4
\$\begingroup\$

in the schematic on page 5 of this document:

enter image description here

The note says:

T2a is connected as a diode (clipping diode) to protect the power source against higher Vcc voltages.

I assume from this I can replace T2A with a simple diode, with one side connected to pins 4/5 of T2A and the other connected to TRKR (pointing towards TRKR) - am I correct, or have I overlooked something?

\$\endgroup\$
13
  • 1
    \$\begingroup\$ Are you also intending on replacing the software they use on the MCU? If so, you will need to fully understand their circuit. Or, if not, how are you planning on buying MCUs without the appropriate circuitry also included? This looks to me like a commercial product from the brochure. And just as a note, that diode-connected BJT has a very different saturation current than "a diode" might have. \$\endgroup\$
    – jonk
    Sep 20, 2017 at 20:46
  • 1
    \$\begingroup\$ Thanks for the reply. I already have a functioning DCC decoder (including the rest of the circuit and my own software on the MCU.) I'm just trying to add the "railcom" part, which is what this schematic focuses on. I understand the rest of the circuit (T1A and T1B provide a ~30mA current source with R1), I just haven't seen a transistor used in the way T2A is before. I'm not sure if its use is because I need a transistor there for some reason, or if it's just using the other transistor in the BC847BV package for convenience. \$\endgroup\$
    – X Codz
    Sep 20, 2017 at 22:00
  • \$\begingroup\$ Thanks. I haven't seen this particular circuit before because I know nothing about DCC (until now.) Is the track voltage a modulated square-wave? This circuit appears to derive it's own voltage using a bridge+regulator circuit, so this does NOT look like a controller. \$\endgroup\$
    – jonk
    Sep 20, 2017 at 22:17
  • 1
    \$\begingroup\$ Similar question: Can I (ab)use a transistor as an ESD protection diode? \$\endgroup\$
    – CL.
    Sep 21, 2017 at 7:55
  • 1
    \$\begingroup\$ @XCodz I think you can just use a diode, after looking all that over. They say it provides diode-clipping, but I really just think it is used to allow current to pass when the track voltage is in only one of the two polarities. It otherwise just blocks. The diode-connected BJT doesn't seem to offer anything special here, unless they are using it for zenering in the rare case where your internal voltage rail is for some reason much higher than their power amplitude. But I've no reason to anticipate that design goal given what I've read. \$\endgroup\$
    – jonk
    Sep 22, 2017 at 7:24

1 Answer 1

2
\$\begingroup\$

The schematic of the TRKI and TRKR signals is a differential output where TRKI have the same polarity of TXD and TRKR have the inverted polarity of TXD. Obviously the TRKI signal must have an external pull-up resistor and the TRKR signal must have an external pull-down resistor. The purpose of this differential driver is to only drive the lines when TXD is 0, and to release the lines (tri-state) when TXD is 1. This is why T2A "diode" is required: it prevent to drive any current on the TRKR line when TXD is 1.

You can safely exchange T2A with a diode if you wants. The schematic take advantage of a single BC847BV package to make two functions into it.

QUCS simulation schematic QUCS simulation result

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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