I have been using an inductor around a 230V AC wire ground-side of the switch to check for current in a wire, it is part of control mechanism.

I have to add a rectifier now, so occasionally the current will be ~400V DC when rectified, and it has to be before the switch.

Will DC current induce voltage in my coil? I think it has to be a varying field, but I cannot check for a few days and its driving me mad!

I cannot put the inductor before the rectifier due to the design of the circuit, and also because it will couple it with the positive mains pin and I get false readings.

Can anyone put my mind at ease?

edit: I have a home made toroidal coil with a ferrite core and a 100ohm resistor across the outputs, that feeds into into a Peak Amplifier,

I am asking if this will work at all/nominally when the current changes from AC to Higher voltage DC.


If you can't put it before the rectifier try this: -

enter image description here

Shown are Current Transformer 1 (CT1), Current Transformer 2 (CT2) and a bridge rectifier.

CT1 is the conventional position for measuring current BUT if you can put two wires through CT2 from the top two diodes shown you will also get the AC measurement. Note the reverse direction the 2nd diode from the top feeds CT2 compared to how the top diode feeds it.

No, DC cannot be measured by a current transformer but if you alternate the DC pulses of current going into the load as shown (by reversing the 2nd diodes current direction through the toroid) then you are feeding it AC current.

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    \$\begingroup\$ That's a very clever solution +1 from me \$\endgroup\$ – JIm Dearden Jul 17 '13 at 8:45
  • \$\begingroup\$ wow! that was outside the box! I cant wait to try it. Shouldn't the field from Diode 2(second from top) cancel out the field from Diode 1 (top) in CT2, or is my electromagnetic field theory flawed? (Very Likely!) Do D1 and D2 alternate on and off due to the Rectifier forcing current here and there, or does D2 alternate with the AC canceling out the field of D1? \$\endgroup\$ – Oliver Kellow Jul 17 '13 at 16:54
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    \$\begingroup\$ Top diode conducts (say) the first half cycle of ac through the CT to the load by entering the CT from the left. Nothing flowing in 2nd diode until the ac voltage reverses (every 20msecs or 16.667msecs depending on where you live). AC reverses and nothing flows in top diode but current flows in the 2nd diode entering the CT from the right. This cons the CT into thinking it is measuring current before the bridge rectifier. \$\endgroup\$ – Andy aka Jul 17 '13 at 16:59

An AC current sensor is basically a step up transformer with a primary of 1 turn. It will only produce an output voltage when the magnetic flux is changing with time. (See Faraday's laws). The induced magnetic flux depends upon the current (DC or AC). A DC component will not produce a continuous output but will magnetize the core and could cause it to saturate.

Consider some different senarios.

enter image description here

(1) Sinusoidal AC

The flux will change sinusoidally and the output will be a sinusoidal voltage (i.e. normal, calibrated operation)

(2) Halfwave rectified.

This contains a DC component that will magnetize the core material in one direction. There will be no output during the 25 - 50mS portion because there is no change in the flux (dB/dt = 0). During 0 - 25mS there will be an output but the effect of the DC component may cause saturation (the core flux can't increase no matter how much current flows). The reading for current will be low and uncalibrated.

(3) Fullwave rectified.

Basically the same as halfwave but with the addition of a doubled frequency AC signal. The readings will be higher than halfwave but uncalibrated also the higher DC level would increase problems with core saturation.

(4) Rectified and 'smoothed'

Worst case senario. The very high DC level would push the core towards saturation. The only AC component would be the 'ripple' at the top of the waveform which (if well smoothed) would be very small. The output reading would be very low and very inaccurate.

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  • \$\begingroup\$ bravo! thankyou, very clear and concise. I don't LIKE the answer, but thats Physic's fault :D Cheers!! \$\endgroup\$ – Oliver Kellow Jul 17 '13 at 7:21
  • \$\begingroup\$ @OliverKellow thanks but I think you should look at Andy's solution - it may be the answer your looking for - the physics is the same but by passing each half cycle through the current transformer it will see it as just normal AC with the benefit of you getting DC. If you're going to smooth the DC (capacitor) add another diode after CT2 to prevent its voltage feeding back to the bridge diodes. \$\endgroup\$ – JIm Dearden Jul 17 '13 at 9:00
  • \$\begingroup\$ @JImDearden you don't need the extra diode and thanks for the +1 \$\endgroup\$ – Andy aka Jul 17 '13 at 9:18

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