Averaging matrix is not invertible in state space averaging

I am having a problem with state averaging method. The circuit is a SIDO converter as in the figure below. This circuit is just an example to show the problem. So please don't care much about the structure and its efficiency.

The problem is that the averaing matrix is not invertible (singular or degenerate matrix). For the detailed description of the problem and my calculation, please refer to the PDF file showing my work in detail.

Question:

1.What state variable is redundant in this case?
2.How to solve this problem so I can get conversion gains?

How can this work? I don't see-it anywhere but this should be a DC converter ? There is no DC path to Vo1 and Vo2, the average will always be zero. Are you sure Cf1 and Cf2 are in the right place? Switching to the configuration in Page 2 is not possible, it will lead to infinite current since on the Vin - Cf1 - C1 loop there are no resistive or inductive loads. Also on Vin - Cf1 - C1 loop. It's not about the efficiency, revise the structure of the converter, it cannot work, no wonder it's not solvable.

• I agree with you Dorian. I searched almost everywhere but I wasn't able to find anybody use this type of topology. However, after long time of thinking I got something I am not sure whether it is wright or not, I need to revise it once again. I would like also to see what @anhnha would say about the applicability and the source of this circuit. Commented Mar 29, 2018 at 13:47
• You are missing the main issue. There is no DC path to Vo1 and Vo2. The matrix described by Haz is inveritble just because he added a weak one through a parasitic resistor in parallel with Cf. Commented Mar 31, 2018 at 10:33
• It looks like you may have accidentally created two accounts. See this link for help: electronics.stackexchange.com/help/merging-accounts
– W5VO
Commented Mar 31, 2018 at 16:33
• @Dorian: thank you. I see the point. I would like to give an example later. Hope you can join. I gave bounty to Haz because I don't want to waste 200 point and no one received it. I will find a good example later. Commented Mar 31, 2018 at 22:21

S1a and S2b both bypass the inductor, taking it out of the circuit. If Sn alternates on/off with {S1b, S2a} then you have a charge pump effect.

VO1 = VO2 is greater than Vin/2, using L1 as a switch mode inductor.

If only S1b or S2a toggle with Sn then only V01 or V02 would be greater than Vin.

S1a and S2b are redundant and would make the circuit not work.

I am going to just assume that this is a "mathematical problem" rather than an engineering problem. As @Sparky256 and @Dorian have mentioned, the circuit topology is questionable and not practical.

I think that the main cause of non-invertible matrix is the floating capacitors CF1 and CF2 in state 2 and 3 respectively. To avoid this I considered parasitic of CF1 and CF2. The parasitic of a cpacitor can be modeled as series resistance and inductor and a parallel resistance as shown below, which is taken from [1].

I neglected ESR and ESL and took only into consideration EPR to avoid the floating capacitors(CF1 and CF2 only). I re-derived A1, A2 and A3 and calculated Aavg using following Matlab code, where rp1 and rp2 are corresponding to cf1 and cf2 respectively.

syms c1 c2 cf1 cf2 r1 r2 rp1 rp2 l d1 d2
A1=[0 0 0 0 0; 0 -1/(r1*(c1+cf1)) 0 1/(rp1*(c1+cf1)) 0; 0 0 -1/(r2*(c2+cf2)) 0 1/(rp2*(c2+cf2)); 0 1/(r1*(c1+cf1)) 0 -1/(rp1*(c1+cf1)) 0; 0 0 1/(r2*(c2+cf2)) 0 -1/(rp2*(c2+cf2))];
A2=[0 -1/l 0 1/l 0; 1/c1 -1/(r1*(c1)) 0 0 0; 0 0 -1/(r2*(c2)) 0 0; -1/cf1 0 0 0 0; 0 0 0 0 -1/(rp2*c2)];
A3=[0 0 -1/l 0 1/l; 0 -1/(r1*(c1)) 0 0 0; 1/c2 0 -1/(r2*(c2+cf2)) 0 0; 0 0 0 -1/(rp1*c1) 0; -1/cf2 0 0 0 0];
Aavg=d1*A1+d2*A2+(1-d1-d2)*A3;


This results is following Aavg,

/                            d2                                               d2                       \
|       0,                 - --,                       #5,                    --,             -#5      |
|                             l                                                l                       |
|                                                                                                      |
|       d2       d1 + d2 - 1          d2                                                               |
|       --,      ----------- - #4 - -----,              0,                    #2,              0       |
|       c1          c1 r1           c1 r1                                                              |
|                                                                                                      |
|   d1 + d2 - 1                             d1 + d2 - 1      d2                                        |
| - -----------,             0,            ------------- - ----- - #3,         0,              #1      |
|        c2                                r2 (c2 + cf2)   c2 r2                                       |
|                                                                                                      |
|        d2                                                            d1 + d2 - 1                     |
|     - ---,                #4,                         0,             ----------- - #2,       0       |
|       cf1                                                               c1 rp1                       |
|                                                                                                      |
|  d1 + d2 - 1                                                                                    d2   |
|  -----------,              0,                        #3,                     0,        - #1 - ------ |
\      cf2                                                                                      c2 rp2 /

where

d1
#1 == --------------
rp2 (c2 + cf2)

d1
#2 == --------------
rp1 (c1 + cf1)

d1
#3 == -------------
r2 (c2 + cf2)

d1
#4 == -------------
r1 (c1 + cf1)

d1 + d2 - 1
#5 == -----------
l


Aavg is in this case is invertible and its rank is 5.