking me to increase the pull-up value. This seems wrong
to me, hence my question about limits.


regards

Ian

Hi, Ian. Of course, as you increase the pullup resistance, you have
more susceptibility to noise. Your vendor has put you between a rock
and a hard place by choosing the 1960s refugee 4N25. There are better,
faster optocoupler transistors with much higher CTRs.

In addition to Mr. Kirwan's great advice, I'd like to add that you may
wish to use a 74HC input instead of an HCT. The input will be much
more tolerant of a not-quite-saturated 4N25 (Vl(max) = 0.3*Vcc or 1.5V
instead of 0.8V). That might help a bit.

Also, expecting fast enough transition times out of the 4N25
(especially when turning off) might be a little much. You may want to
have a schmitt trigger buffer at the input (like a 74HC14), then you
can use a 10K pullup along with a small (100pF) cap to GND, which will
help a lot with noise immunity. This shouldn't slow down your input
more than a microsecond, which you imply in your conversation with Mr.
Kirwan is probably OK. Your input could then look something like this
(view in fixed font or M$ Notepad):

| VCC
| +
| |
| .-.
| | |10K
| | |
| '-'1/6 74HC14
| <2m cable> | |\
| .--------------o----| >O->
| | | | |/ To HC(T)573
| V ~ |/ ---
| - ~ -| 4N25 ---100pF
| | |> |
| | ===
| | GND
| | <2m cable>
| '--------------.
| |
| ===
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Sorry about misreading your earlier post.

Good luck
Chris
 
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