Boost Regulators – analysis of switch open


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=iL, and solving for diL/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 diL/dt positive or negative?
Power electronic converter: Boost Regulators – analysis of switch open
The answer to that question depends on whether vo is larger or smaller than Vs. So let’s see if we can relate vo to Vs.
Relation between v0 and Vs
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.
Power electronic converter: Boost Regulators – analysis of switch open
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 Vs 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 vo

{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.
.
Power electronic converter: Boost Regulators – analysis of switch open
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?
image
Without the chopping action, Vs must be greater than E for transferring power from Vs to E. But with the boost chopper circuit, we can transfer power even if Vs<E.
image
At Relation between v0 and Vs we have

Replacing vo 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, Vs-E &lt; 0.
For this to be the case, Vs &lt; E.
So power transfer (positive current) occurs when source voltage is LESS than the battery voltage!

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