so does driving Vce over saturation affect the IC? How would I be able to calculate Vce mathematically, I think is is what I need to know to get my mind opened a little more
The problem with calculating VCEsat is that there are variations between produced units, and is shown by the "Min" "Max" and "Typical" values on the datasheet for the Current Transfer Ratio parameter. But just as a simple example, assume the ratio is the "Typical" value of 300, then for Ib= 1mA, the collector current will try to make 300mA. Wether or not it actually gets to that value depends on if the transistor goes into saturation or not. Let's assume a 100 ohm collector resistor, and .2V as the typical VCEsat. Further assume VCC = 30V and no LED, just to make the calculations easy:
Vc = VCC - IC*RC = 30V - .3*100 => 0V.
Since the result is less than .2V, the transistor is in saturation, and VCE will be about .2V, the collecter current is then:
IC = (30 - .2)/100 = .298 Amps.
Now, if you want ~24mA, then come at the problem from the other angle:
.024 = (VCC - .2)/RC; and rearrange:
RC = (VCC - .2)/.024 => 1.2K - use closest standard value.
Check to qualify satruation condition:
VC = VCC - 1.2K*.3 = -.06; this result is < .2V, so saturation is satisfied. If it had not been, then vary the diode current to achieve it.
Now, since we used "typical"values for Current Ratio, you have to readjust, or "fudge" the numbers to account for unit vairability. Make sure your diode current gives you lots of margin. Just use good judgement here, and do some experimenting. Good luck