I have a non invasive current sensor SCT 013-030 and I connected it to an AVR mcu using this circuit below (circuit 1).

Circuit 1.


simulate this circuit – Schematic created using CircuitLab

Circuit 2.


simulate this circuit

The questions are:

  1. What does the C1 on the circuit 1 do? What's its purpose on the circuit?
  2. What's the difference about circuit 2? Is it better? Why someone would chose circuit 2 instead of circuit 1?

I understand that: Circuit 1 is way cheaper. Circuit 2 provides much more safety to microcontroller adc input and it's got low output impedance. Also the phase of the signal is inverted which I think it doesn't bother much. Am I missing something more important than that?

  • 1
    \$\begingroup\$ In the first circuit, C1forms a RC filter with R1 || R2, with RC = 0.500 sec \$\endgroup\$ – Huisman Jun 8 at 11:03

first circuit

The purpose of the first one is to build a stable virtual ground at 3.3 V / 2, which you use as reference potential for your sensor.

Your ADC senses the voltage over the sensor. It's important that the ADC has a high input impedance (>> 100 kΩ, which is relatively much for low-noise ADCs), otherwise it doesn't work, because the sensor then starts to shift the virtual ground instead of the ADC input.

second circuit

Opamp is in a negative feedback configuration, so that the voltage difference at the inputs becomes 0 V. Since no current can flow into an Opamp input, that means that the inverting input is pulled to 3.3 V / 2.

From that directly follows that on the left hand side of R4, there's V_sensor, on the right hand side always 3.3 V / 2.

Therefore, R4 conducts a current of


Since that current can't go anywhere, the current through R3 is \$I_3=I_4\$ and hence the voltage at the right hand side of R3 is the 3.3 V / 2 on its left end, plus the voltage across R3, so

\begin{align} V_\text{ADC} &= 3.3 \,\text{V} + I_3 \cdot R_3 \\ &\overset{I_3=I_4}= 3.3 \,\text{V} + I_4 \cdot R_3 \\ &= 3.3 \,\text{V} + \frac{V_\text{sensor}-3.3\,\text{V}/2}{R_4}R_3\\ &\overset{R_3=R_4}=3.3 \,\text{V} + V_\text{sensor}-3.3\,\text{V}/2\\ &=3.3 V /2 + V_\text{sensor} \end{align}

which is the same as for the first circuit, but with an important difference:

The internal source impedance of that \$V_\text{ADC}\$ voltage source is no longer defined by the virtual ground and the current sensor, but by the output impedance of the opamp!

Hence, you can use opamps with much lower input impedances; in fact, what you often do is even have a termination resistor really close to the ADC input, and use the ADC to measure the voltage across that, instead of terminating the line into the ADC itself.
That has the effect that noise voltage (whose sources we typically model as high-internal impedance voltage sources) "breaks down" over the strong termination resistor, whereas your signal of interest is "strong enough" and can be, without degradation, be measured.

Also the phase of the signal is inverted which I think it doesn't bother much.

Not quite sure what you want to measure, but it would bother me very much if my signal's phase was changed. You need to account for anything like that!

  • \$\begingroup\$ I'm afraid your math is all wrong. You need to pay attention to the sign of the currents you're talking about, and your initial statement \$V_{\text{ADC}} = 3.3\text{ V} + I_3 \cdot R_3\$ is simply incorrect. \$\endgroup\$ – Dave Tweed Jun 8 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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