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I'm trying to understand the function of the transistors on the input to this optocoupler. Is this a standard/common transistor configuration (that has a name)? I'm having trouble finding a match in the searching I've done, most links are talking about compound or Darlington pairs but I don't think that's what this is as the base of the second transistor is connected back to the first? I've tried looking up standard optocoupler input circuits but this doesn't seem to be a common one. It was on an arduino board which i'd like to recreate with regular (non-smd) components as the board only had 4 such inputs and I'd like 6. Any insight appreciated.

All components are smds

U5 is labelled as L 1539 817 C - I'm struggling to find a match and datasheet for this 4 pin component

R42 is 103 (10k ohm)

R43 is 121 (120 ohm)

(I may have those backwards)

The transistors are 6CW which I think makes then 45V NPN 500mA types

transistors

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2 Answers 2

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It's a current limiting circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Here's how it works:

At steady state, Q9 will be on by the base current coming through R42.

  • The load current flows through R43 and drops a voltage across it.
  • If the voltage across R43 reaches to Vbe of Q10 (could be 0.5, 0.6, or even 0.7V) then Q10 will try to draw base current.
  • Once Q10 starts to draw a base current its collector-emitter resistance starts to decrease.
  • This decrease leads the Q9's base-emitter voltage to decrease.
  • Once Q9's base-emitter voltage starts to decrease, its collector-emitter resistance starts to increase.
  • This increase leads the load current, and therefore the drop across R43, and ultimately base-emitter voltage of Q10 to decrease.
  • Once the base-emitter voltage of Q10 starts to decrease, its collector-emitter resistance starts to increase.
  • This increase leads the Q9's base-emitter voltage to increase.
  • Once Q9's base-emitter voltage starts to increase, its collector-emitter resistance starts to decrease.
  • This decrease leads to load current to increase.
  • And the whole cycle starts afresh.

All of these happen in micro- or even nanoseconds.

Therefore the limit current can be calculated as

$$ \mathrm{I_L=\frac{V_{BE-Q9}}{R43}=\frac{0.6 \ ... \ 0.7}{R43}} $$

This type of limiters are preferred when there's no strictly defined VCC. For example, if VCC can be something between 9V and 24V in different applications then this circuit can be useful. It also brings a natural short-circuit protection. But, as Spehro stated in his answer, power dissipation for Q9 is a limiting factor.

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  • \$\begingroup\$ What's the advantage of this circuit versus replacing connecting Q10 up as a diode or replacing it with a forward-biased diode? \$\endgroup\$
    – DKNguyen
    Commented Jul 12, 2022 at 16:00
  • \$\begingroup\$ @DKNguyen replacing Q10 with a forward-biased silicon diode won't work properly. \$\endgroup\$ Commented Jul 12, 2022 at 18:58
  • \$\begingroup\$ Oh right, the two forward drops are too close. What if you used two series diodes for a 2Vf emitter follower then? \$\endgroup\$
    – DKNguyen
    Commented Jul 12, 2022 at 19:13
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    \$\begingroup\$ @DKNguyen Load current deviation will be more for different supply voltages. In the configuration above the load current will change with supply voltage as well but, relative change will be more for the other. This also brings higher losses for the BJT (not significantly, but still higher) at higher supply voltages. \$\endgroup\$ Commented Jul 12, 2022 at 19:41
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It is a constant-current source for the LED. It provides approximately 5mA (Vbe of Q9 divided by R43) to the LED for input voltages from a few volts up to tens of volts. Q9 steals base current from Q10 (which is fed via R42) when the voltage across the sense resistor R43 increased to 600-700mV.

It's not a very precise current source, but very simple and more than good enough for this application.

Here is the response for inputs from 0 to 30V:

enter image description here

I used two diodes in series to simulate the IR LED in the optocoupler, which has a Vf of about 1.2V.

At 30V in, power dissipation in Q10 is about 153mW, probably the limiting factor, especially if SOT-23 or SC-70 parts are used.

Note that this kind of circuit is common when you don't know the drive voltage (for example, for the input circuit on a general-purpose SSR). If you know the drive voltage will be (say) 5V, you can just use a series resistor and be done with it.

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    \$\begingroup\$ It also makes the circuit less vulnerable to overvoltage damage. \$\endgroup\$ Commented Jul 13, 2022 at 4:59
  • \$\begingroup\$ Wish I could accept 2 answers, thank you for taking the time to go through this - very insightful \$\endgroup\$
    – madz
    Commented Jul 27, 2022 at 10:23

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