I have the following problem: I use to perform electrical experiments in a small room in an apart where other persons repair computers. It sometimes happens that I make the main differential switch break the power, causing inconvenience to the other persons. The cause of the failure is almost always a wire connected to the device under test that touches the box of the device, which is itself connected to the main ground. Of course, I could call an electrician to install a separate main lines for my room, but this is expensive and exaggerated for what I do. Instead, I've imagined to replace the main ground by a "virtual ground" that will play the same role, but without turning on the main differential. Here is a picture of the circuit, which works in simulations, and where the 2A fuse and additional LEDs have been omitted (the picture can be zoomed to see the details).virtual AC ground
Let me examine if the operator is well protected:

1) If there is a short or an over-current flow, the (non represented) fuse will break, and even if it not breaks, the short circuit breaker of the main power line will still stop the current flow;

2) if some current flows from one of the wires to the earth (not the virtual ground labelled "GND" in the picture), the differential of the main power line will stop the current as usual;

3) finally, if one of the wires touches the virtual ground labelled "GND" in the picture (even through a reasonably small resistor), the solid state relay will stop the current too, and the main differential will see nothing.

My question is: have I missed something ? (I prefer to ask professional regarding safety).

EDIT: details regarding the operation of the circuit: First, the whole circuit is inside a box, the input is the phase and neutral of the main power line. The output is 3 terminals labelled "phase", "neutral" and "GND" in the picture: the user use these 3 terminals as if they were the 3 terminals of the main power line. Regarding the operation, the (opto) solid state relay input is powered by the 5V independent floating voltage source and is normally on because the 2P4M thyristor is not conducting. So, the current flows normally. If now "phase" or "neutral" come into contact with "GND" (through a failure of a connected device), then the thyristor is fired, bringing the voltage of the SSR input near ground. This causes the relay to shut down and the current is stopped.
NOTE: the 4k resistor in the circuit simulated a load in my simulations, and is actually not part of the circuit (I forgot to delete it before posting this image; sorry if this has confused some readers).

  • \$\begingroup\$ It is not clear at all how this offers any protection. How do you think this provides safety? (Edit your question rather than sprinkling details through the comments.) \$\endgroup\$
    – Transistor
    Commented Jul 18, 2017 at 20:51
  • \$\begingroup\$ Transistor, how I think this provides safety is written in the 1), 2) 3) points of the question. Please, be specific. \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:02
  • \$\begingroup\$ Is it necessary to add a detailed description of the circuit operation ? (I thought it is rather simple). \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:07
  • \$\begingroup\$ (a) ... the "virtual ground circuit breaker" in the picture above .... There is nothing labeled "virtual ground circuit breaker" in the schematic. (b) You can't rely on a solid-state relay for safety. (c) If phase and neutral are swapped - you don't have your location entered in your profile so we don't know what mains plugs you use - you have no protection. \$\endgroup\$
    – Transistor
    Commented Jul 18, 2017 at 21:07
  • \$\begingroup\$ I've edited the question, hoping this will clarify it for you and other readers. \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:21

2 Answers 2


It is difficult to know what you think your circuit will do and your explanation is very poor. However some misunderstandings need to be corrected:

  • The whole point of grounding equipment is that if a live to earth incident occurs that you will trip the mains breaker or blow a fuse to remove power from the device and make the circuit safe. Anything you do to prevent this presents a danger to you and your roommates.
  • The fact that this is a regular problem indicates that you are careless in your work.

If you insist on working on live equipment then you should use an isolating transformer.


simulate this circuit – Schematic created using CircuitLab

Figure 1. Bench test mains power supply.

How it works

  • The transformer XFMR1 is a 1:1 mains power transformer. You need to figure out what's the most powerful appliance you need to test and size accordingly. e.g., for 1000 W use a 1000 VA transformer.
  • The transformer will isolate the load from the mains and if there is an earth leakage fault on the appliance it will prevent your main RCD and the one on this setup from tripping. (More later.) It will also protect you while the EARTH TEST switch is open.
  • The lamp serves as a current limiting test device. Open the FULL ON switch before you plug in the device. The brightness of the lamp will give an indication of the current draw. If it goes to full brightness on a small appliance, for example, you may have a short circuit but current will be limited to 0.5 A (230 V AC lamp) or 1 A (120 V AC lamp). If all appears well then switch FULL ON.
  • If all is well then close the EARTH TEST fault switch. Now if there is an earth fault on the appliance some of the current will return on the earth-wire instead of the neutral (N). The RCD will detect a difference between the L and N current and trip out. This would usually only trip if there's an earth fault to a metal-cased appliance.
  • Size the RCD (current rating) to protect the transformer.

That should help you sort out most problems on the bench.

Forget your virtual earth circuit. It is not safe.

  • \$\begingroup\$ 1 point for your detailed answer, but apparently you have not answered the question (you only presume my circuit is unsafe without understanding it). Of course, it is probably better to have an isolation transformer, but this is expensive (at least for me). You said my question is unclear, so, you can ask in comments about specific points and I will be happy to answer or even edit the question. \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:00
  • \$\begingroup\$ @Transistor I think I would choose to work in another building or even on the other side of town... \$\endgroup\$
    – Solar Mike
    Commented Jul 18, 2017 at 21:10
  • \$\begingroup\$ Ha Ha Ha - Very fine. \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:11

You should look at the problem from the other side : why not have some Uninterruptible Power Supplies for the computer testing so if the main breaker goes out everyone is still safe.

  • \$\begingroup\$ Well, I don't see myself demand to these persons to buy an uninterruptible power supply; furthermore this turns the lighting and all the other apparatus off. \$\endgroup\$
    – MikeTeX
    Commented Jul 18, 2017 at 21:37

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