There is a transistor characteristic called "Collector−emitter saturation voltage" (VCEsat). This is the voltage drop accross collector-emitter when the transistor is saturated. So if you bias the transistor correctly and it goes deep into saturation, the lowest voltage you can get is VCEsat. I don't know if someone ever used special circuits to drop this voltage from VCEsat to 0V, but I don't think that you really need it anyway.
Normally and talking about a general purpose transistor, you increase Ib so that VCE will decrease, until it reaches VCEsat. Here we have a phototransistor, so the only thing you could test is to increase photodiode's current and maybe the phototransistor will go deeper into saturation. But again I think that 300mV is OK, it is clearly a '0' state.
yassin.kraouch said:You are talking abour JGBT transistor but i used MOSFET transistor !!
connect the resistor will limite the current that pass throught diode12V are enough for the photodiodes to conduct, but you also need to connect ground to U2 and U3. I suggest you place electrolytic capacitors to the collectors of the phototransistors, 10uF should be OK.
If you plan to implement this circuit on the real world, then R1 consumes over 500mW, so you need at least 1W resistor. However I don't see any practical use of it, since you produce the same signal three times.
If you mean in relation to the output then yes, the current through the diode is the equivalent of a base current in a normal transistor, for each specific optocoupler you will find in the datasheet the current relation between diode and transistor to saturate the outputis there any condition related to the diod ? for example current or voltage ?
What is the voltage U1A?and the voltage U3A is there any condition ?