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:

enter image description here


enter image description here

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.

  • \$\begingroup\$ No. It means that 150mA is the absolute maximum collector current for the optoisolator - notice "Absolute Maximum Ratings" in really big, really dark letters at the top of your image. For safe operation you will need to derate well below this threshold. You need to look at the current transfer ratio (CTR) for your opto to see how much collector current to expect at 10mA forward current on the diode. \$\endgroup\$ Jul 14, 2016 at 14:27
  • \$\begingroup\$ Please include a link to the optocoupler you're considering in your question.. \$\endgroup\$
    – The Photon
    Jul 14, 2016 at 16:01
  • \$\begingroup\$ Here it is:produktinfo.conrad.com/datenblaetter/1200000-1299999/… \$\endgroup\$ Jul 14, 2016 at 16:29

2 Answers 2


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:

MOC8020 and 8021 CTR

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:

Normalised CTR vs. If

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.

  • \$\begingroup\$ So to get the actual CTR I should multiply the normalised CTR with 1000 for Vce = 5V for any given If? \$\endgroup\$ Jul 14, 2016 at 15:25
  • \$\begingroup\$ So to get the actual CTR I should multiply the normalised CTR with 1000 for Vce = 5V for any given If? 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) emitter is meant as a IR LED? \$\endgroup\$ Jul 14, 2016 at 15:34
  • \$\begingroup\$ So at If = 10mA and Vce = 5V I would get 120mA Ice current if we look at the graph from my datasheet? If so I need to provide Vce=5V with the right setup, for example voltage divider? \$\endgroup\$ Jul 14, 2016 at 15:54

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.

  • \$\begingroup\$ This is the problem, I can't wrap my head around these graphs and see how I could get for example 100mA collector current. There has to be a way if there is absolute maximum rating of 150mA in the datasheet. By the way, I am looking at:produktinfo.conrad.com/datenblaetter/1200000-1299999/… \$\endgroup\$ Jul 14, 2016 at 14:42
  • \$\begingroup\$ As you can see from the datasheet, a current of 150mA can be obtained only with a high CE voltage, far from ideal working conditions. \$\endgroup\$
    – CasaMich
    Jul 14, 2016 at 14:56
  • \$\begingroup\$ If normalised CTR/If graph was for Vce=5V would I get approximately 80mA collector current at If=5mA? If so, how do I get 5V across Vce? And why is high Vce bad? Don't I want to be in non-saturated region (high Vce) if I want high Ic? \$\endgroup\$ Jul 14, 2016 at 18:49

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