Example of MNA - Diodes

A diode is an example of a nonlinear component. When connected between two nodes A and B, we can use the Shockley diode model to model the voltage-current relationship.

\(i_D = I_{SS}\left(e^\frac{v_A-v_B}{\eta V_T}\right)\)

The diode current flows out of node A, and into node B, so we have contributions to two current equations.

$$\begin{aligned} f_A(...,v_A,...,v_B,...) &= +i_D = I_S\left(e^{\frac{v_A-v_B}{nV_T}}-1\right)\\ f_B(...,v_A,...,v_B,...) &= -i_D=-I_S\left(e^{\frac{v_A-v_B}{nV-T}}-1)\right) \end{aligned}$$

From this we can calculate the contributions to the Y-matrix. Since the current equation contributions only depend on nodes A and B, we only get contributions to four elements.

$$\begin{aligned} Y_{A,A} &= \left.\frac{\partial f_A}{\partial v_A}\right|^{(k)}=\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &= &+g_D\\ Y_{A,B} &= \left.\frac{\partial f_A}{\partial v_B}\right|^{(k)}=-\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &= &-g_D\\ Y_{B,A} &= \left.\frac{\partial f_B}{\partial v_A}\right|^{(k)}=-\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &= &-g_D\\ Y_{B,B} &= \left.\frac{\partial f_B}{\partial v_B}\right|^{(k)}=\frac{I_S}{nV_T}e^{\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}} &= &+g_D \end{aligned}$$

We then calculate the contributions to the RHS-vector:

$$\begin{aligned} RHS_A &= \pmb J_A\pmb x^{(k)} - f_A(...,v_A^{(k)},...,v_B^{(k)},...)\\ &= g_D\cdot (v_A^{(k)}-v_B^{(k)}) - I_S\left(e^\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}-1\right)\\ &= +c_D\\ RHS_B &= f_B(...,v_A^{(k)},...,v_B^{(k)},...)-\pmb J_B\pmb x^{(k)}\\ &= -\left(g_D\cdot (v_A^{(k)}-v_B^{(k)}) - I_S\left(e^\frac{v_A^{(k)}-v_B^{(k)}}{nV_T}-1\right)\right)\\ &= -c_D \end{aligned}$$

We note that this time the RHS-vector contributions are not 0 for the current equations. This is again typical for nonlinear components. The solution will need to be found in multiple iterations.