The plan is to add in an auxiliary circuit(in red) for the capacitor to charge and discharge when power imbalance occur between the AC and DC side. I could not figure out a way to control the newly added circuit in order to achieve the function above. Can anyone give some advice? Thank you.
Please read my book
The American University Laboratories For Electrical Engineering Part 1
You will find
a good example in it.
Hi, Wuxin Su , I have not personally tried this, but this is my suggestion on what you can try:
1. Let's say the dc-link voltage is Vdc and the battery voltage is Vbat, and the S=0 when lower IGBT is turn on (upper IGBT off) while S=1 when lower IGBT is off (upper is on).
2. The voltage on the left end of the inductor (mid point of the two IGBT) is hence
V1 = Vdc when S=1, and V1 = 0 when S=0. This means that switching the IGBTs controls the voltage on the left end of the inductor, V1 = S*Vdc
3. we can see that the red circuit can be simplified as two voltages across an inductor, such that the inductor current IL flowing into the battery is given by:
dIL/dt = V1-Vbatt = S*Vdc - Vbatt.
4. Since Vdc is higher than Vbatt, we can easily see that switching S=1 will increase IL ( dIL/dt >0) while switching S = 0 will decrease IL (dIL/dt
Hi Wuxin Su , I would suggest the following:
1. build the circuit & control for the AC-DC-AC converter (without the red circuit) first and make sure it works as what you expect. You can use RL load for the DC-AC side to simplify the scenario.
2. once part 1 is done, add the Battery circuit ( red circuit )part.
3. You can first control the Battery circuit in open loop, i.e. give a fixed duty cycle D and observe the inductor current. Then change D and see the effect. You should see that the relation "dIL/dt = D*Vdc - Vbatt " is valid. (if not valid, need to reconsider the approach)
4. Then use a PI controller to control the inductor current. the output of the PI controller will be D. The inductor current should follow the reference current you set.