Operational amplifier

The ideal operational amplifier, as shown in fig. 10.1, is determined by the following equation which introduces one more unknown in the MNA matrix.

Figure 10.1: ideal operational amplifier

$$V_{1} - V_{3} = 0$$ (10.1)

The new unknown variable $I_{out}$ must be considered by the three remaining simple equations.

$$I_{1} = 0 \quad I_{2} = I_{out} \quad I_{3} = 0$$ (10.2)

And in matrix representation this is (for DC and AC simulation):

$$\begin{bmatrix}.&.&.& 0\\ .&.&.& 1\\ .&.&.& 0\\ 1 & 0 & -1 & 0 \e... ... I_{out} \end{bmatrix} = \begin{bmatrix}I_{1}\\ I_{2}\\ I_{3}\\ 0 \end{bmatrix}$$ (10.3)

The operational amplifier could be considered as a special case of a voltage controlled current source with infinite forward transconductance $G$. Please note that the presented matrix form is only valid in cases where there is a finite feedback impedance between the output and the inverting input port.

To allow a feedback circuit to the non-inverting input (e.g. for a Schmitt trigger), one needs a limited output voltage swing. The following equations are often used to model the transmission characteristic of operational amplifiers.

$$I_1 = 0 \qquad\qquad I_3 = 0$$ (10.4)

$$V_2 = V_{max}\cdot\dfrac{2}{\pi}\arctan \left( \dfrac{\pi}{2\cdot V_{max}}\cdot G\cdot (V_1-V_3) \right)$$ (10.5)

with $V_{max}$ being the maximum output voltage swing and $G$ the voltage amplification. To model the small-signal behaviour (AC analysis), it is necessary to differentiate:

$$g = \dfrac{\partial V_2}{\partial (V_1-V_3)} = \dfrac{G}{1+\left( \dfrac{\pi}{2\cdot V_{max}}\cdot G\cdot (V_1-V_3) \right)^2}$$ (10.6)

This leads to the following matrix representation being a specialised three node voltage controlled voltage source (see section 9.19.3 on page ).

$$\begin{bmatrix}.&.&.& 0\\ .&.&.& 1\\ .&.&.& 0\\ g & -1 & -g & 0 \... ... I_{out} \end{bmatrix} = \begin{bmatrix}I_{1}\\ I_{2}\\ I_{3}\\ 0 \end{bmatrix}$$ (10.7)

The above MNA matrix entries are also used during the non-linear DC analysis with the 0 in the right hand side vector replaced by an equivalent voltage

$$V_{eq} = g\cdot \left(V_1 - V_3\right) - V_{out}$$ (10.8)

with $V_{out}$ computed using eq. (10.5).

With the given small-signal matrix representation, building the S-parameters is easy.

$$(\underline{S}) = \begin{bmatrix}1 & 0 & 0 \\ 4g & -1 & -4g\\ 0 & 0 & 1 \end{bmatrix}$$ (10.9)