Can anyone explain what is doing this diode-capacitor branch on a Boost converter?

Question

I've analized the circuit through capacitor currents to find the transfer function, but I keep getting to the exact same transfer function of the typical Boost converter, is the Diode-Capacitor branch in parallel to the inductor actually doing something?

Switch and diode are ideal, capacitors and inductors are large and on steady state and the converter is operating in continuous conduction mode. Thanks in advance!

P.S. Sorry for any posible mistake on my english, I'm not a native speaker

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Idealized everything (including Vf of diodes = 0), and continuous conduction mode in equilibrium:

Simple boost: $$ {duty\ cycle} = \frac{switch\ on}{total\ period} = D = \frac{V_o - E}{V_o} $$

This circuit: $$ D = \frac{V_o - 2 E}{V_o - E} $$ The difference is that this circuit, the \$V_o\$ is raised by \$E\$ by the switched capacitor. Solve for \$V_o\$: $$ V_o = \frac{E(2-D)}{1-D} $$

Answer 2

When the switch closes, the bottom of C1 is connected to ground, and D1 charges C1. When the switch opens, the inductor causes the votage at the bottom of C1 to rise. This allows C1 to discharge through D2 into the load. Note that, in the DC case, C1 never charges up, since the inductor is a perfect short circuit, and draws infinite current. But since E is an ideal voltage source, it can simultaneously put out infinite current AND produce a voltage across zero resistance.

Answer 3

As you noted, this has to be a boost converter. But operating in continuous mode.

Qualitatively, I'd establish (-) terminal of \$E\$ as "ground reference" and then tend to assume the steady state process is something like this: The switch closes and applies voltage \$E\$ across inductor \$L\$ causing the current in \$L\$ to rise from \$I_L{min}\$ to \$I_L{max}\$. It also pulls \$C\$ to ground, causing \$C\$ to charge rapidly up through \$D_1\$ to voltage E. The switch then opens and forces inductor \$L\$ to reverse its voltage and to start declining its current from \$I_L{max}\$ down to \$I_L{min}\$. Capacitor \$C\$ now sits on top of \$E\$, plus \$V_L\$, making the output voltage developed on \$C_o\$ equal to \$2\cdot E + V_L\$. Current in this phase proceeds through \$D_2\$ to reach \$C_o\$.

Without more information, I can assume that in the continuous steady state it is true that \$I_L{min}=I_L{max}\$. Therefore, there is no change in current in \$L\$ and therefore no voltage developed across it. So the output voltage is twice the input voltage, in steady state.

There are issues, such as the fact that \$C\$ will be charged infinitely fast in the first phase and will discharge itself also infinitely fast in the second phase. And I've no idea the value of \$R\$, which will remove charge on \$C_o\$ that must be made up for by dumped charge from \$C\$ in each cycle. But this is all ideal stuff. So... I guess that's okay.

Answer 4

IN this hypothetic continuous mode, the circuit is merely an AC voltage doubler.

C1 is a switched Cap charged at E thus pulses D2 with peak voltage 2E while D2 rectifies that to 2*E output, all the while L feeds continuous current to supply the load.

In real life it is a discontinuous Boost converter that can be much greater than 2*E or just the same as E depending on the proper choices of components to supply the load. ( minus diode drops)