optocoupler trouble

T

ted

Jan 1, 1970
0
I'm getting strange results from an optocoupler (aka optoisolator, and
this is the kind with transistor output). In brief: nothing I do
seems to make the transistor "conduct", except very feebly.

Here's my setup:

Diode anode: 0V/1.07V, 0.4mA
Diode cathode: ground

Transistor collector: connected to a 330 Ohm resistor which is
connected to 4.92V.
Transistor emitter: ground
Transistor base: open

When the diode anode is at 0V, the transistor collector is, as one
would expect, at 4.92V.
But when I give the diode anode 1.07V, the transistor collector drops
to just 4.88V. I want it to go down to 0V and I don't understand why
that's not happening.
 
T

ted

Jan 1, 1970
0
Thanks. I'm going to try answering your question about the datasheet,
though I'm not quite sure what I'm looking for. Is it the "DC Current
Transfer Ratio"? For that it says 20%, at I_F=10mA and V_CE=10V.
 
E

Eeyore

Jan 1, 1970
0
ted said:
I'm getting strange results from an optocoupler (aka optoisolator, and
this is the kind with transistor output). In brief: nothing I do
seems to make the transistor "conduct", except very feebly.

Here's my setup:

Diode anode: 0V/1.07V, 0.4mA

You're only putting 400uA through it ?

There's your problem. Try something more like 10-20 mA depending on the
precise device.

Graham
 
E

Eeyore

Jan 1, 1970
0
ted said:
Transistor collector: connected to a 330 Ohm resistor which is
connected to 4.92V.

330 ohms ? You're hoping for high speed operation are you ?

If you don't need that use 4k7.

Graham
 
E

Eeyore

Jan 1, 1970
0
ted said:
Thanks. I'm going to try answering your question about the datasheet,
though I'm not quite sure what I'm looking for. Is it the "DC Current
Transfer Ratio"? For that it says 20%, at I_F=10mA and V_CE=10V.

It means if you put 10mA into the LED the transistor will conduct 2mA with
its collector-emitter voltage held at 10V.

It also means (assuming the CTR hasn't wildly dropped off) if you put
400uA into the LED the transistor will conduct 80uA.

Why are you using such a low CTR ?

Graham
 
E

Eeyore

Jan 1, 1970
0
Tim said:
Yes. That says that the collector current will be 1/5 of the diode current.

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.

With 10mA LED current NOT 400uA.

Graham
 
B

Bob Eld

Jan 1, 1970
0
ted said:
I'm getting strange results from an optocoupler (aka optoisolator, and
this is the kind with transistor output). In brief: nothing I do
seems to make the transistor "conduct", except very feebly.

Here's my setup:

Diode anode: 0V/1.07V, 0.4mA
Diode cathode: ground

Transistor collector: connected to a 330 Ohm resistor which is
connected to 4.92V.
Transistor emitter: ground
Transistor base: open

When the diode anode is at 0V, the transistor collector is, as one
would expect, at 4.92V.
But when I give the diode anode 1.07V, the transistor collector drops
to just 4.88V. I want it to go down to 0V and I don't understand why
that's not happening.

The diode in an optocoupler is an Infrared LED. It takes more than 1.07
volts to turn on. You can see this with the feeble current of 400uA you
measure. Run the diode on 5 to 10 mA. The voltage should be nearly two
volts.

If this thing has to run fairly fast, place a 10 K resistor from the base of
the transistor to ground. The open base will slow operation appreciably.
 
T

ted

Jan 1, 1970
0
You should check the maximum diode current, too -- that 10mA is
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...)
 
W

Wim Lewis

Jan 1, 1970
0
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 sheet
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 sheet 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).
 
E

Eeyore

Jan 1, 1970
0
ted said:
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...)

No, it is NOT 'ON' or 'OFF' as in infinite or zero ohms.

A transistor will conduct a CURRENT determined by its input drive (in this
case the LED current).

This is fundamental basics and you clearly need to read up about how
transistors operate and how to use them.

Graham
 
E

Eeyore

Jan 1, 1970
0
Wim said:
" In between "on" and "off", the transistor is in
what's called the "active" region, and it acts like a controllable
current-limiter.

Like a 'transfer resistor' in fact from which the word trans_istor is derived.

Transistors were not originally designed as switches but as linear devices.

Graham
 
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