# Constant current limiting circuit

Can someone help me understand the working of this circuit:

My questions :

1. Please explain me how this circuit will work in its entire operation range.
2. I am not able to understand the purpose of the Zener diode and the 10K parallel combination. What purpose do they actually serve in this circuit?
3. The Q2101 transistor does not have a base resistor to limit its current. Isn't this circuit wrong?
• 3. where would the base current come from? Oct 22, 2020 at 17:20
• This is not a constant current source. In this circuit, the output current is a function of a Vce_Q2101 voltage (because of a Zener diode and R2102, R2110). A SOA protection?
– G36
Oct 22, 2020 at 18:43
• If the Vce_Q2101 is lower than 4.7V the Zener diode is OFF and the maximum load current will be limited to 0.7V/R2104 ≈ 47mA. But as Vce_Q2101 increases the Zener diode turns-on and starts to "inject" additional current into Q2102 the base. This lowers the Q2102 Vce voltage and the Q2101 base current a swell drop. All this will decreases the voltage drop across the sense resistor (R2104), hence lowers the maximum allowed load current.
– G36
Oct 22, 2020 at 19:12
• Thank you. But I am not understanding this - When the Zener turns ON, how does the increase base current of Q2102 lower the Vce voltage? And what do you mean by Q2101 base current a swell drop? Could you please explain this working in a little more simpler and briefly as an answer, please
– user220456
Oct 23, 2020 at 3:07
• Could you also please tell on how does the Vce of both the transistors behave during the normal operation?
– user220456
Oct 23, 2020 at 3:47

This is an interesting solution, not a constant current source. An ordinary 2.7k resistor in a given input voltage range.

It will be much more interesting to watch at lower supply voltages. The characteristic will have a phase with a negative dynamic resistance between 5 V and 14 V. It's a two-terminal device that exhibits an area of differential negative resistance much like a slow and high voltage tunnel diode.

If this is indeed the case, then in this range we can also build an oscillator with it. The oscillator already works with a supply voltage of 6V.

With a supply voltage of 12 V, we are still in the negative range and the oscillator is working.

At 17V, we already leave the negative resistance range and the 2.7k linear positive resistance phase begins.

• Could you please explain the working of the circuit and the purpose of the components and then help me understand these simulations?
– user220456
Oct 23, 2020 at 3:38
• Lovely discussion of the behavior! +1. The OP wants analysis now. But you've done enough, already, I think.
– jonk
Oct 23, 2020 at 4:06
• @jonk I would like to see your analysis of this circuit. A similar circuit was used in the audio power amplifier as SOA protection.
– G36
Oct 23, 2020 at 14:23