optocoupler trouble

How much current to the diode? Applying a fixed voltage to a diode will get you a wide range of currents.

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.

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.

You're only putting 400uA through it ?

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

Graham

If you don't need that use 4k7.

Graham

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

Why are you using such a low CTR ?

Graham

With 10mA LED current NOT 400uA.

Graham

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.

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).

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

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

Because with a 0.2 current transfer ratio and 10mA to the diode, the transistor will only sink 2mA. If you want to drop 5V, you need at least 2500 ohms of resistance to do it.

As pointed out elsewhere, transistors are never really "on" in every sense. They are only fully on in the sense that they are voltage limited, which depends on the circuit they're in. In this state any more drive to the transistor will not result in any more voltage drop at it's output -- but in your case that means you can't ask the transistor to sink more than 2mA.