Half-Wave Rectifier

Generated by SMCSim November 21, 2012

Simulation of ckt/HWRectifierFilter.ckt

Circuit Diagram:

NetList:

* Half-Wave Rectifier V1 1 0 sine (5 50) D1 1 2 mymodel (1e-8 0.026) R1 2 0 10000 C1 2 0 10e-3 .tran 0 100 0.5 .plot v(1) v(2) .end

System of Equations representing the electrical circuit:

i_{V_{1}} + D_{1f}(v_{1}, v_{2}) = 0  (1)
(R_{1})v_{2} + (C_{1})\frac{dv_{2}}{dt} + −D_{1f}(v_{1}, v_{2}) = 0  (2)
v_{1} = V_{1}  (3)

D_{nf}(v_{a}, v_{b}) = Is_{n}(1 − e^{(v_{a}−v_{b})/vt_{n}})

where Is_{n}=reverse saturation current and vt_{n}=threshold voltage of diode n

Matrix form: The system of equations Ax + D_{f}(x̂) + C(dx/dt) = b (Symbolically) Where A, D_{f} and C represent matrices corresponding to linear, nonlinear and time dependent electrical elements respectively. b represents the vector corresponding to sources.

A = [ \begin{matrix}0 & 0 & 1 \\ 0 & R̂_{1} & 0 \\ 1 & 0 & 0 \end{matrix} ]  (4)

b = [ \begin{matrix}0 \\ 0 \\ V_{1} \end{matrix} ]  (5)

D_{f} = [ \begin{matrix}D_{1f} \\ −D_{1f} \\ 0 \end{matrix} ]  (6)

C = [ \begin{matrix}0 & 0 & 0 \\ 0 & C_{1} & 0 \\ 0 & 0 & 0 \end{matrix} ]  (7)

x = [ \begin{matrix}v_{1} \\ v_{2} \\ i_{V_{1}} \end{matrix} ]  (8)

x̂ = [ (v_{1}, v_{2}) ]  (9)

Note that the matrix contains R̂ entries (corresponding to resistors) whose values are equal to 1/R

The number of equations are 3 Unknowns: Node potentials: 2 Current Variables: 1

Operating Point (DC) Analysis: All capacitors are open circuited and inductors are short circuited.

System of Equations representing the electrical circuit:

i_{V_{1}} + D_{1f}(v_{1}, v_{2}) = 0  (10)
(R_{1})v_{2} + −D_{1f}(v_{1}, v_{2}) = 0  (11)
v_{1} = V_{1}  (12)

D_{nf}(v_{a}, v_{b}) = Is_{n}(1 − e^{(v_{a}−v_{b})/vt_{n}})

where Is_{n}=reverse saturation current and vt_{n}=threshold voltage of diode n

Application of Newton-Raphson method: Nonliner models:

See linearized model for diode D_{1} in diode D1.eps

Figure 1: linearization of diode D_{1}

System of Equations representing the electrical circuit:

(R_{D_{1}})v_{1} + (−R_{D_{1}})v_{2} + i_{V_{1}} = −i_{D_{1}}  (13)

(R_{D_{1}})v_{1} + (R_{D_{1}} + R_{1})v_{2} = i_{D_{1}}  (14)
v_{1} = V_{1}  (15)

Transient Analysis: Results:

Figure 2: plot