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Some explanations for the circuit below:

  1. This is a CT with 1000 turns secondary. It has high sensitivity down to mA levels so that's why I put an amplifier given that the burden is small (35 ohms).
  2. The CT has 1000 additional turns for the test winding. So the ratio is 1:1 between secondary and test winding.
  3. The reference voltage IC is configured to give Vcc/2 = 1.65V which is the DC shift. I'm planning to use a square wave from the micro followed by a high pass filter to feed the test winding. It just need a small current during a quick moment. The switch is to open the winding when no testing. The general purpose of the test winding is, of course, to test the system, so I don't necessarily need to put current in the primary.

My question is: Do I need to connect the test winding to the DC shift like the sense winding? Or do I need to connect it to ground instead?

enter image description here

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Do I need to connect the test winding to the DC shift like the sense winding? Or do I need to connect it to ground instead?

You can connect it directly to ground because, with your drive circuit using a series capacitor, it fulfils the AC requirements when driving a winding of any transformer. But, you could connect it to any DC voltage within reason but there is no advantage so, connect it to ground is my advice.

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The L1 current just runs around in a loop between L1 and R16 so it doesn't load the VCC/2 reference.

The current from the micro has to return to the micro through ground and if you reference L2 to the VCC/2 point then the return current may affect the reference. Connect it to ground.

Have you considered what voltage will appear across L2 when SW1 is open and the primary current shows up?

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  • \$\begingroup\$ As I have a small burden across L1 I know the reflected voltage across L2 will be small when SW1 is open. Am I right? \$\endgroup\$ – user115094 Apr 23 '20 at 21:20
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    \$\begingroup\$ I'm not sure. I always treat CTs as complimentary to VTs. A VT with two secondaries wouldn't be happy with one winding short circuited, even if there was a reasonable load on the other. The short-circuit current would be too high. A CT with two secondaries, in my logic, wouldn't like an open circuit secondary. The open-circuit voltage would be too high and cause flashover. Someone with more knowledge should be along shortly. \$\endgroup\$ – Transistor Apr 23 '20 at 21:54
  • \$\begingroup\$ @Andy aka do you have any comment on this? \$\endgroup\$ – user115094 Apr 23 '20 at 22:56
  • \$\begingroup\$ @Blue_Electronx assuming C18 has low AC impedance, the current into the extra winding is determined by the drive voltage, R29 and the burden resistor on the proper winding. Given that these windings are 1:1, in effect, the burden resistor is seen as a direct load to the drive voltage feeding via R29. \$\endgroup\$ – Andy aka Apr 24 '20 at 9:05
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    \$\begingroup\$ Both have the same turns hence, the burden resistor will be "seen" at the test winding as if it were a proper resistor placed there. That's what transformers do - they turn an impedance on one winding to be seen on the other via the turns ratio squared. Given that N = 1000 on both, the ratio squared is still unity. \$\endgroup\$ – Andy aka Apr 24 '20 at 14:10

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