Originally, my plan was to just get the voltage output of a current transformer (obviously with the burden resistor circuit stuff and DC biasing circuit stuff) before ADC'ing it, averaging it’s output per second to get the average current per second. However, I will be using all 6 analog pins in the uC to get 6 outputs from a 60hz current transformer, and the uC only has one ADC, which will make my readings kind of inaccurate. Or is it really inaccurate? Please enlighten me.

enter image description here https://openenergymonitor.org/forum-archive/node/156.html

I was thinking of rectifying the output of the transformer first and using a capacitor to smooth out the output to get a more stable reading, but since I will be dealing with small voltages, I don’t think that's a viable option. Is there some kind of trick in rectifying small signals with as little drop out as possible?

enter image description here Source: https://www.electronics-notes.com/articles/analogue_circuits/diode-rectifiers/full-wave-bridge-rectifier.php

Also, I was wondering if it is possible to use a voltage doubler circuit to boost the voltage output of this transformer, since the highest voltage that can be attained by this circuit is 2Vp, but is this really 2Vp? Or is there also a voltage drop from the diodes that needs to be taken into consideration?

enter image description here https://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html

  • \$\begingroup\$ Enlightenment comes when you present your proposal as a circuit. \$\endgroup\$ – Andy aka Sep 12 at 9:11
  • \$\begingroup\$ The circuits aren't really that complicated as they can be seen on Google, but I will add it. \$\endgroup\$ – thisjt Sep 12 at 9:29
  • \$\begingroup\$ A voltage doubler needs the input voltage to be higher than the forward voltage of the diodes to work. \$\endgroup\$ – JRE Sep 12 at 9:30
  • \$\begingroup\$ That's the reason why it’s a bad idea to use a rectifier. My question is does anyone have some kind of clever way of rectifying while eliminating the voltage drop that a diode introduces. \$\endgroup\$ – thisjt Sep 12 at 9:34
  • \$\begingroup\$ Have you heard of "precision rectifiers?" That would at least get you a DC representation of your AC current. \$\endgroup\$ – JRE Sep 12 at 9:54

A CT can be generally regarded as having a current output that is proportional to the current flowing through the primary. This is converted to a voltage by the burden resistor. So, if you apply a bridge rectifier between CT secondary winding and burden resistor you get a pretty good approximation to a full-wave precision rectifier: -

enter image description here

A word of warning; don't push it too far - you should be able to get a few volts peak-to-peak but don't aim to get a peak voltage higher than what would be obtainable with the burden placed conventionally because you might suffer a little bit of inaccuracy and, if you take it too far, you might start to overly saturate the CT's core. In other words, don't choose a burden resistor value that is significantly higher than that recommended by the data sheet.

  • \$\begingroup\$ I was thinking of doing it like this. However, won't the two diodes provide a voltage drop of 0.7? Let’s say that I get a reading of 2 Volts from the CT. Due to the voltage drop that the diode will provide, won’t that change my value to 0.6 Volts? (0.7 V times 2 diodes) \$\endgroup\$ – thisjt Sep 12 at 15:59
  • \$\begingroup\$ No, a CT provides a constant amplitude current AC waveform on the output and that constant current waveform will automatically forward bias the diodes into conduction without losing any diode drop on the burden resistor. The voltage on the input to the bridge will look a little strange but that's unimportant. Try using a simulator and see. \$\endgroup\$ – Andy aka Sep 12 at 17:37
  • \$\begingroup\$ Ohhhh. Makes perfect sense now. Didn't realize it until you pointed out that a constant current will drive a diode, not caring about the voltage drop. Thank you. \$\endgroup\$ – thisjt Sep 12 at 17:50

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