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Low side configuration

I have a circuit design as above. I'm trying to sense the voltage difference accross the \$ 0.1 \Omega \$ resistor. I physically connected the sensory circuit to the low side of the AC line, and saw that it is working alright.

I'm going to make a boxed circuit of this, and I'm going to connect it to the wall outlet with a standard plug, seen as below:
Standard AC plug
The plug is symmetric. I can accidentally connect it reversed (on the other hand, i don't want to check the polarity every time I'm connecting it to wall outlet). In that case, the sensory circuit is connected to the high side as seen below:

High side configuration

My question is, What happens if the sensory circuit is connected to the high side? Would a terrible thing happen? Do I need to do any modification on the overall circuit for making it connectable to the high side?

Supose that,
RMS of the AC is 230V.
VCC is 5V, and it is electrically isolated from the AC by a transformer.
All circuit elements are rated to work between -5V to +10V. The electrical device drains small current (below 10A RMS).

This question is not about personal usage risks. Please focus your answers on the circuit level, about the risk on the circuit elements.

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  • \$\begingroup\$ If you do not know the answer to the question already, you should not be connecting anything directly to the mains. You will kill somebody. \$\endgroup\$ – markrages May 7 '12 at 18:35
  • \$\begingroup\$ At least tell us that VCC and GND and all electrical connections are completely isolated from the end user, with no exposed ground screws or connectors. Because VCC and GND have a 50% chance of having 230VAC superimposed on them. This is still a dangerous (possibly illegal) construction method. We don't have AC/DC radio sets anymore for a reason. \$\endgroup\$ – markrages May 7 '12 at 18:38
  • \$\begingroup\$ @markrages This is not going to be a commercial product. I'm just making a project in my home-lab. If I were to make a product for other people, of course, I would take required safety cautions. \$\endgroup\$ – hkBattousai May 7 '12 at 22:25
  • \$\begingroup\$ @hkBattousai I hope you will not sneeze and accidently kill yourself by touching somewhere you shouldn't. \$\endgroup\$ – abdullah kahraman May 8 '12 at 8:18
  • \$\begingroup\$ @hkBattousai by the way, why don't you introduce some circuitry to lessen the death risk? It will be trivial and you will learn, if you do not know. If you know, then I think your are too lazy to work with high-voltage :) And yes, I know nothing about high voltage.. \$\endgroup\$ – abdullah kahraman May 8 '12 at 8:24
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If VCC and GND are electrically isolated from Phase, Neutral and Earth, it doesn't matter the direction with which you connect the plug.

Your sensory circuit doesn't even know there's 230 VRMS somewhere else. It will just see the voltage across the current-sense resistor.

Update: It is hard to tell whether your sensory circuit is in fact electrically isolated or not from your mains, because the text in your figure says they are, but the connection between your "mains side" and the GND of your sensory circuit, according to the schematic, says they are not. I have to think that what is right is the schematic, and that you think they are electrically isolated, but they are not. You wrote "This question is not about personal usage risks. Please focus your answers on the circuit level, about the risk on the circuit elements", and I gave you an answer according to that. The one-wire (or two-wire, with little volts between them) connection between the mains and your sensory circuit won't do any harm to your circuitry, and won't distort any reading, but just remember: YOUR SENSORY CIRCUIT DOES NOT SEEM TO BE ELECTRICALLY ISOLATED FROM THE MAINS. If you touch any node of your sensory circuit, you may die. So, be very careful if you have experience with such circuits, and do not build anything if you don't have such experience.

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    \$\begingroup\$ Given that the way he has drawan it and the way he has described it do not match, there is no way to be sure tha your answer is correct, or safe. \$\endgroup\$ – Russell McMahon May 7 '12 at 15:05
  • \$\begingroup\$ @RussellMcMahon I don't understand. The OP wrote "VCC and GND are electrically isolated from Phase, Neutral and Earth" in his figures. I assume he knows what "electrically isolated" means. Otherwise, answers are too long, and don't focus on the point. \$\endgroup\$ – Telaclavo May 7 '12 at 15:33
  • \$\begingroup\$ His diagrams show a connection from one pin to the "ground of his circuit - whatever that means. I assume that given he is asking a fundamentally very simple question here he may NOT "know" what electrically isolated means in practice. eg if the 0.1 ohm sense resistor goes O/C - as well it may - so that 230 VAC is then connected across the sense input terminals - as it would be - is his circuit still "isolated"? Would this destroy the sense circuitry. Will the sense circuitry conflagrate and break down the isolation? Will ...? \$\endgroup\$ – Russell McMahon May 7 '12 at 15:43
  • \$\begingroup\$ @RussellMcMahon You are right, they are not electrically isolated, but a one-wire connection between the two "worlds" won't do any harm to the circuit. It may do to him, if he touches any node in the sensory circuit. Of course, if the current-sense resistor breaks, he'll have problems (I've pointed that out in other answers). \$\endgroup\$ – Telaclavo May 7 '12 at 15:54
  • \$\begingroup\$ I'm hoping that his diagram is wrong and that in fact he has genuine isolation despite the diagram. If not then "all bets are off". Allowing mains in to an "isolated" circuit by a single wire gives Murphy license to do about anything at all. \$\endgroup\$ – Russell McMahon May 7 '12 at 17:32
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The answer depends on your "sensory circuit". As you have provided zero information about how the sensor works and have not included the utterly obligatory actual circuit diagram then it is not possible to answer your question with certainty.

You say to ignore safety but the fact that you are asking this question at all strongly suggests that for a competent person to answer this question and to ignore safety would be dereliction of duty.

Indications are that the circuit will work as well with the sense element in either lead. Your statement that "it is working alright" should be treated with great care. The connection between sense circuit ground and one input pin does not make sense if the unit is electrically isolated as stated.

If the 0.1 ohm sense element goes open circuit then some of all of the whole input side of your interface goes to phase in either configuration shown. This may be a safety issue but in the absence of a circuit we cannot be sure.

Your transformer must, of course, be rated to handle worst case voltage input (say 250 VAC+) and inductive spikes and general surges or dips.


Added:

What has happened here so far is extremely unsatisfactory and potentially dangerous and life threatening. A member who is keen to learn but who has a limited grasp of electronic and electrical fundamentals (based on prior questions and comments) asks people to trust him over his ability to do things safely with mains voltages. Based on the information given , which is inadequate when dealing with mains, there is a significant chance that events well within the range of normal may kill the questioner or a friend or family member or random stranger. Odds are this won't happen. It is even likely that the precautions taken are adequate. But this is by no means certain and competent annswerers need to be more discerning than has been the case so far. @Kortuk - your thoughts on this may be of value.

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AC isolation aside as requested, although it may cause certification issues.

In the first circuit, input 1 sees +/- 1.41 V, and input 2 sees 0 V. In the second circuit, input 1 sees 0 V, and input 2 sees +/- 1.41 V. You say your sensor can withstand these voltages. If it also produces the correct results with these voltages then you're in luck.

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    \$\begingroup\$ No, in the second circuit, input 2 also sees +/- Vx volts (you called 1.41 V to that Vx). \$\endgroup\$ – Telaclavo May 7 '12 at 10:59
  • \$\begingroup\$ @Telaclavo I had given the current value in RMS. I think he mentioned the peak values by saying 1.41V. \$\endgroup\$ – hkBattousai May 7 '12 at 12:46
  • \$\begingroup\$ @hkBattousai Yes, but my point is that if input 1 of the first circuit sees +/- 1.41 V, then input 2 of the second circuit will also see +/- 1.41 V, not 0 to -2.83 V. The current between Phase and Neutral is bidirectional (AC) in all cases. \$\endgroup\$ – Telaclavo May 7 '12 at 12:48

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