I need to generate a current signal between 12V and GND, which has to have galvanic isolation. For this, I want to use an optocoupler.

  • Maximum permittable rise / fall time: < 1µs
  • Output side Vcc: 12V-14V
  • Output current: > 28 mA

Because of the high output current, an external transistor is required, since most optocouplers that are fast enough are limited at around 16 mA. This allows the use of open collector optocouplers, driving a PNP transistor.

The PNP transistor then pulls a line to ground, which bypasses a resistor, increasing the current signal.

Can I use this open collector output to drive a PNP transistor (with an appropriate base resistor) without using a dedicated pull-up resistor?

There are also biased PNP transistors which include a pullup, but those are mostly (only?) available as SMD parts, while I would like to keep using THT.

  • 1
    \$\begingroup\$ OnSemi's FJN430xR are THT. \$\endgroup\$
    – CL.
    Commented Jun 12, 2019 at 13:44
  • \$\begingroup\$ It depends entirely on what you are hoping to achieve with regards to performance and functionality and also depends on power rails, none of which you have told us about. \$\endgroup\$
    – Andy aka
    Commented Jun 12, 2019 at 13:44
  • \$\begingroup\$ @CL. Those look very promising, thank you. \$\endgroup\$
    – towe
    Commented Jun 12, 2019 at 13:54
  • \$\begingroup\$ @Andyaka I expanded on that a bit, sorry. \$\endgroup\$
    – towe
    Commented Jun 12, 2019 at 13:54
  • \$\begingroup\$ It's normally an NPN transistor that pulls a line to ground. \$\endgroup\$
    – Andy aka
    Commented Jun 12, 2019 at 14:23

1 Answer 1


For a project requiring galvanic isolation, I would like to use an optocoupler.

Several fast optocouplers are available that offer an open collector output.

Can I use this output to drive a PNP transistor (with an appropriate base resistor) without using a pull-up resistor?

The PNP transistor and base drive resistor effectively ARE the pullup.

The only addition which MAY be useful is an extra resistor from PNP base to emitter (V+ usually).
This passes base leakage current when the transistor is off. In most cases it is not needed. In some cases it is possible for the PNP to turn on partially without this resistor.

Along these lines.
R1 optional as above.
R2 value to suit application.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you for your answer, the PNP to turn on partially part is what I was worried about, and R1 is what I hoped to omit. Can you expand on when this resistor would be required? I suppose some minimal charge would pass from Collector to Base, energizing the Base, which in turn turns on the transistor - albeit only slightly? \$\endgroup\$
    – towe
    Commented Jun 12, 2019 at 13:57
  • \$\begingroup\$ @towe what are : PNP load current / resistance, voltage, It's usually when you are dealing with very low currents and leakage and drive are closer than usual. When you get down to having to not use resistors if possible you MAY be trying too hard :-). It happens, but what is the limitation you are fighting? \$\endgroup\$
    – Russell McMahon
    Commented Jun 12, 2019 at 14:04
  • \$\begingroup\$ PNP load current: 14 and 28 mA (I need this twice); Voltage at the Collector: 9.4 V (for the 28 mA PNP) and 6V (for the 14 mA PNP) \$\endgroup\$
    – towe
    Commented Jun 12, 2019 at 14:07
  • \$\begingroup\$ @towe Should be good (in many cases). R1 is "insurance". If using TH then you haven't said why you need no R1 and it can be tiny - but without it should be OK depending on what off collector current is acceptable. | \$\endgroup\$
    – Russell McMahon
    Commented Jun 12, 2019 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.