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I've been working with an opto-isolated output for a high-speed camera I'm interfacing with and was a little puzzled by the circuit they were using for an opto-isolated open collector output. As far as I can tell, the NPN photo-transistor is configured in a darlington pair with two parallel NPN BJTs. I assume they've done this to increase the current the circuit can sink. According to other posts R67 provides a current path to switch off the second stage quicker and prevents leakage current turning the BJTs on.

What I can't figure out is why the designers of this circuit have included R66 in their design. Does anyone know why a base-collector resistor would be included when the BJTs should be acting as a switch? Or what the advantage of this configuration is?

Unfortunately I don't know the part-numbers or values for any components since the only schematic I have is from their documentation. Any thoughts would be helpful, I've ammended the original schematic with an external pull-up resistor and power supply.

FLIR blackfly camera opto-isolated GPIO EDIT I have improved the circuit diagram because it had been laid out weirdly. Improved figure

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    \$\begingroup\$ No Darlington pairs are used in your circuit. \$\endgroup\$
    – Andy aka
    Commented Apr 12, 2021 at 12:22
  • \$\begingroup\$ The divider R66/R67 may create voltage close to base-emmiter drop. So the switch will open faster. Check the resistors values. \$\endgroup\$
    – user263983
    Commented Apr 12, 2021 at 13:38

4 Answers 4

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Perhaps to limit maximum voltage, in case of arbitrary application. The PTCs are suggestive of short-circuit protection or current limiting function as well.

The C-B, B-E resistor divider motif is called a VBE multiplier, which draws little current for VCE less than threshold, but acts as a shunt limiter above there.

Why two PTCs are shown, in the same series loop, is also a mystery. I can't say the circuit makes a whole lot of sense in general.

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A possible reason for the odd BC resistor is so some small current is drawn when connected to the plug .This could allow plug disconnect sense now or in the future.

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Those resistors are there for maximizing speed - they prevent the transistors from saturating which would limit the bandwidth of the high-speed application. Similar techniques are exploited in bipolar comparator design.

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The voltage divider bias in this paralleled transistor circuit sets the threshold. Since you wanted to know the function of R66 & R67.

The transistor, regardless if it is paralleled like above or a single transistor, must be biased so that when the base goes in positive voltage it saturates the transistor. So basically, it insures saturation of the transistor from the input signal, by setting a threshold bias.

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    \$\begingroup\$ Usually there is no 'trolling' when someone downvotes, its a suggestion to improve answers and it is a fundamental part of SE to give feedback for answers. 1) Don't take it personally 2) Improve your answer (which you did so it looked like it worked). IMO the answer does look improved so I'll vote up. \$\endgroup\$
    – Voltage Spike ♦
    Commented Apr 12, 2021 at 16:39
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    \$\begingroup\$ In addition TonyM's first comment was actually nice, they asked nice and said 'please', I don't know how you can get nicer than that. I've cleared all the comments because they are not relevant after the edit. \$\endgroup\$
    – Voltage Spike ♦
    Commented Apr 12, 2021 at 17:01
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    \$\begingroup\$ Thanks David. That does makes sense. I guess I am still puzzled as to why you would do it this way, instead of putting a resistor in series with the base and that allows enough current flow to ensure saturation. Do you know of any good resources that might take me though how to calculate R66 and R67? \$\endgroup\$
    – JulianEE
    Commented Apr 13, 2021 at 9:25
  • \$\begingroup\$ This is sometimes used instead of a darlington when a darlington would draw too much current from the power supply. Thermistors R47 and R11 regulate the current flow. You use the correct transistor formula, instead of the bogus one academia adopted. There is no trans-conductance formulas for transistors in reality. Only the applied circuit is the trans-conductance element. That is even why the inverse formula they adapted even works in the first place. Your BetaDC is added, and the r'e is halved (because of two transistors). Use the old transistor formulas, \$\endgroup\$
    – David Mikeska
    Commented Apr 14, 2021 at 11:28
  • \$\begingroup\$ For an (NPN) BJT to be saturated, the collector voltage must be less than the base voltage. Put another way, the base voltage must be greater than the collector voltage. In this circuit, that is not possible. Therefore, the transistors will never saturate. \$\endgroup\$
    – Math Keeps Me Busy
    Commented May 24, 2023 at 11:47

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