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# Diode Clamped Multilevel Converter

#### peterwarzeck

##### New Member
Hi Guys, i am stuck with this problem:
The image below shows a single Inverter Leg (phase a) of a Diode Clamped Multilevel Converter with M=4 capacitors.
If a three phase inverter has 3 such identical legs (as shown below) connected in Y with common node at N and line terminals (A,B,C), what is the maximum RMS line voltage possible in terms of E.

Based on my understanding, my answer is:

My reasoning:
The maximum attainable RMS phase voltage without distortion for one leg is given by M*E/sqrt(2), where M is the number of levels minus one and E is the voltage of one DC source. The relationship between the line voltage and phase voltage in a 'Y' configuration is to multiply the line voltage by sqrt(3). Hence i got my solution.

Please let me know if my method is correct or correct me if i am wrong. Thank you

The circuit you've shown should be able drive the voltage at A between 0 V and 4 * E. That would give the peak - peak voltage of each phase as being 4 * E. The peak voltage would be 2 * E. The RMS voltage would be sqrt(2) * E, and the RMS line voltage would be sqrt(6) * E.

The N point on the circuit would not be the the neutral or star point of a 3 - phase output because the circuit can only drive the voltage positive compared to the N point.

W
The circuit you've shown should be able drive the voltage at A between 0 V and 4 * E. That would give the peak - peak voltage of each phase as being 4 * E. The peak voltage would be 2 * E. The RMS voltage would be sqrt(2) * E, and the RMS line voltage would be sqrt(6) * E.

The N point on the circuit would not be the the neutral or star point of a 3 - phase output because the circuit can only drive the voltage positive compared to the N point.
When u said that the peak voltage is 2*E, then shouldn't the RMS voltage be sqrt(2)*2*E?

W

When u said that the peak voltage is 2*E, then shouldn't the RMS voltage be sqrt(2)*2*E?
The peak voltage is sqrt(2) * the RMS voltage, it's larger than the RMS voltage.
The peak to peak voltage is twice peak voltage.

If the peak to peak voltage is 4*E, the peak voltage will be 2*E and the RMS voltage will be 2*E/sqrt(2) which is sqrt(2)*E

I agree with Diver, but your assumptions do not compute.

If N is common to Neutral then how do you generate -ve peak with 3 half bridges connected between A & N to Y{A,B,C} & Neutral?

I think the confusion may be that the 5 state levels are complementary switching.
+1 V/2
+2 V/4
+0 0
-2 -V/4
-1 -V/2 So the filtered Vp is only E and not 2E

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