I'm trying to understand and tweak this smartplug reference design from TI. I'm stumped on the Metrology Analog Filters section on page 1. What I do not understand is why is there a 0R resistor on the Neutral line and more importantly why is this line grounded?

Also since the shunt method is inherently unsafe, can I simply replace the shunt with the primary of the current transformer? In the attached schematic the current source and the terminating resistor essentially mimic the secondary of a current transformer. If I can replace the shunt, can I simply do away with the instrumentation amp in the reference schematic and simply put the attached bit in and feed the signal to the MCU?

Current sense with a CT


1 Answer 1


why is this line grounded?

The line isn't grounded to proper earth - it's just an internal reference node that happens to be called ground - it could easily have been called 0V or some other obscure title. It doesn't mean it should be galvanically connected to AC mains earth. That would be plain wrong.

Also since the shunt method is inherently unsafe....

The shunt method IS NOT inherently unsafe but you do need to take precautions and I wouldn't advise this project for anyone who doesn't understand the implications.

why is there a 0R resistor on the Neutral line

The 0R resistor is probably there to act like a fuse (0603 SMD) should someone do something incorrect (like think they can solidly earth the electronics.).

  • \$\begingroup\$ Thanks Andy. So would it be alright to replace the shunt with the current transformer? \$\endgroup\$ Apr 27, 2016 at 12:17
  • \$\begingroup\$ Yes that should be fine but make sure you calculate the burden to give the same voltage as what the shunt would do. \$\endgroup\$
    – Andy aka
    Apr 27, 2016 at 12:23
  • \$\begingroup\$ Got it. The voltage seen at the MCU should be no more than 1.4V at the maximum current measured (which in this case is 15A). Considering the CT turns ratio of 1:125, Isec = 120mA. To convert this into 0.7V (without the offset) would mean the burden resistor would be 5.8 ohms. Does that sound right? \$\endgroup\$ Apr 27, 2016 at 13:00
  • \$\begingroup\$ And... to further protect the ADC pin, insert 1kohm between pin and shunt resistor. \$\endgroup\$
    – Andy aka
    Apr 27, 2016 at 13:28
  • 1
    \$\begingroup\$ @seetharaman most MCUs with inbuilt ADCs are usually OK with 1kohm impedance and no cap but, as always read the data sheet of the device. \$\endgroup\$
    – Andy aka
    Apr 28, 2016 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.