I was just going through the purpose of the current transformer and its working & using.

I found some schematics where I found a common pattern as below. It is a single-phase connection. Line and neutral is the input of a single-phase supply.

The line and neutral current connectors are connected to a current transformer.

The outputs I2P/I2N & IP/IN go as inputs to an energy-metering IC.

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Can someone tell me what are the purpose of the resistors and capacitors in the above circuits and how the values are selected?

Also, I'm not able to understand the purpose of the line to VP connection.

I searched about the type of connections, but was unable to find it.

Any idea/guidance on this would be greatly helpful.

  • \$\begingroup\$ R1-R12 form a voltage divider, so VP will give you a scaled down version of line voltage. R10+R13 and R14+R15 forms burden resistors. Please look it up. The rest are low pass filters. \$\endgroup\$
    – winny
    Commented Jun 2, 2023 at 13:08

1 Answer 1


R10 and R13 are very small compared to R19 and R20, therefore almost all transformer current flows via R10 and R13. These two resistors therefore convert that current into a potential difference, as per Ohm's law.

Their centre junction is grounded, ensuring that the voltages across those resistors are in antiphase, and relative to (centered about) the measurement system's own ground.

C2 and R19 form a low pass filter. Higher frequency noise and interference is blocked before arriving at I2P. R20 and C3 perform the same function for I2N.

R1..R6 and R11 form a combined resistance of 840kΩ. In series like that, they share the large 340V potential difference equally between them, ensuring that no single resistor ever has the full 340V across it. They also share any power they dissipate equally between them, instead of having a single resistor getting hot.

When considered as single 840kΩ resistance, they form a potential divider with R12 (1kΩ), so that:

$$ V_P = V_{LINE} \times \frac{1}{840+1} $$

That means that a ±340V line voltage will correspond to about ±0.4V at VP, safe for the following measurement system.

C7 forms another low pass filter with R12, doing the same job as C2 and C3.

  • \$\begingroup\$ Thank you for the answer. I have 2 doubts. 1. When you mentions that the small value resistors R10 and R13, help to convert current into voltages, as per Ohm's law, - All resistors do this stuff right, converting current into voltage? Or is it only the small value resistors help to convert current into voltage? 2. So, the purpose of the center tap connection between R10 and R13 to ground is to ensure that the voltages across the resistors are anti-phase? Why is that important? What happens if I don't ground that point? \$\endgroup\$
    – Freshman
    Commented Jun 2, 2023 at 18:42
  • 1
    \$\begingroup\$ @Freshman If you don't ground the centre node, you still have a differential voltage, but it could be centered around any voltage, called "common mode" voltage, some of which may appear in your measurement (study "common mode rejection" in differential amplifiers to understand why). This "common mode" component of the signal can be influenced by any number of environmental factors, all of which are considered noise. \$\endgroup\$ Commented Jun 3, 2023 at 2:20
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    \$\begingroup\$ @Freshman the signals I2P and I2N would be in antiphase regardless; as I2P goes up in potential, I2N goes down, and vice versa, when measured relative to the junction between R10 and R13, My answer is misleading in that respect, sorry. The grounded (0V) mid-point simply means that \$V_{I2N}=-V_{I2P}\$. \$\endgroup\$ Commented Jun 3, 2023 at 2:24
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    \$\begingroup\$ @Freshman It's true that all resistors convert current to voltage, but in this case we want only R10 and R13 to be responsible for that particular task. For the voltages across R10 and R13 to be proportional to transformer current, all current must pass via R10 and R13. Any current drawn away from that path (any "loading", by R19, C2 etc) will change the current through R10 & R13, introducing an error into the signal being measured. So we must ensure anything connected to them draw/inject as little current as possible, by making them comparatively high impedance. That is \$R_{19} >> R_{10}\$ \$\endgroup\$ Commented Jun 3, 2023 at 2:35
  • \$\begingroup\$ Thank you very much for the answer & the clarifications! \$\endgroup\$
    – Freshman
    Commented Jun 3, 2023 at 7:25

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