Gain of a non ideal differential amplifier

waulu

Joined Dec 23, 2016
62
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,

I_{+}+I_{fb+}=I_{L+}

I_{in}+I_{PullUp}=I_{fb+}

I_{in}+I_{fb-}=I_{sig}

I_{fb-}+I_{-}=I_{L-}

After that, defining currents in terms of voltages,

I_{+}=\frac{2.5V+\frac{A_{V}}{2}V_{d}-V_{1}}{R_{o}}

I_{fb+}=\frac{V_{-}-V_1}{Z_{P}}

I_{L+}=\frac{V_1-V_{out+}}{R_{L}}

I_{PullUp}=\frac{2.5V-V_{-}}{R_{P}}

I_{in}=\frac{V_d}{Z_{in}}

I_{sig}=\frac{V_{sig}-V{+}}{R_{P}}

I_{fb-}=\frac{V_{+}-V_3}{Z_{P}}

I_{-}=\frac{2.5V+\frac{A_V}{2}V_d-V_3}{R_{O}}

I_{L-}=\frac{V_3-V_{out-}}{R_{L}}

V_d=V_{+}-V_{-}

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.

V_{-}=(2.5V-V_{1})\frac{Z{P}}{R_{P}+Z_{P}}

V_{+}=(V_{sig}-V_{3})\frac{Z_{P}}{R_{P}+Z_{P}}

V_{out+}=V_1\frac{Z_P}{R_{L}+Z_{CL}} \Leftrightarrow V_1=V_{out+}\frac{R_{L}+Z_{CL}}{Z_{CL}}

V_{out-}=V_3\frac{Z_{CL}}{R_{L}+Z_{CL}} \Leftrightarrow V_3=V_{out-}\frac{R_{L}+Z_{CL}}{Z_{CL}}

ericgibbs

Joined Jan 29, 2010
15,517
Hi waulu,
This PDF may help with the analysis.

E

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