How to transform the line voltages to the phase voltages in control software?

Question

Let's say I am measuring line voltages of the three phase grid 400V/50Hz via three instrument transformers whose secondary windings are connected to the adc of my microcontroller i.e. I have following configuration

The control algorithms inside the microcontroller software require the phase voltages instead of the line voltages. So I need to transform the line voltages onto the phase voltages. Due to the fact that I have been working with the three phase quantities I have decided to use the space vectors to represent them i.e. use the Clarke transformation.

My idea how to do the transformation of the line voltages onto the phase voltages was following

where the block \$abc\to\alpha\beta\$ realizes the Clarke transformation according to the following matrix equation $$ \begin{bmatrix} v_{dg\alpha} \\ v_{dg\beta} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{bmatrix} \cdot \begin{bmatrix} v_{ab} \\ v_{bc} \\ v_{ca} \end{bmatrix} $$ and the block \$\Delta\to Y\$ realizes the transformation of the space vector of phase to phase voltages onto the space vector of the phase voltages according to the following matrix equation $$ \begin{bmatrix} v_{yg\alpha} \\ v_{yg\beta} \end{bmatrix} = \frac{1}{\sqrt{3}} \cdot \begin{bmatrix} \cos{\frac{\pi}{6}} & -\sin{\frac{\pi}{6}} \\ \sin{\frac{\pi}{6}} & \cos{\frac{\pi}{6}} \end{bmatrix} \cdot \begin{bmatrix} v_{dg\alpha} \\ v_{dg\beta} \end{bmatrix} $$

The idea behind the second transformation is based on the phasor diagram

i.e. it is based on the fact that the line voltages lead the phase voltages by the \$\frac{\pi}{6}\$ and their magnitudes are \$\sqrt{3}\$ times greater.

My question is whether the above suggested approach is correct also in case the three phase grid is not ideal i.e. under unbalanced conditions or harmonic distortions?

Answer 1

You cannot get the exact phase-ground voltages because the delta connection removes the zero sequence component. For example, below I show the symmetrical component equation for a & b phase voltages in an abc rotation system,

$$V_{an}=V_0 + V_1 + V_2 $$ $$V_{bn}=V_0 + \alpha^2V_1 + \alpha V_2 $$

So, the a-b voltage will be, $$V_{ab}=V_{an}-V_{bn}$$ $$V_{ab}=(V_0 + V_1 + V_2)-(V_0 + \alpha^2V_1 + \alpha V_2)$$ $$V_{ab}=V_1(1-\alpha^2) + V_2(1-\alpha)$$

So, you can see that the zero sequence is removed by the delta connection, never to return.

If you want to assume the system is perfectly balanced (with no zero-sequence or negative-sequence) then you can just shift by 30° and scale by \$\frac{1}{\sqrt3}\$ as per your phasor diagram.

Also, you could change the transformer to a Ynyn (wye-grounded/wye-grounded) and get exactly what you need.

Doing the Clarke transformation would buy you nothing with your present configuration.

UPDATE: Answering comment question about calculating phase-ground voltages from phase-phase voltages ignoring zero-sequence.

From C.L. Fortesue's definition of his symmetrical components we know (where \$\alpha=1∠120°\$), $$V_{ab} = (V_0+V_1+V_2)-(V_0-\alpha^2V_1-\alpha V_2)$$ $$V_{bc} = (V_0+\alpha^2V_1+\alpha V_2)-(V_0-\alpha V_1-\alpha^2V_2)$$ $$V_{ca} = (V_0+\alpha V_1+\alpha^2V_2)-(V_0-V_1-V_2)$$

which reduces to,

$$V_{ab} = V_1(1-\alpha^2)+V_2(1-\alpha)$$ $$V_{bc} = V_1(\alpha^2-\alpha)+V_2(\alpha-\alpha^2)$$ $$V_{ca} = V_1(\alpha-1)+V_2(\alpha^2-1)$$

Now taking the differences of the above we can create a set of equations to solve,

$$V_{ab}-V_{bc} = V_1(1-2\alpha^2-\alpha)+V_2(1-2\alpha+\alpha^2)$$ $$V_{bc}-V_{ca} = V_1(\alpha^2-2\alpha+1)+V_2(\alpha-2\alpha^2+1)$$

which reduces to,

$$V_{ab}-V_{bc} = V_1(3∠60°)+V_2(3∠{-60°})$$ $$V_{bc}-V_{ca} = V_1(3∠{-60°})+V_2(3∠60°)$$

and in matrix form,

$$\begin{pmatrix}3∠60° & 3∠{-60°} \\\ 3∠{-60°} & 3∠60°\end{pmatrix} \begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} V_{ab}-V_{bc} \\\ V_{bc}-V_{ca} \end{pmatrix}$$

which is easy to solve and subsequently compute the phase-ground phasors.

Example (using per unit and assumes abc rotation system): Let \$V_a=1∠0° pu\$, \$V_b=1.12∠-130° pu\$, and \$V_c=0.95∠122° pu\$

$$V_{ab}=1.924∠26.54° pu$$ $$V_{bc}=1.680∠{-97.46°} pu$$ $$V_{ca}=1.706∠151.81° pu$$ So, $$V_{ab}-V_{bc}=3.18∠52.5° pu$$ $$V_{bc}-V_{ca}=2.78∠{-62.5°} pu$$

$$\begin{pmatrix}3∠60° & 3∠{-60°} \\\ 3∠{-60°} & 3∠60°\end{pmatrix} \begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} 3.18∠52.5° \\\ 2.78∠-62.5° \end{pmatrix}$$

with solution, $$\begin{pmatrix} V_1 \\\ V_2 \end{pmatrix}=\begin{pmatrix} 1.019∠{-3°} \\\ 0.091∠51.3° \end{pmatrix}$$

Plugging these values into the fundamental symmetrical component equations (with assumption \$V_0=0\$) we have,

$$V_{an}=V_0 + V_1 + V_2=1.075∠0.94° $$ $$V_{bn}=V_0 + \alpha^2V_1 + \alpha V_2=1.06∠{-127.5°} $$ $$V_{cn}=V_0 + \alpha V_1 + \alpha^2V_2=0.928∠117.56° $$

Now, as a check, since we know \$V_0=(1∠0°+1.12∠-130°+0.95∠122°)/3 = 0.076∠-166.8° pu\$ we can add it to each of the above 3 values to get the actual original phase-neutral voltages. This shows us the error we could expect if zero-sequence voltage is not negligible (depends on what you are trying to accomplish).

$$V_{an}=V_0 + 1.075∠0.94° = 1∠0°$$ $$V_{bn}=V_0 + 1.06∠{-127.5°} = 1.12∠-130°$$ $$V_{cn}=V_0 + 0.928∠117.56° = 0.95∠122°$$

Finally, if what you are really interested in are the phase-ground voltages on the high-voltage side (left side in your first drawing) then you simply need to translate \$V_1\$ and \$V_2\$ across the transformer first. The transformer you drew has low-side leading by 30° (\$V_{an}\$ in phase with \$V_{AB}\$). So, you would multiply \$V_1\$ by the voltage ratio (HV ph-ph divided by LV ph-ph) and shift it's phase angle by \$-30°\$. You would multiply \$V_2\$ by the voltage ratio and shift it's angle by \$+30°\$. From those 2 quantities you can now calculate the high-voltage side phase-ground voltages (of course, assuming \$V_0=0\$).