Connecting an 'open collector' type data pin to a DC output - result?

Hi!

Here's the problem:


DEVICE #1

.-------------.
.--||---||-----| 2.5 mm | "Tip" = 'open collector' type
< || || | stereo | data port (0...5V)
`--||---||-----| plug |
`-----. | "Ring" = DC output (5V)
T R S | |
`-+---+-' "Sleeve" = GND
| |
| |
| |//
| //
| //
| //|
|// |
// |
// |
//| |
| |
DEVICE #2 | |
.-+---+-. "Tip" = 'open collector' type
| | data port #1 (0...5V)
.-----´ |
.--||---||-----| 2.5 mm | "Ring" = 'open collector' type
< || || | stereo | data port #2 (0...5V)
`--||---||-----| plug |
`-------------´ "Sleeve" = GND
T R S


I intend to connect the devices together (for some remote
control / programming experiments.)

Coincidentally, the connectors and signals are almost identical
on both devices, but Device #2 has two 'open collector' type
data ports (on "tip" and "ring") whereas Device #1 only has
one (on "tip"), and designates the "ring" connection as
a +5V DC output instead.

The second data port is not needed for this application;
I would just like to safely disregard it.

Now, what I'd like to know is what would happen if I dared
to connect these devices together using a simple, directly
connected cable like the above? To recap, both ends have a
common ground, but one has an 'open collector' type data port
(0...5V) on the "ring" whereas the other end would connect
a 5V DC supply there. This "ring" data port is not needed
for anything, but would I fry it or short something with
this kind of connection?

* * *

The devices in question are a LANC camcorder and a TI
programmable/hackable calculator. The intended application
is to use the calculator as a wired remote control device
for the camcorder.

Of course I could simply disconnect the "ring" wire to be
on the safe side. However, I would much rather use the
standard pre-made cable that came with the calculator if
it is at all possible (since this way others could too
easily use the same program without having to build a
special cable for themselves.)

* * *

I would appreciate any ideas and insights. Electronics is
not really one of my strong points, and I have a bit hard
time figuring out how the circuit would behave in this case.

--?
znark

 

I believe OC port number 2, while "not needed for anything"
would effectively be short circuited to +5V when it did
operate. That is not good. While I do not know the equipment
you are using, the potential current through the driver would
most likely exceed 100ma, and would most likely be destructive.

At a minimum, I think you need a series resistor in the "ring"
line. Better not to connect that line through from end to end.
If OC port number 2 could be guaranteed to never operate, then
it could be connected end to end - but it may be an "accident
waiting to happen"

It could also be that you need a (seperate) pullup resistor from
"tip" to "ring" (or more exactly the +5V) signal.
10 kilohm is often good for this. Put a meter on the signal
and see if there is a stable +5V when the OC driver is OFF.
If so, no resistor rqd. If there is much less than 5V, add 10k.

Hope your project goes well.

Steve
http://www.airborn.com.au/circuits/index.html

 

Thanks for your comments. I suspected something like the above.

Actually, I could program the second device (a programmable
calculator with two open collector type I/O ports) to always
pull up the open collector data port #2 to +5V. This, to my
understanding (am I correct?), would effectively prevent
short-circuiting the +5V DC power source at the other end
of the cable to the common ground, as long as everything
else works as intended.

However, even if I can control the pin in my _own_ code, I
still cannot guarantee that it will _always_ stay at +5V.
Someone might leave the cable connected and run some other
program that deliberately pulls the same pin down to the
ground level. (Moreover, the calculator is powered by
batteries, and when they run out, there will not be any
potential left between the OC pin #2 and ground, which
- according to my limited understanding of the matters -
would translate to a short-circuit situation again.)

Yep, I think I will just have to resort to building another
cable for this purpose, and leaving the DC output line
disconnected. (Shame - it would have been nice if I could
have used the standard link cable that comes standard with
the calculator, as it is otherwise physically compatible.)

Thank you for your help.

 

Can you assure that your equipment will _never_ try to output
a low? If that's the case, (output transistor off all the time)
then it essentially doesn't matter what the input is (as long
as it's between 0 and 5).

Hope This Helps!
Rich