Optocoupler paramethers

Question

I need to do something and don't know how to tackle the problem.

I need to use optocoupler to turn on a pnp transistor which will be able to conduct relatively high current, theoretically speaking 400mA (probably a few time higher).

Supply voltage for higher voltage side will be 12V and 3.3V for LED side of the optocoupler. If I look at the datasheet of optocoupler I see:

and

First puzzle I need to solve is next:

Does this mean that at 10mA (If) current 150mA will be able to flow through CE if the supply is capable of outputing that current?

If so is CE voltage drop negligible?

Update: Ok so now I know that I should give more attention to the graphs, but how to get appropriate IF value for desired Ic current is still a mystery to me.

Answer 1

How much current flows in the output depends on the current transfer ratio

For the MOC8021, this is 1000% minimum (at the stated conditions):

The datasheet shows this:

What this means is that at DC, 1mA in the emitter will generate a collector current in the output of 10mA at the stated conditions (If = 10mA, Vce = 5V).

The CTR varies with varying forward current as shown in this normalised graph:

To get the actual CTR, multiply the CTR by the nominal CTR from the tables.

If you are then going to drive a PNP, you need to choose a device where the DC current gain is large enough to keep the required collector current below the 150mA maximum rating.

For a high side load you get something like this:

^{simulate this circuit – Schematic created using CircuitLab}

Note that the collector emitter saturation voltage can be as high as 2V so the emitter of the PNP may be as high as 2.7V

Note that the PNP will dissipate a power of 2.7V * Ic (max)

Using a low side load gets around this:

^{simulate this circuit}

In here, when the optocoupler conducts, it pulls the PNP base down to get the device into saturation (about 0.1V Vce on the PNP), solving the power dissipation and saturation voltage issue.

Answer 2

On the datasheet you might find a curve representing the collector current as a function of the LED current. You will see that the gain of the optocoupler is less than 1, so forget 150mA on the collector with 10mA on the diode. Moreover, you can find another curve with collector - emitter voltage and collector current. On that curve you can find the CE voltage that you can expect for a given collector current. Obviously it is not guaranted that the output current is limited by the optocoupler itself, so you must assure that the absolute maximum ratings are not exceded designing the circuit consequently.