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

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

• As LED voltage increaases available voltage for Vce and V_RE decreaaee. Once the transistor saturate Vce cannot (by definition of saturation) get any smaller. Any attempt to increase current cannot work. So saturation is one extreme of the regulator. – Russell McMahon Dec 15 '19 at 6:40
• The part that I am not able to understand is that, as LED voltage increase, I understand that Vce will decrease. But how does Emitter voltage decrease. Emitter voltage is always 0.7V or 0.6V less than the base right. So, as long as the Emitter voltage stays at 2V, irrespective of the transistor whether its operating in Active or saturation, Ie=Ve/Re = 20mA (2V/100Ohms), the constant current is maintained right? – Newbie Dec 15 '19 at 6:44
• The transistor stops working as a transistor once it enters saturation. In saturation you can connecta wire from C to E with minimal difference (say use a 0.1 or 0.2V source or whatever. to model small Vsat drop) At that stage it dos not matter if you put more voltage on the base - it has no effect. Imagine that you have an LED that drops 3.9V at 100 mA and that the transistor has 0.1V Vce Vsat. | Vresistor = Vcc - V_LED - Vsat = 5 - 3.9 - 0.1 = 1v. Ires = V/R = 1V / 100 Ohm = 100 mA. The circuit is stable with 1V on Re. It does not "CARE" about Vbe if transistor is saturated at 0.1V. – Russell McMahon Dec 15 '19 at 6:52
• Ok. I get it almost. Just a simple question. I found your comment and the below answer to be contradicting in one point. When the transistor is in saturation condition, the voltage at the emitter will be 2V is what the below answer says and , you are saying that the voltage at the emitter will be 1V ? – Newbie Dec 15 '19 at 7:04
• A week ago or so I wrote about a similar problem, try read this forum.allaboutcircuits.com/threads/… – G36 Dec 15 '19 at 7:20

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 VCE goes to zero, so the Base draws more current. In your circuit that will cause greater voltage drop across RB1, 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 RE. But the higher Base current also reduces Base voltage as it draws more current through RB1, which in turn reduces the Emitter voltage. With less voltage across RE (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 VE 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.

• Thank you for the detailed answer. But I want to understand 2 things in general. Excuse if its too basic. 1. In the circuit mentioned in the question, how to calculate the base current? We don't have a base resistor, right? 2. And, when you say, "Collector current dropped to zero and emitter at 1.8V" in the last para, how did you arrive at the Emitter at 1.8V value ? Could you please explain these two questions – Newbie Dec 15 '19 at 14:44

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.

• So, what you are saying is, the current in the collector emitter side, will always be the same 2V/100ohms = 20mA even in saturation, but the forward voltage of the LED will be only 2.8V, which might be sufficient enough to turn the LED ON – Newbie Dec 15 '19 at 6:32
• The current is always decided by the value of the emitter resistor and the emitter voltage, right ? And is my understanding of the above comment, correct? – Newbie Dec 15 '19 at 6:36
• In saturation, the base voltage will actually drop a bit, but I left that effect out to simplify things. – The Photon Dec 15 '19 at 6:48
• Ok. Could you please just edit your answer to explain it simple better with the example, please? It would really help me – Newbie Dec 15 '19 at 6:51