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I am wondering how to properly measure the CMRR vs. frequency of a(n) (audio) transformer circuit. I have found a couple of methods but not sure I understand them entirely.

The main idea I am going off of is from page 7 of this document, with image shown below.

enter image description here

For normal mode, I connect the frequency generator +/- across the primary coil, frequency sweep at a set Vin and record Vout. Simple enough. But for common mode test, I don’t understand where to connect the generator negative.

From this source, we have:

enter image description here

This suggests connecting the gen (neg) to the center tap of the primary. This only works if you have a center tapped transformer, and low ohm transformers load the signal generator heavily (at least in my attempt at this).

On the datasheet of a Jensen JT-11P-1, they use the following circuits:

enter image description here

In this case I don’t understand how this accurately measures common mode, since unlike with a normal differential connection, there is no closed circuit (other than maybe capacitive leakage to the secondary?). And returning to the first schematic where a 10 Ohm imbalance is used for testing, I’m not seeing why either leg of the primary would be at a voltage other than Vin from the generator since no current is flowing.

So what is the proper circuit setup to measure transformer CMRR and how should it be measured?

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2 Answers 2

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Common mode rejection ratio in this configuration is measured by connecting both input terminals to generator positive and DUT grounding, enclosure or whatever that is supposed to be at the zero potential to negative. Hard to give concrete advice without more info about your DUT.

In your first picture negative is connected to DUT enclosure, in later pictures, it is connected to both enclosure and transformer shield winding.

You are correct about coupling capacitance: it is the primary common mode path in transformer decoupling circuits. Your later pictures have shield winding that is meant to reduce mutual capacitance between windings and increase CMRR.

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  • \$\begingroup\$ The circuit I’m trying to measure is actually very similar to the Jensen datasheet images, the only difference being that the transformers in my case are center tapped (although I’m not using the CT), there are no shields and there is a slightly different secondary load. Thinking about your answer, since common mode is by definition relative to earth ground, I see why you would want to connect the generator negative to earth ground. But when dealing with common-mode noise issues, is the main problem that the common-mode voltage turns into a differential voltage at the secondary? (cont) \$\endgroup\$
    – User7251
    Oct 2, 2017 at 23:39
  • \$\begingroup\$ Or is the issue that there is still common-mode voltage at both secondary legs relative to earth? Or both? I’m guessing that if the secondary is earth grounded, that would shunt any leftover CM voltage to ground. Correct? This is how I understand most audio equipment to be wired, so differential noise would be the problem. The way Jensen shows the test circuit, measuring Vout during the common-mode test seems to be a differential measurement, and that is how it is shown in the first schematic from the IEC as well. \$\endgroup\$
    – User7251
    Oct 2, 2017 at 23:40
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    \$\begingroup\$ Yes, the problem is that CM appears at the output together with differential mode signal. It does not matter how it gets there: CMRR simply states that CM signal in input is reduced that many times at the output. You can see in Jensen pictures that you must define how your transformer is connected to do a proper measurement. In this connection CM turns into some amount of DM at the secondary because the secondary is grounded. IEC example has a diff. output and Jensen's - single-ended. It does not matter how you measure the output amplitude, it matters how much CM from input appears there. \$\endgroup\$
    – Jurkstas
    Oct 3, 2017 at 5:31
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Imbalances in capacitances, from primary to secondary and specifically to the upper half and the lower half of the secondary, define your CMR.

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