Boost Regulators – analysis of switch open - Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis

Mode 2: switch open

Now the diode is forward biased, and circuit appears as shown in figure bellow.

${v_L} = {V_S} - {v_o} = L\frac{{di}}{{dt}}$

Observe that i=i_L, and solving for di_L/dt results in

$\frac{{d{i_L}}}{{dt}} = \frac{{{V_S} - {v_o}}}{L}$

Since we know v₀ is constant, we see again that the rate of change of current is constant! Therefore the current must change linearly while the switch is open. But does it increase or decrease, that is, is di_L/dt positive or negative?

Power electronic converter: Boost Regulators – analysis of switch open

The answer to that question depends on whether v_o is larger or smaller than V_s. So let’s see if we can relate v_o to V_s.

Relation between v₀ and V_s

Since current must change linearly while switch is open, we can write:

$\frac{{\Delta {i_L}}}{{\Delta t}} = \frac{{{V_S} - {v_o}}}{L}$

Consider peak-to-peak ripple current. It must be the negative as that given by mode 1 condition.

Now we have 2 different expressions for the peak-peak ripple current:

From mode 1 analysis:

$\Delta I = \frac{{{V_S}}}{L}\Delta t = \frac{{{V_S}}}{L}kT$

From mode 2 analysis:

$\Delta I = \frac{{ - ({V_S} - {v_o})}}{L}(T - kT)$

So equate them!

$\frac{{{V_S}}}{L}kT = \frac{{ - ({V_S} - {v_o})}}{L}(T - kT)$

Bring V_s to the left.

$\frac{{{V_S}}}{L}kT + \frac{{{V_S}}}{L}(T - kT) = \frac{{{v_o}}}{L}(T - kT)$

Multiply by L and collect like terms.

$\frac{{{V_S}}}{L}kT + \frac{{{V_S}}}{L}(T - kT) = \frac{{{v_o}}}{L}(T - kT)$

Express v_o

${v_o} = \frac{{{V_S}}}{{1 - k}}$

Consider the case of k=0. What does this mean in terms of the circuit? It means the moment when the switch opens is $t = 0 \to$ Is always open!

Now what happens as k increases from 0? v₀ gets larger $\to$ It BOOSTS the voltage!

What happens when k→0?

Theoretically, ${{\rm{v}}_0} \to \infty$ , as shown.

Practically, parasitic elements become influential, and the output voltage is unstable for values of k approaching 1.0.

Consideration of Power Transfer

Let’s replace the load in our circuit with a battery.

How to deliver power to the battery?

Without the chopping action, V_s must be greater than E for transferring power from V_s to E. But with the boost chopper circuit, we can transfer power even if V_s<E.

At Relation between v₀ and V_s we have

Replacing v_o by E:

$\Delta I = \frac{{ - ({V_S} - E)}}{L}(T - kT)$

ΔI is a positive number since it was defined that way for mode 1.
For this to be the case, V_s-E < 0.
For this to be the case, V_s < E.
So power transfer (positive current) occurs when source voltage is LESS than the battery voltage!

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Power, Electronic Systems, Applications and Resources on Electrical and Electronic Project-Thesis

Boost Regulators – analysis of switch open

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