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I am Understanding the Behavior of the following circuit with the help of simulation.

Following Circuit Gives High signal at the opto coupler output when the supply voltage goes (V3) below 13.68 V.

Now if i remove the R1 , R2 , R3 and Q2 i Still gate the same result .

then whats purpose using Q2 here if the circuit behaves same with its absence ?

Kindly help me understand the functionality of Q2 .

I believe this is not current limiter circuit for power supply ( V3 ).enter image description here

Fig 1 Actual Circuit enter image description here

Fig 2 Without R1, R2, R3 and Q2

Thank you in advance

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    \$\begingroup\$ Poor man's hysteresis loop with a weak design. \$\endgroup\$
    – D.A.S.
    Commented Nov 7, 2019 at 15:49
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 Could you please briefly explain how the hysteresis action takes place here ? \$\endgroup\$
    – Ronnie
    Commented Nov 7, 2019 at 15:58
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    \$\begingroup\$ @Rohan That's probably good enough. You should now be able to derive the following relationship:$$\begin{align*}I_\text{D}&=\frac{V_\text{B}\left(1+\frac{R_2}{R_3}\right)-V_\text{D}}{R_4+R_5}\\&\therefore\\V_\text{B}&=\left(V_\text{D}+2\,I_\text{D}\,R_4\right)\frac{R_3}{R_2+R_3}\end{align*}$$ where \$V_\text{B}\$ is the base voltage at \$Q_2\$ and \$I_\text{D}\$ is the current in the photodiode and \$V_\text{D}\$ is the photodiode voltage. Does that help you think about the \$Q_2\$ circuit at question? \$\endgroup\$
    – jonk
    Commented Nov 8, 2019 at 7:32
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    \$\begingroup\$ Read my comments again and correct spelling in bold. Generally Hysteresis is used to eliminated multiple or false triggering due to noise, ripple near the threshold. If your supply varies significantly with load this much that 15V is OK even with 2Vpp of ripple, then that’s up to your system design spec and then hysteresis may not be needed. But if it was say -25% then you might want 10% of hysteresis or Pwr OK = 18V+/- 2.4 or 15.6, 20.4V. So it all depends on your system spec. \$\endgroup\$
    – D.A.S.
    Commented Nov 9, 2019 at 18:39
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    \$\begingroup\$ @Tony Stewart Sunnyskyguy EE75 @ jonk . I understood this now. Since the hysteresis is small , even the Q2 is absent , it wont affect the circuit application in my case. \$\endgroup\$
    – Ronnie
    Commented Nov 9, 2019 at 19:56

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