Buck Controller
Contents
Buck Controller#
Find the small signal output voltage to duty ratio \(G_{vd}(s)\) and small signal output current to duty ratio \(G_{id}(s)\) transfer functions. General form defined as
\[\begin{align*}
G_0 \dfrac{1-\dfrac{s}{w_z}}{1+\dfrac{s}{Qw_0}+ \left(\dfrac{s}{\omega_0} \right)^2} .\\
\end{align*}\]
Volt-Seconds
\[\begin{align*}
\langle v_{L}(t)\rangle &= L\frac{d}{dt} \langle i(t)\rangle =
\bigg[\bigg(\langle{v_g}(t)\rangle - \langle i(t)\rangle R_{on} - \langle i(t)\rangle R_L - \langle v(t)\rangle \bigg)d(t) + \bigg( - \langle i(t)\rangle R_{on} - \langle i(t)\rangle R_L - \langle v(t)\rangle \bigg)d'(t) \bigg] \\[1em]
\langle v_{L}(t)\rangle &= L\frac{d}{dt} \langle i(t)\rangle =
\bigg[\langle{v_g}(t)\rangle d(t) - \langle i(t)\rangle R_{on} - \langle i(t)\rangle R_L - \langle v(t)\rangle \bigg] \\[1em]
V_L &= V_g D - I R_{on} - I R_L - V \\[1em]
V_L &= V_g D - \dfrac{V}{R} R_{on} - \dfrac{V}{R} R_L - V \\[1em]
\end{align*}\]
Charge Balance
\[\begin{align*}
\langle i_C (t) \rangle &=
C\frac{d}{dt} \langle v(t)\rangle =
\bigg[\bigg(\langle i(t)\rangle -\frac{\langle v(t)\rangle}{R} \bigg)d(t) +\bigg(\langle i(t)\rangle -\frac{\langle v(t)\rangle}{R} \bigg)d'(t) \bigg] \\[1em]
\langle i_C (t) \rangle &=
C\frac{d}{dt} \langle v(t)\rangle =
\bigg[\langle i(t)\rangle -\frac{\langle v(t)\rangle}{R} \bigg] \\[1em]
I &= \dfrac{V}{R}
\end{align*}\]
System Inputs
\[\begin{align*}
\dot{x} = \dfrac{d}{dt}
\begin{bmatrix}
\langle \hat{i}(t)\rangle \\
\langle \hat{v}(t)\rangle
\end{bmatrix} =
\begin{matrix}
\dfrac{1}{L} \\[1em]
\dfrac{1}{C}
\end{matrix}
\begin{bmatrix}
\langle{v_g}(t)\rangle d(t) - \langle i(t)\rangle R_{on} - \langle i(t)\rangle R_L - \langle v(t)\rangle \\[1em]
\langle i(t)\rangle - \dfrac{\langle v(t)\rangle}{R}
\end{bmatrix}
\end{align*}\]
Linearized Small-Signal Model#
Derive a linearized small-signal model
\[\begin{align*}
\dot{\hat{x}}(t) &\approx A \hat{x}(t) + B \hat{u}(t) \\[0.5em]
\hat{y}(t) &= C \hat{x}(t) + E \hat{u}(t)
\end{align*}\]
where
(9)#\[\begin{matrix}
x(t) = \hat{x}(t) + X && X =
\begin{bmatrix}
I \\
V
\end{bmatrix} \\[0.5em]
u(t) = \hat{u}(t) + U && U =
\begin{bmatrix}
D \\
V_g
\end{bmatrix}
\\[0.5em]
\end{matrix}\]
Find A, B, C, D
\[\begin{align*}
A = \dfrac{d}{d\hat{x}(t)} f(x(t),u(t))\bigg|_{x=X,u=U} &=
\begin{bmatrix}
\ \ \dfrac{\partial f_1}{\partial x_1} & \dfrac{\partial f_1}{\partial x_2} \ \ \\[1em]
\ \ \dfrac{\partial f_2}{\partial x_1} & \dfrac{\partial f_2}{\partial x_2} \ \
\end{bmatrix}\bigg|_{x,u} =
\begin{bmatrix}
\dfrac{\partial f_1}{\partial \langle \hat{i}(t)\rangle} & \dfrac{\partial f_1}{\partial \langle \hat{v}(t)\rangle} \\[1em]
\dfrac{\partial f_2}{\partial \langle \hat{i}(t)\rangle} & \dfrac{\partial f_2}{\partial \langle \hat{v}(t)\rangle}
\end{bmatrix}\bigg|_{x,u} \\[1em]
A &=
\begin{bmatrix}
\dfrac{-(R_{on} + R_L)}{L} & \dfrac{-1}{L} \\[1em]
\dfrac{1}{C} & \dfrac{-1}{RC}
\end{bmatrix}
\end{align*}\]
\[\begin{align*} B = \dfrac{d}{d\hat{u}(t)} f(x(t),u(t))\bigg|_{x=X,u=U} &=
\begin{bmatrix}
\dfrac{\partial f_1}{\partial u_1} & \dfrac{\partial f_1}{\partial u_2} \\[1em]
\dfrac{\partial f_2}{\partial u_1} & \dfrac{\partial f_2}{\partial u_2}
\end{bmatrix}\bigg|_{x,u} =
\begin{bmatrix}
\dfrac{\partial f_1}{\partial \hat{d'}(t)} & \dfrac{\partial f_1}{\partial \langle \hat{v_g}(t)\rangle} \\[1em]
\dfrac{\partial f_2}{\partial \hat{d'}(t)} & \dfrac{\partial f_2}{\partial \langle \hat{v_g}(t)\rangle}
\end{bmatrix}\bigg|_{x,u} \\[1em]
B &=
\begin{bmatrix}
\dfrac{V_g}{L} & \dfrac{D}{L} \\[1em]
0 & 0
\end{bmatrix}
\end{align*}\]
C =
(10)#\[\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}\]
E =
(11)#\[\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}\]
System Outputs in Frequency Domain#
\[\begin{align*}
\dot{\hat{x}}(t) &= A \hat{x}(t) + B \hat{u}(t) \\[0.5em]
\dot{\hat{y}}(t) &= C \hat{x}(t) + D \hat{u}(t) \\[0.5em]
\text{Laplace Transform } &\\[0.5em]
s\hat{x}(s) &= A \hat{x}(s) + B \hat{u}(s) \\[0.5em]
sI\hat{x}(s) - A \hat{x}(s) &= B \hat{u}(s) \\[0.5em]
\hat{x}(s) &= \left(sI-A\right)^{-1}B \hat{u}(s) \\[0.5em]
\hat{y}(s) &= \bigg(C(sI-A)^{-1}B +E \bigg)\hat{u}(s) = G(s) \hat{u}(s)
\end{align*}\]
Solver
G(s) =
(12)#\[\begin{bmatrix}
G_{id}(s) & G_{ig}(s) \\[1em]
G_{vd}(s) & G_{vg}(s)
\end{bmatrix}\]
(13)#\[\begin{bmatrix}
V_g \dfrac{RCs+1}{RLCs^2 + (R R_L C + L)s + R +R_L + R_{on}} &
D \dfrac{RCs+1}{RRLCs^2 + (R R_L C + L)s + R +R_L + R_{on}} \\[1em]
R V_g \dfrac{1}{RLCs^2 + (R R_L C + L)s + R +R_L + R_{on}} &
R D \dfrac{1}{RLCs^2 + (R R_L C + L)s + R +R_L + R_{on}}
\end{bmatrix}\]
Small signal output voltage to duty ratio transfer function
\[\begin{align*}
G_{vd}(s) &= G_0 \dfrac{1-\dfrac{s}{w_z}}{1+\dfrac{s}{Qw_0}+ \left(\dfrac{s}{\omega_0} \right)^2} \\
G_{vd}(s) &= R V_g \dfrac{1}{RLCs^2 + (R R_L C + L)s + R +R_L + R_{on}} \\
\end{align*}\]
Small signal output current to duty ratio transfer function
\[\begin{align*}
G_{id}(s) &= G_0 \dfrac{1-\dfrac{s}{w_z}}{1+\dfrac{s}{Qw_0}+ \left(\dfrac{s}{\omega_0} \right)^2} \\[1em]
G_{id}(s) &= V_g \dfrac{RCs+1}{RLCs^2 + (R R_L C + L)s + R +R_L + R_{on}} \\[1em]
\end{align*}\]