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Newbie question, would really appreciate some help.

I am modelling a simple full-bridge inverter in Matlab. I've used the supplied example file of a full-bridge inverter that was developed by the Matlab engineers.

The DC current appears to be wildly fluctuating / inverting. (see image below)

I have always thought that DC-AC inverters had (by definition) a DC and an AC side and that thus naturally the current on the DC side could only flow in one direction. Also, when I look at the DC input voltage, it is constant and positive.

My questions:
1. How does a DC-AC inverter's input current change with time? (ie. Is this actually how DC-AC inverters behave?)
2. Is it at all possible to have a positive voltage and a negative current, as we see in the model? (or does that break physics....)
3. If this model is insufficient, any suggestions on how I can model a DC-AC inverter?

The matlab model, straight from the built in examples but with a couple of extra scopes for monitoring: The Matlab Topology

The strange-looking DC input current and voltage:DC Input Monitor

The inverter AC output current and voltage:AC Output Monitor

Thanks!

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Yes, it's all about right. The 400 volt supply is an ideal DC voltage source, so its output shows absolutely no variation, even though its current has horrendous spikes.

You'll notice that the input current spikes have an approximate sine envelope, but modulated, And you'll notice that the crossover point is not at 0 volts amplitude. This is because of the 5 mH inductor on the output.

The output voltage has "bunches" of voltage spikes which are all positive or negative, and which correspond to the positive and negative divisions of the current waveform. Those spikes are the direct output of the bridge, so of course they are spikes. If you expand the waveform, you'll see that the pulse width of those output spikes corresponds to the amplitude of the current waveform.

What you want to do in order for things to seem to make more sense is increase the 1 ohm load to 10 ohms, and swap it with the inductor, so that one end is now grounded. If you put a scope across the resistor, I think you'll find it closer to your expectations.

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  • \$\begingroup\$ Thank you so much for your super-fast and really helpful response, I'm going to go try that now :) You made my first stack exchange experience really great. \$\endgroup\$ – Luke Jan 30 '15 at 2:58

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