I have a setup in which several (3 or 4) microcontroller boards (locally, they are within 3 to 5 meters away from each other) need to be "perfectly" synchronized, so we're broadcasting a "master clock" signal; a very slow clock (in the order of 1 Hz).

Since we are pretty sure that we'll have to deal with noise, ground loops and related nasties, I'd like to send out that signal differentially (possibly just two wires, and then all of the boards' grounds going to a common "earth" point).

My first thought was: designate two GPIO pins of the processor (we have lots of available pins), and drive them from a timer-interrupt procedure such that they are always the complement of each other. That way, we get a differential signal that swings from -3.3V to +3.3V. For the receiver, I'm thinking an analog differential circuit --- a rail-to-rail op-amp powered with 3.3V and GND (single supply) driven to saturation so that it produces a digital signal.

Then, there is the option of using a line driver / receiver such as the DS90LV019, or the combo SN65LVDS1 + SN65LVDS2; then, I notice that these all produce a differential voltage in the order of 300 to 500 mV. I'm a little bit paranoid, and such a low level, even being differential, scares me.

I thought of a hybrid: I could transmit with two straight outputs from GPIO pins that are always the complement of each other, and at the receiving end I use an SN65LVDS2 receiver --- but this (and a couple other receivers I'm seeing) specify as absolute maximum ratings a differential input voltage of +/-1V (i.e., not to exceed +/-1V).

Am I over-complicating things? My guess is that I should trust the driver/receiver combo; but I get this bad feeling that I'm going to regret taking that route (plus, devoting two pins and keeping them as the complement of each other seems soooooo easy).

Any experience with these? Any tips or advice on what you think should work best?


  • 1
    \$\begingroup\$ hm, might be a good idea to define "perfect" synchronization in terms of a timing offset. also, what is it that you need to build? Feels a lot like yes, if you can work with a 1s granularity, there might be easier ways. \$\endgroup\$ Commented Dec 17, 2017 at 0:26
  • \$\begingroup\$ If you're worried about ground loops, why not single ended opto coupler? \$\endgroup\$
    – Wesley Lee
    Commented Dec 17, 2017 at 0:29
  • \$\begingroup\$ @MarcusMüller Not sure if we mean the same thing when you say 1s granularity --- a 1Hz signal means that once a second we synchronize; but it's the edge of that clock signal that we use. I put "perfect" in quotes to mean that we're not happy with things like NTP or the like. A physically-shared clock signal gives us that "perfect" sync; where "perfect" in our context means: any deviations within a few microseconds. \$\endgroup\$
    – Cal-linux
    Commented Dec 17, 2017 at 0:56
  • 1
    \$\begingroup\$ how about using fiberoptics .... microcontrollershop.com/… ........... pub.ucpros.com/download/ifd91.pdf ...... pub.ucpros.com/download/ife97.pdf \$\endgroup\$
    – jsotola
    Commented Dec 17, 2017 at 2:08
  • 1
    \$\begingroup\$ Suppose you have 2.5 nanosecond edges, over 2.5 volt levels. Suppose you have 0.1volt RMS noise. What is the jitter? Tj = Vnoise/SlewRate = 0.1v/(1volt/nanosecond) = 0.1 nanosecond Time jitter. \$\endgroup\$ Commented Dec 17, 2017 at 2:48

2 Answers 2


The first thing you should do is specify the acceptable timing error. The solutions to sync within 100ns or 10ps are very different. If you just need to trigger an interrupt within a few cycles in each processors, then differential signaling is unnecessary, a coaxial cable will do the trick.

The two GPIOs solution is risky. If you cannot guaranty that they will start their transitions at the same time, then they will have the same output and the difference will be 0. If the difference is 0, the output of the line receiver is undetermined, and could be rapid oscillation due to noise.

The line driver / receiver will give you much better timing performance, if required. You don't have to be paranoid, differential signals can be very robust. for exemple, LVDS is used extensively in the automotive industry. However, differential signals can tolerate only a certain amount of voltage shift, so if your ground is very noisy you could still have issues.

For synchronization down to a few nanoseconds, over a short distance, and with frequency below 1MHz, an LVCMOS signal over coaxial cable can give good results:


simulate this circuit – Schematic created using CircuitLab You can add other protections as needed.

You also need to look at where you are going to input that signal to the microprocessor if you need synchronization better than what interrupts can give you.

If you are expecting ground loops, maybe work at the root of the problem instead and try to minimize the area between your ground connections.


An LVDS driver is not much different from a pair of GPIOs; it's designed to drive the specified LVDS current into the specified LVDS impedance.

The main difference is at the receiver: a plain CMOS input has almost infinite input impedance, while LVDS uses a 100 Ω resistor. The low impedance increases noise resistance (you'd need a much stronger electromagnetical field to induce an error voltage high enough to affect the signal).

When using GPIOs, just put a resistor between the receiver's pins (for a current of about 3.5 mA), and your signal will be as robust as LVDS.

You should replace the opamp with a comparator; the latter is optimized for fast switching and saturating its output. (Note: an LVDS receiver is a comparator.)

  • \$\begingroup\$ Good points. I was actually planning to use a termination resistor --- not only higher immunity to EMI pickup, but also to avoid reflections due to impedance mismatch (infinite input impedance in a sense is like a mirror). As for the op-amp, I figured I could add some hysteresis and/or low-pass filtering at a few MHz cutoff to minimize spikes/glitches due to skew between the two signals. But I guess I can also find a comparator with schmidt-trigger input --- that is, if I decide to go with the pair of GPIO pins. \$\endgroup\$
    – Cal-linux
    Commented Dec 17, 2017 at 17:08

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.