Buck Regulators


Two modes:
1. Mode 1: switch closed


2. Mode 2: switch open


Power Electronic Converter: Buck Regulators
Power Electronic Converter: Buck Regulators
Mode 1: switch closed, i=i1
{V_s} = R{i_1} + L\frac{{d{i_1}}}{{dt}} + E
Mode 2: switch open, i=i2
0 = R{i_1} + L\frac{{d{i_1}}}{{dt}} + E
Power Electronic Converter: Buck Regulators-mode
Mode 1:
For i=i1
{V_s} = R{i_1} + L\frac{{d{i_1}}}{{dt}} + E
{V_s} - E = R{i_1} + L\frac{{d{i_1}}}{{dt}}
Laplace Transform
\frac{{{V_s} - E}}{s} = R{I_1}(s) + Ls{I_1}(s)
\frac{{{V_s} - E + Ls{i_1}(0)}}{s} = {I_1}(s)(R + Ls)
{I_1}(s) = \frac{{{V_s} - E + Ls{i_1}(0)}}{{s(R + Ls)}}
Partial Fraction Expansion
{I_1}(s) = \frac{{{V_s} - E + Ls{i_1}(0)}}{{s(R + Ls)}} = \frac{{{V_s} - E + Ls{i_1}(0)}}{{sL(s + R/L)}}
               = \frac{{\frac{{{V_s}}}{L} - \frac{E}{L} + s{i_1}(0)}}{{s(s + R/L)}}
               = \frac{{{V_s} - E}}{{sR}} - \frac{{{V_s} - E - R{i_1}(0)}}{{R(s + R/L)}}
Inverse Laplace Transform
{i_1}(t) = \frac{{{V_s} - E}}{R} - \frac{{{V_s} - E}}{R}{e^{ - tR/L}} + {i_1}(0){e^{ - tR/L}}
Rearrange to get-
{i_1}(t) = {i_1}(0){e^{ - tR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - tR/L}})
Use initial cdt: I1=i1(0)
{i_1}(t) = {I_1}{e^{ - tR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - tR/L}})
Power Electronic Converter: Buck Regulators
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
Power electronic converter: Equivalent circuits and waveforms of Buck Regulator
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
{i_1}(t) = {I_1}{e^{ - tR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - tR/L}})
We have
{i_1}(t = {t_1} = kT) \equiv {I_2}
 \Rightarrow {I_2} = {i_1}(t = kT) = {I_1}{e^{ - kTR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - kTR/L}})               (5.15)
From Mode 2 current expression & “boundary” condition at switching
{i_2}(t) = {I_2}{e^{ - tR/L}} - \frac{E}{R}(1 - {e^{ - tR/L}})
We have {i_2}(t = {t_2}) = {I_3} and also note {t_2} = (1 - k)T
 \Rightarrow {I_3} = {i_2}(t = (1 - k)T) = {I_2}{e^{ - (1 - k)TR/L}} - \frac{E}{R}(1 - {e^{ - (1 - k)TR/L}})               (5.16)
From previous elastration, observe that, in steady-state, I3=I1….
 \Rightarrow {I_1} = {i_2}(t = (1 - k)T) = {I_2}{e^{ - (1 - k)TR/L}} - \frac{E}{R}(1 - {e^{ - (1 - k)TR/L}})               (5.16a)
Substitute (5.16a) into (5.15):
{I_2} = {i_1}(t = kT) = \left( {{I_2}{e^{ - (1 - k)TR/L}} - \frac{E}{R}(1 - {e^{ - (1 - k)TR/L}})} \right){e^{ - kTR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - kTR/L}})
Now solve for I2
{I_2} = {I_2}{e^{ - (1 - k)TR/L}}{e^{ - kTR/L}} - \frac{E}{R}(1 - {e^{ - (1 - k)TR/L}}){e^{ - kTR/L}} + \frac{{{V_s} - E}}{R}(1 - {e^{ - kTR/L}})
 = {I_2}{e^{ - TR/L}} - \frac{E}{R}({e^{ - kTR/L}} - {e^{ - TR/L}}) + \frac{{{V_s} - E}}{R}(1 - {e^{ - kTR/L}})
 \Rightarrow {I_2}(1 - {e^{ - TR/L}}) =  - \frac{E}{R}({e^{ - kTR/L}} - {e^{ - TR/L}}) + \frac{{{V_s} - E}}{R}(1 - {e^{ - kTR/L}})
{I_2} =  - \frac{E}{R}\frac{{({e^{ - kTR/L}} - {e^{ - TR/L}})}}{{(1 - {e^{ - TR/L}})}} + \frac{{{V_s} - E}}{R}\frac{{(1 - {e^{ - kTR/L}})}}{{(1 - {e^{ - TR/L}})}}

{I_2} = \frac{{[ - E{e^{ - kTR/L}} + E{e^{ - TR/L}} + {V_s} - E - {V_s}{e^{ - kTR/L}} + E{e^{ - kTR/L}}]}}{{R(1 - {e^{ - TR/L}})}}
 = \frac{{\left[ {E{e^{ - TR/L}} + {V_s} - E - {V_s}{e^{ - kTR/L}}} \right]}}{{R(1 - {e^{ - TR/L}})}}
 = \frac{{\left[ { - E(1 - {e^{ - TR/L}}) + {V_s}(1 - {e^{ - kTR/L}})} \right]}}{{R(1 - {e^{ - TR/L}})}}
 = \frac{{{V_s}(1 - {e^{ - kTR/L}})}}{{R(1 - {e^{ - TR/L}})}} - \frac{{E(1 - {e^{ - TR/L}})}}{{R(1 - {e^{ - TR/L}})}}
 = \frac{{{V_s}(1 - {e^{ - kTR/L}})}}{{R(1 - {e^{ - TR/L}})}} - \frac{E}{R}
{I_2} = \frac{{{V_s}(1 - {e^{ - kTR/L}})}}{{R(1 - {e^{ - TR/L}})}} - \frac{E}{R}                  (5.18)
Substitute eq. 5.18 into 5.16a:
{{I_1} = \left[ {\frac{{{V_s}({e^{ - kTR/L}} - 1)}}{{R({e^{ - TR/L}} - 1)}} - \frac{E}{R}} \right]{e^{ - (1 - k)TR/L}} - \frac{E}{R}(1 - {e^{ - (1 - k)TR/L}})}
 = \frac{{{V_s}({e^{ - kTR/L}} - 1)}}{{R({e^{ - TR/L}} - 1)}}{e^{ - (1 - k)TR/L}} - \frac{E}{R}{e^{ - (1 - k)TR/L}} - \frac{E}{R} + \frac{E}{R}{e^{ - (1 - k)TR/L}})
 = \frac{{{V_s}({e^{ - TR/L}} - {e^{ - (1 - k)TR/L}})}}{{R({e^{ - TR/L}} - 1)}} - \frac{E}{R}
 = \frac{{{V_s}(1 - {e^{kTR/L}})}}{{R(1 - {e^{TR/L}})}} - \frac{E}{R}
 = \frac{{{V_s}({e^{kTR/L}} - 1)}}{{R({e^{TR/L}} - 1)}} - \frac{E}{R}
Summary-
Define z=TR/L
{I_2} = \frac{{{V_s}({e^{ - kTR/L}} - 1)}}{{R({e^{ - TR/L}} - 1)}} - \frac{E}{R} \Leftarrow Eq.5.18 \Rightarrow {I_2} = \frac{{{V_s}({e^{ - kz}} - 1)}}{{R({e^{ - z}} - 1)}} - \frac{E}{R}
{I_1} = \frac{{{V_s}({e^{kTR/L}} - 1)}}{{R({e^{TR/L}} - 1)}} - \frac{E}{R} \Leftarrow Eq.5.17 \Rightarrow {I_1} = \frac{{{V_s}({e^{kz}} - 1)}}{{R({e^z} - 1)}} - \frac{E}{R}
What is z?
Recall τ =L/R is the time constant.
 \Rightarrow z = T/\tau
So z is the ratio of chopping (or switching) period to load time constant.
Now let’s consider the peak-to-peak ripple current.
Power electronic converter: Equivalent circuits and waveforms of Buck Regulator
Figure: Equivalent circuits and waveforms for RL loads.
Therefore the peak-to-peak ripple current is given by
\Delta I = {I_2} - {I_1}
Plug in eq. 5.17 and 5.18 and will get:
\Delta I = \frac{{{V_s}}}{R}\frac{{1 - {e^{ - kz}} + {e^{ - z}} - {e^{ - (1 - k)z}}}}{{1 - {e^{ - z}}}} - \frac{E}{R}......Eq.5.19
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.
\frac{d}{{dk}}\Delta I = \frac{d}{{dk}}\left\{ {\frac{{{V_s}}}{R}\frac{{1 - {e^{ - kz}} + {e^{ - z}} - {e^{ - (1 - k)z}}}}{{1 - {e^{ - z}}}} - \frac{E}{R}} \right\} = 0
And you will get k=0.5, which produces the maximum peak-peak ripple of
\Delta {I_{\max }} = \frac{{{V_s}}}{R}\tanh \frac{R}{{4fL}}.......Eq.5.21
For 4fL >> R, tanhθ≈θ, and we may approximate 5.21 as
\Delta {I_{\max }} = \frac{{{V_s}}}{{4fL}}.......Eq.5.22
But only for continuous current!!!!
Discontinuous Operation
{I_1} = 0
{i_1}(t) = \frac{{{V_s} - E}}{R}(1 - {e^{ - tR/L}})
{i_1}(t = {t_1} = kT) = {I_2}
At{\rm{ }}t = kT{\rm{ }}{I_2} = \frac{{{V_s} - E}}{R}(1 - {e^{ - kz}})
{t_2} = \frac{L}{R}\ln \left( {1 + \frac{{R{I_2}}}{E}} \right)
{\rm{ = }}\frac{L}{R}\ln \left| {1 + \left( {\frac{{{V_S} - E}}{E}} \right)\left( {1 - {e^{ - kz}}} \right)} \right|
For a continuous load current
For{\rm{ }}{I_2} > 0{\rm{  }}\frac{{{V_S}}}{R}\left( {\frac{{{e^{kz}} - 1}}{{{e^z} - 1}}} \right) - \frac{E}{R} > 0
\left( {\frac{{{e^{kz}} - 1}}{{{e^z} - 1}} - \frac{E}{{{V_S}}}} \right) > 0
Which gives x = \frac{E}{{{V_S}}} \le \frac{{{e^{kz}} - 1}}{{{e^z} - 1}}
Power electronic converter: Discontinuous Operation of Buck Regulator
previous Control of DC-DC converters by PWM
next Boost Regulators

No comments:

Post a Comment

Please wait for approval of your comment .......