I have this circuit:

enter image description here

I understood the operation:

  1. On power-up Q1 and Q2 are off. There is no collector current so L1 is off.
  2. If the digital control input on the left is brought high (5 V) Q1 will turn on. Current will flow through L1, Q1 and R2.
  3. As the voltage drop across R2 increases to about 0.6 V Q2 will start to turn on and shunt some of the base current away from Q1.
  4. The result is that the circuit will settle at whatever Q1 emitter current will drop 0.6 V across R2.


  • Are both the transistors (Q1 & Q2) in the active region at all times during operation?
  • Will the load current through the load resistor and the LED will always remain constant and not exceed the limit regardless of the situation?
  • Is there any scenario when either of the transistors will go into saturation or cut-off during powered condition?

2 Answers 2


Q2 is approaching saturation as the collector is one diode drop above the base. The saturation region begins where the collector base diode becomes forward biased.

Just what constitutes forward bias in a transistor is somewhat subjective; the junction will start conducting at quite a low voltage.

If the programmed current was high enough to cause Q1 to heat up significantly and lower the base emitter voltage significantly, (the device has a negative temperature coefficient of -2.1mV per degree C) then Q2 (which would not heat up as much unless it was in thermal contact with Q1) would more closely approach the saturation region.

It is highly unlikely that would have any significant effect on the operation of the circuit unless Q1 got really hot (where you would choose a different transistor anyway).

  • \$\begingroup\$ I read some where that when Vc > Vb, the transistor is in Active region. But in the above circuit also, Vc > Vb. So, Q2 is in Active region only right? How are you saying that Q2 is approaching saturation and could you show the collector base diode of the transistor image when it becomes forward biased and what will happen to the circuit? \$\endgroup\$
    – user220456
    Oct 22, 2020 at 16:07
  • \$\begingroup\$ So, apart from the scenario you mentioned, both the transistors will always be in active region only , right? \$\endgroup\$
    – user220456
    Oct 22, 2020 at 16:43
  • \$\begingroup\$ Updated the answer regarding forward bias. Other than that I cannot see the transistors being other than in the active region provided the power sources are sufficient. \$\endgroup\$ Oct 22, 2020 at 16:56
  • \$\begingroup\$ Thank you for the answer. Accepted \$\endgroup\$
    – user220456
    Oct 22, 2020 at 16:58

If the input is at or near 0V then the transistors will be cut off regardless of the presence of the supply voltage Vbb.

If the LED forward voltage is high relative to Vbb - 0.7V then Q1 can saturate, and at some point (maybe when Vce for Q1 falls below 100mV or so) the current will no longer be regulated.

It is not possible for Q2 to saturate under normal conditions (maybe if Q1 is very leaky and R2 is very high value, but that would require strange values and operating conditions for this circuit- eg 500K and close to 200°C Tj).

BTW, the current is only roughly constant- it will vary somewhat with input voltage, supply voltage and with temperature as well as with transistor variations from unit to unit.

P.S. Please do not use "L" for an LED. That is reserved for inductors.


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