Turning on an LED based on light and darkness?

I know this question has been asked a lot of times and there are a lot of tutorials but please bare with me for one second. I understand the basic concept but based on my simulated results, one thing is not matching my expectation. Please see circuit below.

I expect that when I increase the base voltage very high, the collector voltage will decrease to 0, as it does. But why doesn't this put the transistor into cutoff/saturation? Why doesn't the current decrease to zero? No matter what Vbe is, all Vce curves go to zero.

Here is a link to the VCE curve: https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html

[EDIT] Also, note that the 2N3904 has a saturation collector-emitter voltage of 0.2. So really, as soon as the voltage decreases below 0.2, it should decrease.

• Saturation doesn't mean the collector current goes to zero. It means whatever is supplying current to the collector can't provide enough current to keep the transistor forward-active. (Also think about this: if collector voltage is 0, then voltage across R1 and D1 is 5 V, so there must be current flowing through those parts. Where else can it go but through the BJT then?) Feb 27, 2016 at 5:02
• To say it another way, saturation in BJT's means the BJT doesn't control the collector current any more, instead the bias circuit (Vdd, R1, D1 in this case) controls the current. Feb 27, 2016 at 5:04
• And that is why it flattens out. Thanks a lot. Can you post an answer? Feb 27, 2016 at 5:18