The circuit I made basically looks like this:
http://embedded-lab.com/blog/wp-content/uploads/2011/03/Lab11_Circuit_SevenSegmentMultiplexing.jpgI want to get 10mA per segment. This would be 70mA for a 7-segment display as I am not using the decimal point. I have left the decimal point floating which should be acceptable and not cause latch up or any other anamoly since this is not a microelectronic IC.
Assuming I aim to get 100mA max Ic, this would be no less than 10mA of Ib for saturation, which does seem a lot. If I do use Rb, this would be R=V/I=(3.3-0.6)/0.001=2.7kohm or less. For double the base current, this shall become 1.35kohm. With 0.4V VCE as per the datasheet, then 0.6V across the LED itself, we have 3.3-0.6-0.4=2.3V across the Rs of the LED segment. This gives Rs=2.3/0.001=230ohm.
Now, with each segment having Rs=230ohm, when the transistor is in saturation mode, it should only be able to force 10mA per segment, thus when more segments are on, the Ic will increase to accomodate more segments, 10mA with just one segment and 20mA with two segments and so on. Thus, the Ib can remain same and put transistor in saturation, but the actual Ic will be limited by the Rs and how many segments are being activated, each segment will only let 10mA through due to Rs.
In other words, it is not that the transistor will "force" the 100mA (calculations for saturation) of Ic resulting from 10mA base current through regardless of how many segments are on. It will not happen that if one segment is on, 100mA will be forced through it, and if two segments are on then 50mA will be forced through each. No. Because of limit imposed by Rs.
I hope that my above understanding is correct. Correct me if I am wrong please. Thank you very very very very much, Ian.M said that the VCE rises sharply once the Ic becomes more than 50mA, I am not sure how exactly that is as per the datasheet.
And another transistor is perhaps better to get even though when multiplexing the microcontroller will only drive 10mA from one pin for a period of time. I will consider FET as well which I have not done yet. I shall think about it.