3
\$\begingroup\$

I had a PCB made for a homebrew SBC but while waiting for the boards I discovered a bug. I have a 74HC138 output (I believe this is just normal push-pull) that I need to hook up to a CPU pin but I need it to be open-collector so other chips can drive it low as well (basically wire-AND.) Because this is going to be a kludge, I want to reduce the part count as much as humanly possible.

I have what I believe will do the job in 2 transistors: a PNP transistor acting as an inverter and an NPN transistor acting as the open-collector:

Non-inverting open collector using 2 transistors

Can this be done with a single transistor?

\$\endgroup\$
3
  • \$\begingroup\$ You can probably do a PNP emitter follower with an open emitter, but the answer suggesting the Schottky diode is a better solution. \$\endgroup\$
    – Ste Kulov
    Commented Feb 9 at 16:47
  • 1
    \$\begingroup\$ Is this a CPU or a microcontroller? They're different kinds of devices and simple CPU's are quite rare on the market these days. 1. Do you control the "CPU" program? Just program to output the opposite value from what you want to come out of the transistor buffer. 2. Have you checked that your "CPU"'s output can't be set for open-collector (or open-drain) output mode? Either option costs nothing, not even a single external transistor. \$\endgroup\$
    – The Photon
    Commented Feb 9 at 18:01
  • \$\begingroup\$ It's an original motorola 68000 and I want to connect this to the /VPA pin. I basically have button that triggers an NMI but it's not a real device so it can't acknowledge the interrupt. Without getting into detail, I have an expansion slot that could want to legitimately drive /VPA that I want to isolate this kludge from. \$\endgroup\$ Commented Feb 10 at 11:58

2 Answers 2

8
\$\begingroup\$

Just use a low-current SMT Schottky diode in series with the output. Cut the trace and splice the diode in, easy-peasy.

Do it carefully and nobody will notice!

\$\endgroup\$
2
  • \$\begingroup\$ How does this work exactly? I assume it the diode direction is from the CPU pin to the '138? So while the '138 output is high, the diode blocks the high signal, but when it's low, it will sink current? \$\endgroup\$ Commented Feb 9 at 16:48
  • 1
    \$\begingroup\$ Yes, exactly. It adds a bit of voltage drop so the noise immunity is slightly less, but with normal CMOS the threshold is around 2.5V typically and no lower than 1.5V, so a few hundred mV won't make much difference. \$\endgroup\$ Commented Feb 9 at 16:53
5
\$\begingroup\$

I wouldn’t bother with transistors. Single logic gates come in packages the same size as SMT transistors. Just use a single open drain buffer gate. It’s guaranteed to work correctly, it’s a single part, only needs a decoupling capacitor.

You can also buy double transistors in the same 4,5, or 6-pin package, even with biasing resistors built-in. But what’s the point of that if you can get a properly characterized gate in the same package?

\$\endgroup\$

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.