Hello,

I am trying to write the equation of the gain for the non-ideal differential amplifier.



I start by using the Kirchhoff's laws,

\begin{equation}
I_{+}+I_{fb+}=I_{L+}
\end{equation}

\begin{equation}
I_{in}+I_{PullUp}=I_{fb+}
\end{equation}

\begin{equation}
I_{in}+I_{fb-}=I_{sig}
\end{equation}

\begin{equation}
I_{fb-}+I_{-}=I_{L-}
\end{equation}

After that, defining currents in terms of voltages,

\begin{equation}
I_{+}=\frac{2.5V+\frac{A_{V}}{2}V_{d}-V_{1}}{R_{o}}
\end{equation}
\begin{equation}
I_{fb+}=\frac{V_{-}-V_1}{Z_{P}}
\end{equation}
\begin{equation}
I_{L+}=\frac{V_1-V_{out+}}{R_{L}}
\end{equation}
\begin{equation}
I_{PullUp}=\frac{2.5V-V_{-}}{R_{P}}
\end{equation}
\begin{equation}
I_{in}=\frac{V_d}{Z_{in}}
\end{equation}
\begin{equation}
I_{sig}=\frac{V_{sig}-V{+}}{R_{P}}
\end{equation}
\begin{equation}
I_{fb-}=\frac{V_{+}-V_3}{Z_{P}}
\end{equation}
\begin{equation}
I_{-}=\frac{2.5V+\frac{A_V}{2}V_d-V_3}{R_{O}}
\end{equation}
\begin{equation}
I_{L-}=\frac{V_3-V_{out-}}{R_{L}}
\end{equation}
\begin{equation}
V_d=V_{+}-V_{-}
\end{equation}

And now I solve it for voltages in terms of resistors,
While writing this post I realized that the equations for V- and V+ are wrong. I leave it here just to show what I did and maybe someone can help me to rewrite it. In the meantime, I will try to rewrite it myself.

\begin{equation}
V_{-}=(2.5V-V_{1})\frac{Z{P}}{R_{P}+Z_{P}}
\end{equation}
\begin{equation}
V_{+}=(V_{sig}-V_{3})\frac{Z_{P}}{R_{P}+Z_{P}}
\end{equation}
\begin{equation}
V_{out+}=V_1\frac{Z_P}{R_{L}+Z_{CL}} \Leftrightarrow V_1=V_{out+}\frac{R_{L}+Z_{CL}}{Z_{CL}}
\end{equation}
\begin{equation}
V_{out-}=V_3\frac{Z_{CL}}{R_{L}+Z_{CL}} \Leftrightarrow V_3=V_{out-}\frac{R_{L}+Z_{CL}}{Z_{CL}}
\end{equation}

Thank you for your patience.

 

Hi waulu,
This PDF may help with the analysis.

E

 

Hello ericgibbs,

Thank you for the PDF, I will take a look.