# Neutral Point Diode Clamped Inverter Switching States

Can someone explain to me the 3 main switching states of the following inverter? I understand the +Vd/2 and 0 output states but I do not understand how there is a (-)VD/2 state unless it has to do with the flyback diodes. 1 is considered actuated. If Sa1 is 1 nSa1 is 0 according to the text this was pulled from for clarification.

You are being confused by the rather cryptic drawing and logic table.

Note the output is termed $V_{aN}$ That implies the measured output is the voltage difference between node $a$ and node $N$. Since node $N$ is already biased at $V_{dc}/2$ when the output is high $V_{aN} = V_{dc}/2$, when the output swings to the other rail $V_{aN} = -V_{dc}/2$.

However the truth table seems a tad messed up. From my understanding output low should be 0,0 on the inputs. Further the state 1,0 should produce open circuit, or high-impedance, output.

• I also thought 0,0 would be the low state which is why I was confused originally. 1,0 is not causing a short because of the diode orientation correct? I did pull this directly from an IEEE review so i'm not sure if the table is actually incorrect. – Paul Feb 14 '18 at 15:12
• @Paul correct. re source. mistakes happen, even in the most prestigious papers. – Trevor_G Feb 15 '18 at 4:05

For connecting the output (a) to (+Vdc/2) you can put Sa1 and Sa2 to on-state, for connecting (a) to the neutral point (N) you might put Sa2 and nSa1 into on-state, and for connecting (a) to (-Vdc/2) you might have the two lower switches nSa1 and nSa2 in on-state (while the not-mentioned switches are always off). This will work for both directions of (inductive) load current, output voltage at (a) will be current-independent.

• I should've added that the complimentary switches are always the opposite of what is on the trtuh table for their paired switch. – Paul Feb 14 '18 at 15:04
• Your table was a little bit confusing, I tried to be more clear. – UweD Feb 14 '18 at 18:30