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I'm simulating the circuit below in LTspice and there's something I don't get regarding C3 and C4. Suppose the input signal is a 32 uArms 60Hz current as shown. This value multiplied by the burden resistor gives (32uA * 1.91k) = 61.12 mVrms. However, in the simulation if I measure the voltage at the burden resistor, I get about 30mVrms signal. What is the effect of C3 and C4 that is causing the input signal to be reduced?

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    \$\begingroup\$ Since U1 input is a virtual earth, consider C4 and R6 are in parallel with R5 (and C3). At 60Hz, you do the math... Also, what is the point of setting U1 gain to 250ish? At 61mV rms input, the amplifier will be well into clipping. \$\endgroup\$ – Brian Drummond Feb 22 at 22:47
  • \$\begingroup\$ This circuit is pretty old and I'm trying to modify it for a microcontroller using an ADC 0-3.3V. It detects currents from 30mA to 30A varying the resistors R5 and R7. The CT is 950 turns. It seems a complex approach. I don't the point of some things. \$\endgroup\$ – Blue_Electronx Feb 22 at 22:56
  • \$\begingroup\$ For 30 mA primary current, (~30 uA secondary), R5=1.91k and R7=511k) \$\endgroup\$ – Blue_Electronx Feb 22 at 22:57
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    \$\begingroup\$ @Blue_Electronx Are you asking for a complete discussion about this circuit? Or just a summary of why you observe about half of the signal amplitude you expected? One is more time consuming that the other. Brian's point is that the (-) input of the opamp is basically a ground (virtual ground) because of the opamp's behavior and where its (+) input is tied. So C4 and R6 (series) are effectively in parallel to R5 and C3. C4's impedance is quite small compared to R6 at 60Hz, so you basically have a 2k resistor in parallel. As R5 is about that value, the parallel combination is about half. \$\endgroup\$ – jonk Feb 23 at 3:43
  • \$\begingroup\$ I'd like a complete discussion on this circuit. However, as per the post, the question has been answered. \$\endgroup\$ – Blue_Electronx Feb 23 at 3:47

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