Re: optocoupler trouble

In article <eaab71c3-4815-4de7-9af7-c974c8d1222d@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
ted <strnbrg59@xxxxxxxxx> wrote:
You should check the maximum diode current, too -- that 10mA is
certainly suggestive of what you ought to be able to do. Figure 10mA
in, 2mA out, means you'll need to use at least a 2500 ohm resistor to
pull the collector all the way down.

Thanks, and would you please explain why 2500 ohm is the correct
resistance, indeed why the resistance at that point makes any
difference to what the voltage is going to be at the collector pin?
(I was under the impression that when the transistor is "on", it has
no internal resistance...)

It has a *pretty low* internal resistance, as long as the current flowing
through it is less than some value. That value depends on (for a normal
transistor) the voltage/current on its base, or (for a photo-transistor)
the amount of light hitting it.

"On" and "off" are just shorthand. "On" is a short way of saying
"there's plenty of base current (or light), which means that the
transistor will allow as much current as I'm likely to give it".
"Off" is a short way of saying "there's not much base current (or
light), so the transistor will allow so little current through that
I can ignore it." In between "on" and "off", the transistor is in
what's called the "active" region, and it acts like a controllable
current-limiter.

That's what the "current transfer ratio" on the optocoupler data ***
is talking about: it's the ratio of input current (through the LED) to
output current (through the output transistor). If you put 1 mA through
the LED, it emits X amount of light; part of that light hits the
transistor; as a result the transitor allows Y amount of current ---
the ratio 1mA:Y is the current transfer ratio.


The 2500 ohms Tim Wescott mentions is because there's a limit to
how much current the LED in the optocoupler will take: if you max
it out at 10 mA (which is an educated guess at what the limit is;
the data *** will say for sure), then the transistor half will
be allowing 2 mA (that's 10 mA multiplied by the current transfer
ratio). If you can only draw 2mA of current, the resistor has to
be at least 2500 ohms in order to swing 5V (2500 ohms * 2 mA = 5V).


--
Wim Lewis <wiml@xxxxxxxx>, Seattle, WA, USA. PGP keyID 27F772C1
.