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I am trying to calculate the equivivalent capacitance to ground of a 3 phase LCL filter with capacitors to ground. Inductors removed from circuit as I assume they will be ignored by the meter...

schematic

simulate this circuit – Schematic created using CircuitLab

Ignoring C3, it`d be simply 2 series networks in paralell with C4. Specifically:

C4 + 1/(1/C1 + 1/C5) + 1/(1/C2 + 1/C6)

This almost fits with measurements of actual circuit (3% less), small enough that could be explained by the capacitors +-10% tolerance. However what effect does C3 have in such a calculation?

Update Inductors are back in the schematic from comments suggestions, at the cost of a much uglier schematic. I think this is not necessary as the probe should only send a very low frequency current and then measure the change in voltage. And as frequency approaches zero, reactance of inductors also approaches zero, while reactance of capacitors approaches infinity. In effect making inductor values negligible. Therefore I think this simplified circuit is ok to use:

schematic

simulate this circuit

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    \$\begingroup\$ The "equivalent capacitance" should probably be calculated from the negative reactance. So you should not remove the inductors unless you find that their effect is negligible at your operating frequency. \$\endgroup\$ – The Photon Feb 15 '16 at 13:05
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    \$\begingroup\$ Use impedance values for components, then use a wye-delta equivalent to get your final result. Can you do it, or should I expand this into an answer? \$\endgroup\$ – Vicente Cunha Feb 15 '16 at 15:43
  • \$\begingroup\$ @The photon - The calculation is not for designing an LCL filter in practice, but to verify if caps still hold their original value. And tested with a fluke capacitance meter function, which I presume input a controlled DC current and measure the changed voltage over x amount of time. As such I assume any inductors should not change measurements. Please correct me if I`m wrong. \$\endgroup\$ – Imbrondir Feb 15 '16 at 22:11
  • \$\begingroup\$ @Vicente - Thanks for the wye delta clue. This is new to me, I`ll see if I can do it myself. \$\endgroup\$ – Imbrondir Feb 15 '16 at 22:12
  • \$\begingroup\$ @Imbrondir The inductors you removed should be represented in the second circuit as short-wires, and not as open circuit branches. \$\endgroup\$ – Vicente Cunha Feb 16 '16 at 11:25
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Your circuit, eliminating inductors due to low frequency, should look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Let's redraw it in a more clear fashion.

schematic

simulate this circuit

The capacitor pairs in parallel have equivalent capacitance given by the sum of their values. So this simplifies to this pretty little circuit:

schematic

simulate this circuit

Now I will implement a delta-wye transformation on capacitors C1, C2 and C3, in the shape of new capacitors Cx, Cy and Cz, ending up with a circuit like this:

schematic

simulate this circuit

Capacitance seen from probe is then:

$$ C_{probe} = C4 + \frac{1}{\frac{1}{Cx} + \frac{1}{C_{eq}}} $$

Where Ceq is given by:

$$C_{eq} = \frac{1}{\frac{1}{Cy} + \frac{1}{C5}} + \frac{1}{\frac{1}{Cz} + \frac{1}{C6}}$$

Ok, now down to calculating Cx, Cy and Cz. Capacitor impedance Z is inversely proportional to capacitance:

$$Z = \frac{1}{sC}$$

Therefore, wye-delta equations for capacitance must be "flipped" upside down (that is, from here, where one reads R, change it to 1/C). This, in turn, has the same effect as observed in parallel-series duality. Resistors wye-to-delta is equivalent to capacitors delta-to-wye. Since we have performed a delta-wye, the equations are such:

$$C_x = \frac{C_1 C_2 + C_2 C_3 + C_3 C_1}{C_3} = \frac{3*204^2}{204} = 612\mu F$$

$$C_y = \frac{C_1 C_2 + C_2 C_3 + C_3 C_1}{C_2} = \frac{3*204^2}{204} = 612\mu F$$

$$C_z = \frac{C_1 C_2 + C_2 C_3 + C_3 C_1}{C_1} = \frac{3*204^2}{204} = 612\mu F$$

Finally, coming to our answers:

$$C_{eq} = \frac{1}{\frac{1}{612} + \frac{1}{25}} + \frac{1}{\frac{1}{612} + \frac{1}{25}} = \frac{30600}{637} \mu F$$

$$C_{probe} = 25 + \frac{1}{\frac{1}{612} + \frac{637}{30600}} = \frac{15925}{229} = 69.54 \mu F$$

Hope this helps.

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  • \$\begingroup\$ It did. Great work on an answer. I learned a lot. Thanks :) \$\endgroup\$ – Imbrondir Feb 16 '16 at 18:34
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I`ll try to answer my own question with Vicents suggestion. Applying wye-delta transformation gets an analyzable circuit:

Ca = Coposite/(C1*C2 + C1*C3 + C2*C3)
Or since C1 = C2 = C3 = Cd (delta):
Cy = Cdelta*3

Further simplifications
Cg = C4 = C5 = C6
leads to 
Ctotal = Cg + 1/(1/Cy + 1/(2/(1/Cg + 1/Cy)))

Which coincidentally gives exactly the result with my values as my naive approach of pretending the original C3 didnt exist.

Hopefully somebody in the comments will correct me if the inductors will skew my results either way.

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    \$\begingroup\$ The only reason why C3 can be excluded from your equation is because of the same principle as "Wheatstone bridge-simmetry": the ratio C1/C5 equals the ratio C2/C6. If you were to reattempt your calculations with new values that don't follow this, C3 will show up. \$\endgroup\$ – Vicente Cunha Feb 16 '16 at 11:18
  • \$\begingroup\$ Heads up though: you have excluded your inductors in a wrongly manner! They should be considered as short-wires in low frequencies, not as open circuits! \$\endgroup\$ – Vicente Cunha Feb 16 '16 at 11:23
  • \$\begingroup\$ Thanks, but they are short circuited. The caps to ground now become 2 caps in parallel. I realize I didnt make that clear though. \$\endgroup\$ – Imbrondir Feb 16 '16 at 18:31

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