Hi
im not the best when it comes to these types of schematics, but i think your diode is the wrong way around
The diode on the ic output to the transistor base is a LED and it is the right way round.
, and the transistor you have drawn is a NPN with your collector connected to +V ?
A NPN transistor does have its collector to a +V supply.
and you may be able to use a single input inverter instead of the IC you are using,because one of you inputs is always connected to +V
He dosn't need any invertor or 7486 if he reconfigures the photodiode circuit.
with an inverter the output is high(1) with the input been low, and the output is low (0) when the input is high (1)
This is correct.
And as for your resistor values, well the one for your base can be anything low i suppose, try 10K, 100K ?
Remember the forward voltage drop across the LED, say 1.8V and the Vbe drop in the transistor, you finish up with +5V - ( 1.8 + 0.7) =2.5V across the resistor, so a 330R or 470R would be suitable.
And for the collector, it just has to suit your emitter load i think, so in your case a LED.
I dont follow what you mean.?