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

### Buck Regulators

Two modes:
1. Mode 1: switch closed

2. Mode 2: switch open

Mode 1: switch closed, i=i1
Mode 2: switch open, i=i2
Mode 1:
For i=i1
Laplace Transform
Partial Fraction Expansion

Inverse Laplace Transform
Rearrange to get-
Use initial cdt: I1=i1(0)
Design features of Buck Regulator

1. Good design requires load time constant > switching time and τ=L/R > T
2. τ : time required to reach 63.2% of final value.
3. When τ<T, we see the full exponential change in i
4. 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
5. 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 i1 and i2
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, I3=I1….
(5.16a)
Substitute (5.16a) into (5.15):
Now solve for I2

(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.
And you will get k=0.5, which produces the maximum peak-peak ripple of
For 4fL >> R, tanhθ≈θ, and we may approximate 5.21 as

But only for continuous current!!!!
Discontinuous Operation