Two modes:

1. Mode 1: switch closed

2. Mode 2: switch open

Mode 1: switch closed, i=i

_{1}
Mode 2: switch open, i=i

_{2}
Mode 1:

For i=i

_{1}
Laplace Transform

Partial Fraction Expansion

Inverse Laplace Transform

Rearrange to get-

Use initial cdt: I

_{1}=i_{1}(0)**Design features of Buck Regulator**

- Good design requires load time constant > switching time and τ=L/R > T
- τ : time required to reach 63.2% of final value.
- When τ<T, we see the full exponential change in i
- When τ>T, we see just a part of the exponential change in i before the switch, and so we assume that I increases/decreases linearly between switching times
- Ex: T=.001sec; L=7.5mH, R=5Ω and τ=L/R=.0015sec

**Equivalent circuits and waveforms**

Figure: Equivalent circuits and waveforms for

*RL*loads.**But what is I1, I2?**

Let’s look at previous equations for i

_{1}and i_{2}
From Mode 1 current expression & “boundary” condition at switching

We have

(5.15)

From Mode 2 current expression & “boundary” condition at switching

We have and also note

(5.16)

From previous elastration, observe that, in steady-state, I

_{3}=I_{1}….
(5.16a)

Substitute (5.16a) into (5.15):

Now solve for I

_{2}
(5.18)

Substitute eq. 5.18 into 5.16a:

Summary-

Define z=TR/L

What is z?

Recall τ =L/R is the time constant.So z is the ratio of chopping (or switching) period to load time constant.

Now let’s consider the peak-to-peak ripple current.

Figure: Equivalent circuits and waveforms for

*RL*loads.
Therefore the peak-to-peak ripple current is given by

Plug in eq. 5.17 and 5.18 and will get:

It is interesting to identify the duty ratio for maximum ripple. To do this, differentiate eq 5.19 with respect to k and set to 0.

But only for continuous current!!!!

**Discontinuous Operation**

_{}

For a continuous load current

_{}

Which gives

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