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One wishes to measure the RMS current (12A max.) drawn by an AC motor using a PIC16F887 at intervals of 1 second. One cost effective solution I am considering is using a current transformer, a bridge rectifier and then a large-ish capacitor to filter the signal to be fed into the ADC pin. Like this perhaps:

current measurement using CT

  1. Are these calculations correct?

    For max. primary current: 12A RMS
    Turns ratio for this CT: (1/70)
    Secondary current = Primary Current * Turns Ratio = 12*(1/70) = 0.186A RMS
    Burden Resistor: 33 ohms
    Secondary Voltage = Secondary Current X Burden = 0.186 X 33 = 6.14 V RMS
    Therefore Secondary peak-to-peak voltage = 6.14 X 1.414 = 8.68V

    The DC voltage after the capacitor should then be close to 8.68V? Why does the above Proteus simulation show otherwise?

  2. The diode D1 is supposed to avoid over voltage on the ADC pin due to inrush currents of about 40A that happen about every half an hour. Would that be adequate?

  3. Is there any reason this method should not be used at all/won't work?

Edit:
Simulation with the burden before the bridge:

Current measurement using CT

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  • \$\begingroup\$ I haven't used current transformers before but it will be interesting to see a good answer. This might be wrong but as you're only simulating at the moment try putting the burden resistor before the bridge and don't forget the bridge will have a diode forward voltage drop (x2). \$\endgroup\$
    – PeterJ
    Jul 23 '14 at 12:37
  • \$\begingroup\$ @PeterJ The way he has it means the diode drops don't introduce much error or nonlinearity since the output of a properly specified CT is (to first order) a constant output current for a given input current. \$\endgroup\$ Jul 23 '14 at 13:02
  • \$\begingroup\$ @SpehroPefhany, ahh yes that makes sense now I think about it further. \$\endgroup\$
    – PeterJ
    Jul 23 '14 at 13:08
  • \$\begingroup\$ @PeterJ I was hoping current measurement and CTs would be more commonplace but this is turning out to be a less traveled path. I tried putting the burden before the bridge but still couldn't justify the voltage at the ADC pin (which has increased slightly) through any calculations. About the diodes, from what I understand about CTs and as points out, the CT pushes out the rated stepped down current independent of the burden. \$\endgroup\$
    – Sohail
    Jul 24 '14 at 6:04
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That CT only is rated to be accurate down to 50kHz. If you have 50/60Hz, it won't work. A CT rated for accurate operation at 50/60 Hz will be much bulkier.

The bridge should give you peak 8.0V without the capacitor (12A/70 = 0.172A = 0.242A pk * 33R = 8.0V) minus 5% or so average ripple at 50Hz. The CT may not be capable of the burden including the bridge - but looking at it- the bridge introduces two diode drops so it adds to the burden, but it's specified for 70R so 33R + 2 diodes should be okay.

D1 will just ensure you raise up Vcc- you're asking Vcc to sink around 200mA.. It might work with a series resistor from the RC depending on the resistor value your ADC can tolerate and what other loads are on Vcc.

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    \$\begingroup\$ I completely neglected that part about the frequency. The 50Hz ones are bulkier indeed; about three times as big. But about the calculations, here veduprophysics.blogspot.in/2012/12/… and also elsewhere, the peak current (after it is flipped) for full wave rectification is given by Ip = Irms * 1.414. The peak voltage then turns out to be 0.186 * 33 * 1.414 = 8.68V which accounting for ripple could be say 8.2V. Could you point out how the peak is 4.34V \$\endgroup\$
    – Sohail
    Jul 24 '14 at 6:17
  • \$\begingroup\$ Okay, working from scratch.. 12A/70 = 0.172A = 0.242A pk * 33R = 8.0V, minus ripple. So yes, something is wrong-- I'll correct my answer- maybe your simulation is using peak for the input. \$\endgroup\$ Jul 24 '14 at 12:41
  • \$\begingroup\$ Hmm... I'm sceptical about simulations anyways so I wouldn't give it much importance as long as there is a sound manual calculation. I'd have verified this setup physically on a proto but as it turns out, there are hall effect sensors (Allegro ACS712) with analog output that cost about as much as a CT. I think I'll try that instead. Thanks a lot! \$\endgroup\$
    – Sohail
    Jul 26 '14 at 5:21

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