Edit: I think I've figured out WHY this is happening (see my answer below), but I still would love to hear ways to mitigate this issue.

I've got two microcontrollers that are powered by the same dc power source, but are powered through separate bridge rectifiers and LDOs. Can these two microcontrollers communicate over UART using a single metal wire (no optoisolation) without sharing a common ground wire?


simulate this circuit – Schematic created using CircuitLab

The rectifiers in this circuit are CDBHD140L-G.

To be clear, these circuits do not share a common ground, they only have the DC voltage source V1 in common (yeah, I know rectifiers don't make much sense here with a DC voltage source; it's a UX decision that I'm not looking to change right now). Also, there are LDOs after each bridge rectifier, I've just left them out for simplicity.

My intuition tells me that

  • The absolute voltage levels for ground will be the same in both circuits relative to V1's negative side, so I'd expect this to be fine for DC signal transmission.

  • The return path from RX to TX might be impeded in a weird way by the rectifiers.
    I'd expect this to affect the signal integrity of the transmitted signal for things higher than a few MHz.

I've already done a small experiment, and the results didn't make sense to me. In my setup, I powered a dev board through a bridge rectifier and grounded my oscilloscope through a different bridge rectifier. I basically got the opposite of what I was expecting: the signal edges were clean and fast, but the voltage swing only went from ~400mV to Vcc - 400mV. Even worse, the range of measured voltages fluctuated by ~200mV at a frequency of ~100kHz (no switching regs on my tiny FPGA dev board afaik).


simulate this circuit

enter image description here

(This isn't really an oscillator pin; my dev board has a camera attached. Cause this was a quick and sloppy experiment, I just scoped one of the data pins on the camera bus hence the irregular signal.)

My questions:

  • Why exactly is this happening?
  • Do I need to provide a second ground wire between RX and TX? Would some cap-based galvanic isolation work instead?
  • Are there any other sneaky ways around this issue without adding a ground wire between these 2 circuits? Capacitively-coupled FM modulation / manchester encoding, etc? Some fancy bridge rectifier substitute like the LM74670?
  • \$\begingroup\$ Q!: Where is the scope ground connected to? | Q2: Why use 2 bridges? Why not use 1 bridge and 2 x LDOs? \$\endgroup\$ – Russell McMahon Mar 21 '20 at 6:09
  • \$\begingroup\$ Are you aware that your scope is capable of storing screenshots on USB storage devices...? \$\endgroup\$ – jusaca Mar 21 '20 at 9:48
  • \$\begingroup\$ @RussellMcMahon Q1: The scope ground is connected to BR2, as shown in the second circuit diagram; if it was connected to the ground in BR1 or the negative side of , there would be nothing interesting happening. Q2: I'm looking for these circuits to be part of a kit, and I can't change the design for UX reasons. \$\endgroup\$ – John M Mar 21 '20 at 15:41
  • 1
    \$\begingroup\$ Why does it need to be a single wire, rather than a pair of wires? You could even use a preterminated cable if your concern is UX. \$\endgroup\$ – Hearth Mar 21 '20 at 15:47
  • 1
    \$\begingroup\$ After a quick scan of the question, what is the turn-off time of the bridge rectifier diodes at 100 kHz? \$\endgroup\$ – Transistor Mar 21 '20 at 16:22

Unless your two circuits, the TX block and RX block are galvanically isolated from each other you need to have a ground (signal) return path between the two circuits. The currents that flow into your RX device (and there will be some, even if they are small) need to be able to find their way, so to speak, back to the TX block and its power supply. Remember that current always flows in a loop.

There may be a sneak return path back through the blocks labeled BR, but without knowing the details of what's in those blocks it's hard to say.

Even if such a path exists, it may suffice for DC purposes, but may not work for high speed, even moderately high speed, signals.

EDIT 1 - Added current flow graphic

Current Flow?

  • 1
    \$\begingroup\$ The BR blocks are bridge rectifiers, simple as that. Four diodes. \$\endgroup\$ – Hearth Mar 21 '20 at 15:57
  • \$\begingroup\$ If you look at the scope trace, the signal issue isn't the sort of thing traditionally described as "Bad SI" in SI books. There's no ringing, just the vascillating of V_high and V_low. \$\endgroup\$ – John M Mar 21 '20 at 16:10
  • \$\begingroup\$ I wasn't thinking of this so much as an SI problem (though that may come into play), but more of a DC issue. I was trying to figure out if there may be a (DC) return current path without having the extra wire OP wants to avoid. \$\endgroup\$ – SteveSh Mar 21 '20 at 17:58

After the discussion here and spending some quality time with LTSpice, I can answer my first question: "Why is this (the changes in v_high and v_low in my scope picture) happening".

The FPGA dev kit consumes different amounts of current as it does different things. When it consumes more current, the V_f drop over its bridge rectifier will be larger, increasing the potential of its ground. On the other hand, the scope grounds a negligible amount of current, so the V_f drop over its bridge rectifier diode will be smaller.

This will change the measured voltage level relative to the ground of BR2.

I put together a circuit that demonstrates this in Falstad. You can find it here.

  • 1
    \$\begingroup\$ There are mosfet circuits you can use to form a "zero forward voltage" diode. Perhaps you can use these to reduce the ground impedance between your circuit blocks and the power distribution. \$\endgroup\$ – Drew Mar 21 '20 at 20:52
  • 1
    \$\begingroup\$ Or perhaps if the data rate is low, you can modulate your signal into AC, and then pass it through a cap? Then demodulate it. That would eliminate any ground loops, and reduce the effect of ground voltage differences. \$\endgroup\$ – Drew Mar 21 '20 at 20:55
  • \$\begingroup\$ @Drew You can also just buy "zero forward voltage" diodes, with a typical forward voltage less than 50 mV at several amperes. But I'm not sure this will solve the problem. \$\endgroup\$ – Hearth Mar 23 '20 at 13:46
  • \$\begingroup\$ @Drew I'm sorry to be so picky and shoot down good solutions, but all of the ideal diodes I could find (as in IC's that control a mosfet) cost too much. \$\endgroup\$ – John M Mar 23 '20 at 13:57
  • \$\begingroup\$ @Drew As far as the AC modulation goes, do you have any suggestions for how I could do it? I've been thinking about using a VCO to do FM modulation or just manchester encoding my data at a high enough rate and throwing it on the wire. \$\endgroup\$ – John M Mar 23 '20 at 14:00

Q3: Isolate with a logic optocoupler, like a H11L1M.

  • \$\begingroup\$ Unfortunately, I need these subcircuits to be connected by a single wire, for UX reasons. Unless that wire was an optical fiber (which would make connector solutions too confusing and expensive for the target audience of this project), an optocoupler wouldn't work. \$\endgroup\$ – John M Mar 21 '20 at 15:43
  • 1
    \$\begingroup\$ @johnny_boy Have you considered using toslink? Pretty cheap as optical communication goes (also pretty slow for optical fiber, but you haven't said anything about needing high speed communication) \$\endgroup\$ – Hearth Mar 21 '20 at 15:56
  • \$\begingroup\$ @Hearth I love the idea, and it totally fits my spec for data rate (1Mbps-ish), but I need connectors and cables that are on the order of 10 cents, not 10 dollars. \$\endgroup\$ – John M Mar 21 '20 at 16:00
  • \$\begingroup\$ @johnny_boy Fair enough, though connectors on the order of ten cents is going to be a bit hard to find unless you're buying in quantity! \$\endgroup\$ – Hearth Mar 21 '20 at 18:06
  • 1
    \$\begingroup\$ @johnny_boy I looked into that, and found that RCA cables are a cheaper option, at least if you're getting your parts from digi-key. \$\endgroup\$ – Hearth Mar 23 '20 at 13:45

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.