Curiosity: NPN Opto Coupler with no base

[bxdobs]

Even though a discrete NPN transistor can be easily biased so the emitter Voltage is above ground potential ... it isn't clear how this is supposed to work for most opto couplers when they don't provide a base Connection. The device I was using is for this test is a Fairchild FOB817A

From my experience the typical usage for an NPN opto coupler is just as an open collector switch to ground with emitter ground ... the other day I thought I should be able to offset the Emitter voltage by just adding a resistor between the emitter and ground and still keeping one on the collector to Rail ... I randomly selected both resistors to be 1K ... the expectation was that once the transistor was turned on by the opto led that I should have a voltage at the emitter of around 2 volts instead this circuit would only give me a Ve of .6 volts ... this wouldn't change regardless of trying several combinations of the 2 resistors in the 1K or less range. 1000 to 2000 ohms at 5V is a max of 5 mA which is well below the 50mA max for this device (Note the Led currents for testing were 5mA and 2.5mA)

The expected result based on:: Ie = Ic + Ib ... if we assume Ib is zero then Ic should equal Ie ... with Rc and Re equaling 1000 then Ve = (Vcc - Vb) / (Rc + Re) * Re = (5 - .6) / (1000 + 1000) * 1000 = 2.2 V

Looking at some online descriptions of Photo Transistor this "failure" in my mind was due to the value of resistors I chose ... I reran the same test with 2 10 K resistors and this provided the desired result.

My question is how can one model this design without doing a trial and error approach ... there must be some formula that will predict what values to use if one wants to have an effective offset Ve. ... what parameter in the spec sheet https://www.fairchildsemi.com/datasheets/FO/FOD814A.pdf will predict this behaviour?

The practical application of this Circuit is to directly drive an N Power Mosfet instead of having to add another (PNP) Transistor

 

[Bluejets]

Think about how the base current originates.
Perhaps you will see an answer to your question.

 

[KrisBlueNZ]

You're right that the current into the collector pin and the current out of the emitter pin should be (will be) equal. Anything else would mean that the optoisolator is not properly isolated.

So if you have a 1 kΩ collector resistor and a 1 kΩ emitter resistor, you should expect toget the same voltage across each one. Assuming the phototransistor is receiving enough light, that should mean at least 2V at the emitter, and less than 3V at the collector, assuming a 5V supply.

If you got the expected result using 10k resistors but not using 1k resistors, your phototransistor was not conducting heavily enough. You can confirm this by measuring the collector voltage. You will see the same voltage drop across each 1k resistor. So if you were getting 0.6V at the emitter, you will have 4.4V at the collector, and there will be 3.8V across the optocoupler transistor. You can also try using a higher LED current to confirm that this was the problem.

You can avoid the inversion stage and drive a MOSFET gate directly from the emitter of an optocoupler. Especially if it's a fast optocoupler, or if the MOSFET doesn't need to switch quickly. If you tell us more about the project we can give more specific advice.

 

[Laplace]

Use the characteristic curves for the transistor and draw the load line. Vce=5 when Ic=0. For Rc+Re=2K, Ic=2.5mA when Vce=0. That is the top load line (in blue) on the diagram. For Rc+Re=20K, Ic=0.25mA when Vce=0. That is the bottom load line (in blue) on the diagram. Because If<5mA the operating region is highlighted in yellow. This is the region you are trying to analyze. My only question is, Why?