What happens when transistor operates in saturation region for constant current source

Question

I am learning about the constant current sources and I found this video

What I dont understand is that, when the forward voltage of the LED is increased from 1.7V to 3.3V, and our Emitter voltage is maintained at 2V, the voltage across the Collector Emitter Junction is decreased. This makes the transistor to work in the saturation region.

Why does the transistor in saturation region does not help in providing the constant current. In all the cases mentioned in the screenshot, Ie=Ve/R which is always 2V/100ohms, 20mA. Since, in all cases (different LED Forward voltage and same Vdd supply voltage), we are getting the same load current. So, what's wrong with the transistor operating in saturation region in this case?

Please clarify if I am wrong

Answer 1

when the forward voltage of the LED is increased from 1.7V to 3.3V, and our Emitter voltage is maintained at 2V...

In saturation β reduces towards zero as V_CE goes to zero, so the Base draws more current. In your circuit that will cause greater voltage drop across R_B1, so the Emitter voltage won't be maintained at 2V. Instead it will go down, causing the Collector current to also go down. The LED will get 3.3V, but at lower current.

Here's the result of simulating your circuit in LTspice, with LED voltage varying from 1.7V to 4.5V:-

Between 1.7V and 2.9V the transistor does a good job of keeping the LED current (red line) constant.

Above 2.9V the the transistor goes into saturation as the voltage between the Collector (green line) and Emitter (blue line) drops below 0.2V, causing β to reduce and requiring more Base current (magenta line) to maintain current through R_E. But the higher Base current also reduces Base voltage as it draws more current through R_B1, which in turn reduces the Emitter voltage. With less voltage across R_E (as well as more current coming from the Base instead of from the Collector) the Collector current also reduces. In this region the transistor is acting more like a resistor than a constant current source.

If the voltage divider was 'stiffer' and held the Base voltage constant despite the increased Base current then the transistor would go into hard saturation, with the Base supplying enough Emitter current to keep V_E close to 2V even if the Collector current dropped to zero. With a 5V supply and the Emitter at 1.8V there would not be sufficient voltage left to light a 3.3V LED.

Answer 2

What's missing from your description is that \$V_{CE}\$ won't actually go to -0.3 V. If it did, that would mean the transistor is generating power and delivering it to the circuit somehow.

Instead, \$V_{CE}\$ will be stuck at the saturation voltage, typically taken as ~0.2 V or so. That means the cathode of the LED will be at ~2.2 V, giving only 2.8 V across the LED, so it will not produce much current, and won't light up visibly.