I am using shunt method to measure AC mains Current form circuit, attached image explain the systems. I am using a SMPS to isolated 120 VAC main form other control circuit. By i have some doubt, As i am measuring current using shunt, is i am breaking isolation somehow as per attached image. Or Do i need op-amp with higher common mode voltage rating ( greater than 120 VAC)

enter image description here

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    \$\begingroup\$ Why did you choose shunt resistor instead of a current transformer or Hall effect sensor? \$\endgroup\$ – AltAir May 9 '17 at 7:15
  • \$\begingroup\$ Reason of cost of current transformer \$\endgroup\$ – Short Circuit May 9 '17 at 8:05
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    \$\begingroup\$ Creating a galvanic isolation when using a resistive shunt is really cheaper? \$\endgroup\$ – AltAir May 9 '17 at 8:47

Yes, this will be a problem. Unless there is galvanic isolation between the "earthed" controller and the op-amp there will be instant smoke. The 3.3 volt supply (and its 0 volts) also need to be galvanically isolated from any electrical connection to the "earthed" controller.

Remember that live and neutral are not floating above earth - neutral is usually earthed and this means "live" rises and falls around earth at hundreds of volts peak to peak.

Do i need op-amp with higher common mode voltage rating ( greater than 120 VAC)

The input common mode range of an op-amp is largely determined by its supply rails and so to achieve +/- 170 volts (peak) you would need to find a supply with very large power rail capabilities. However, you would still have a safety problem feeding the op-amp output to the controller - you ought to consider using either a CT (current transformer) or something like ADI's range of digital isolation products (typically ADuM5401 and an ADC): -

enter image description here

The ADuM5401 provides kV of isolation but, remember, when working on anything that connects to live voltages, death or injury are just around the corner for those who are not wary and careful.

  • \$\begingroup\$ There is no earth in circuit, that symbol is for circuit return ground. No earth coming in circuit. \$\endgroup\$ – Short Circuit May 9 '17 at 8:06
  • \$\begingroup\$ digital isolation method look more costly than CT. look like measuring AC current using shunt is not possible in low cost because if you need isolator and other stuff , the cost goes more than CT method. \$\endgroup\$ – Short Circuit May 9 '17 at 8:09
  • \$\begingroup\$ So you are quite happy that the "controller and other low power circuit" is at the same potential as your live wire? That's the implication of what you are saying. \$\endgroup\$ – Andy aka May 9 '17 at 9:25
  • \$\begingroup\$ No, they are isolated , since i am using SMPS. which has 4KV isolation. so my circuit ground and Line or neutral are not on same potential. \$\endgroup\$ – Short Circuit May 9 '17 at 9:45
  • \$\begingroup\$ @ShortCircuit OK, but after you make the connection from the op-amp "shunt measure" circuit to your "controller", your circuit ground will acquire a potential that is equivalent to your live wire. This is "usually" a big deal but not always. It means that your controller has a live ground and is therefore unsafe for any connections not shown in your diagram unless those connections use isolation circuits. \$\endgroup\$ – Andy aka May 9 '17 at 10:08

I'd have done this in comment, but it's too long.

You have badly misunderstood your circuit. From a comment

No, they are isolated , since i am using SMPS. which has 4KV isolation. so my circuit ground and Line or neutral are not on same potential.

Well, they were isolated - right up until you connected them to your current sense amplifier. Since the voltage across the shunt is very small, it can be ignored. Then your line voltage connects to your op amp + input through a single resistor. If your reference voltage does not float with the line voltage , then you'll get line voltage at the input of a 3.3 volt op amp, and that will not end well. If the reference voltage does float, then your entire circuit is, by definition, riding on the line voltage. You can't have it both ways. The only limitation on current through the op amp is the value of the input resistor.

To make things worse, your voltage will only manifest itself when the circuit actually connects to power ground. It's perfectly possible to have a test setup, using isolated power supplies, which itself floats, and so you don't see the voltage. Then some poor customer or test engineer is actually using it in the wrong setup and makes the connection, and ZAP!, there goes trouble.

  • \$\begingroup\$ Look like its not easy to measure current through shunt without providing isolation on Op-amp side. If So Current transformer will be cheaper that using this. Wbo say now shunt method is cheaper, how Multi-meter do this. strange hmm. \$\endgroup\$ – Short Circuit May 10 '17 at 5:37
  • \$\begingroup\$ @ShortCircuit - Shunt method is cheaper with multimeter. Multimeter is electrically isolated, in plastic case, and readout is visual display. But that's not what you are trying to do. \$\endgroup\$ – WhatRoughBeast May 10 '17 at 12:04

It won't be easy finding a shunt amplifier with such a high common mode voltage.

Instead, you'd have to take one side of the shunt and short that to your ground or VCC. Or Ref Volt in your image. This depends on the common mode voltage of your amplifier, and other circuitry. Eg: is the circuit also measuring the L-N voltage? It isn't clear what Ref Volt is from your information.

With 10A and a 10mOhm shunt you have 0.1 AC, thus +-140 mV DC. If you use an INA199 with a -0.3V to 26V common mode voltage you can either use the top or the bottom of the shunt as ground or Vcc since the specification will be within your +- 0.14V range.

If you have an amplifier that has no common mode voltage outside of it's supply range, then Ref Volt has to be within 0+0.14V to 3.3-0.14V.
Also, Ref Volt has to connect before the filter resistor.

The 10A is taken as absolute maximum here.

Depending on how you wire it up you're either measuring high-side or low-side. Look at TI's Application Notes for in depth information about shunt placement.

So yes, the shunt method is indeed breaking isolation. They call it invasive for a reason. If you do not want that you'd have to trade price/resolution/accuracy and get magnetic measurement. From an ACS-712 sensor, a transformer or LEM*. Although, you could also use analog isolators like AMC1100.

Mandatory safety warning: your circuit is live (CAT II or CAT III). This means the ground of the oscilloscope, the USB chassis of your debugger, and thus your laptop, can be live too. Be aware of this. Otherwise sparks will occur, with death of equipment as a result.

*Magnetically shield your lem modules if they are not already!

  • \$\begingroup\$ This is a really dumb thing to advise: "Instead, you'd have to take one side of the shunt and short that to your ground or VCC. " \$\endgroup\$ – Andy aka May 9 '17 at 7:37
  • \$\begingroup\$ @Andyaka you have to, otherwise you'd have to build a high voltage differential amplifier. Which is very difficult at these signal levels. \$\endgroup\$ – Jeroen3 May 9 '17 at 7:55
  • \$\begingroup\$ Ref voltage use to shift the line AC measure waveform , because my Controller ADC will take any negative side voltage. \$\endgroup\$ – Short Circuit May 9 '17 at 8:11

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