With respect to how to reduce the effect of ground currents on sensitive analog portions of a circuit, The Art of Electronics (3rd Edition), chapter 8 includes this suggestion:

Decoupling the dc rail supplying low-level stages

Figure 8.125. Decoupling the dc rail supplying low-level stages.

Supply-voltage fluctuations caused by load currents at the high-level stage are impressed on the low-level supply voltages. Unless the input stage has very good supply rejection, this can lead to oscillations. Even worse, the load current returning to the supply makes the case “ground” fluctuate with respect to power-supply ground. The input stage ties to this fluctuating ground, a very bad idea. The general idea is to look at where the large signal currents are flowing and make sure their 𝐼𝑅 drops don’t wind up at the input. In some cases it may be a good idea to decouple the supply voltages to the low-level stages with a small 𝑅𝐶 network (Figure 8.125).

My question: what is the purpose and effect of the 10 Ω resistor connecting the two grounds?

I understand how this allows the low-level section's ground to float a little separately from the high-level ground, which might come about if the two sections were each attached to different external systems (not shown in the diagram), each with their own disparate grounds, while still allowing the high and low level sections to receive power from the same PSU.

However, usually the low-level section will need to communicate signals to the high level section (or vice versa), and, for those signals, any current in the 10 Ω ground resistor produces a voltage that adds unwantedly to the analog signals between the two sections. After all, this excerpt follows a discussion of reducing the impedance of ground networks (along with deliberately designing the current paths) for just that reason, not increasing them.

For example, the 10 Ω resistors in the positive and negative rails will split the voltage drop attributable to the current consumption of the low-level section, thereby adding that half amount to any signals from one section to the other. That's already an issue, and made worse if that current consumption fluctuates (more of a problem at low frequency, since higher frequencies are absorbed by the reservoir capacitor).

No doubt there are techniques that ameliorate the mentioned problem, such as communication between the two sections using an analog optocoupler or differential signals. But for plain old individual analog signals, I'm at a loss to understand how that 10 Ω ground resistor helps, or in what scenarios, compared to a zero-resistance connection.


  • \$\begingroup\$ Not the complete solution, but I think this circuit is relying on power supply margin. If the lowside amp is only going to swing to +5V and you have a +10V rail, then you can afford to lose some of it in the 10Ω resistor, but I would probably choose a ferrite bead. \$\endgroup\$
    – Aaron
    Sep 4, 2020 at 23:04
  • \$\begingroup\$ I once had an RF FAE suggest using a 1kΩ on the power supply pin to their low power cmos antenna switch. Why? For noise decoupling. It worked like a charm. \$\endgroup\$
    – Aaron
    Sep 4, 2020 at 23:07
  • \$\begingroup\$ @Aaron yes -- agreed. However my question is specifically about the 10-ohm resistor on the ground side. \$\endgroup\$
    – gwideman
    Sep 4, 2020 at 23:34
  • \$\begingroup\$ @gwideman Look at what's between the high-level supply and the low-level supply sides. Two resistors, both sides of the rails, with a bypass cap on the low-level supply side. What's difficult to follow about that? (And yes, you'd want both Rs. Think about it. And while doing so, please note that the low-level input shows two inputs -- so its differential mode.) Good question though. +1. (In past, I've used as much as a 5-pole isolation, high and low side, using beads and caps and resistors to isolate a low-level input with very high gain, guard rings, metal can, etc.) Others have done more. \$\endgroup\$
    – jonk
    Sep 5, 2020 at 1:31
  • \$\begingroup\$ @jonk OK, so what is the actual function of the 10 ohm resistor in the ground path, or even more, your multi-pole filter between the two grounds. At best it causes the two grounds to differ due to the DC and AC current consumption of the low-level section. At most it causes the grounds to further vary due to some external connections to the two sections. And sure, differential input is suggested by the two inputs to the low level side. So the single input to the high-level side suggests singled ended signals from low level to high level? Making the ground offset a problem. \$\endgroup\$
    – gwideman
    Sep 5, 2020 at 20:45

1 Answer 1


A lock-in amplifier accept extremely small input signals, often below thermal noise levels. Groundloop currents can be a dominant noise source.
A low-noise JFET preamplifier and its BNC front panel input jack are isolated from ground with a small resistor of about 10 ohms. Ground-loop currents on input coax cables must flow through this 10-ohm resistor. The resistance is far larger than coax shield (braid) resistance.
Ground currents develop a voltage across the 10 ohm resistor, while those same ground currents develop a much smaller voltage along the coax braid. The preamp rejects the voltage across the 10-ohm resistor, but must cope with the much smaller voltage drop along coax braid.


simulate this circuit – Schematic created using CircuitLab

A differential amplifier rather than a single-ended JFET preamp could also be used, where the voltage difference between hot input and BNC ungrounded shell is amplified. But a differential amplifier is likely noisier than a discrete single JFET.

  • \$\begingroup\$ The AofE 3rd edition diagram shows a differential mode amplifier on the low-level side. \$\endgroup\$
    – jonk
    Sep 5, 2020 at 1:26
  • \$\begingroup\$ @DAS I added that gnd as an afterthought, because we're talking about ground loop currents. Without it, there's no loop. \$\endgroup\$
    – glen_geek
    Sep 5, 2020 at 12:49
  • \$\begingroup\$ Your example is about coupling two separate systems together, where I guess we assume that the two triangle ground symbols are at the same potential (due to a not-shown connection), but somehow a voltage is induced in the the ground braid, perhaps inductively or capacitively from some other influence? Your R1 10-ohm resistor is not in the ground path. You've essentially created a JFET differential input stage where the two inputs are the gate and the source, and R1 is the source resistor. I do see how this solves a problem, but I don't think it's quite the case I'm asking about. \$\endgroup\$
    – gwideman
    Sep 5, 2020 at 20:52
  • \$\begingroup\$ Yes, triangle symbols connect, likely via power-line ground...many signal sources are unavoidably grounded (if the source floats above gnd, we have no problem with ground loops). Reducing gnd path resistance only increases groundloop currents due to magnetic coupling. And adding another ground adds loop area. That 10 ohm resistor decreases groundloop current because it is in series with the loop. I believe it is exactly the scenario in AofE. Noise at amplifier input not amplified doesn't appear at amplifier output. One should take care that no other currents flow through than 10ohm. \$\endgroup\$
    – glen_geek
    Sep 5, 2020 at 22:13
  • \$\begingroup\$ @glen_geek I appreciate your efforts, but thiis seems off-track from the AoE example, that's about a single system within a board or chassis, given that power is supplied to the low-level system via an R-C from a shared power supply. "That 10 ohm resistor decreases ground loop current". If you mean the AoE example, it may reduce GL current (from some unspecified source), but it won't cut down the corresponding offset voltage, which will add to analog signals crossing from low-level to high-level. (If you meant your R1, then as previously noted, you've essentially made a differential input). \$\endgroup\$
    – gwideman
    Sep 5, 2020 at 23:57

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