Answer:
- VX = V1·R2/(R2 +R1(1 +AVOL))
- VX ≈ -99.0001 μV
- Vout = -V1·R2·AVOL/(R2 +R1(1 +AVOL))
- Vout ≈ 0.990001 V
Explanation:
It is useful to consider the voltage at X to be the superposition of two voltages divided by the R1/R2 voltage divider. It is also helpful to remember that OUT = -VX·AVOL.
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a)
[tex]V_X=\dfrac{V_1\cdot R_2+V_{out}\cdot R_1}{R_1+R_2}=\dfrac{V_1\cdot R_2-V_X\cdot A_{VOL}\cdot R_1}{R_1+R_2}\\\\V_X(R_1(1+A_{VOL})+R_2)=V_1\cdot R_2\\\\\boxed{V_X=V_1\dfrac{R_2}{R_2+R_1(1+A_{VOL})}}[/tex]
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b)
Filling in the given numbers, we find VX to be ...
[tex]V_X=-0.01V\cdot\dfrac{400000}{400000+4000(1+10000)}=-0.01V\cdot\dfrac{100}{10101}\\\\\boxed{V_X=\dfrac{-1}{10101}V\approx-99.0001\mu V}[/tex]
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c)
As we said at the beginning, OUT = -VX·AVOL. Multiplying the expression for VX by -AVOL, we get ...
[tex]\boxed{V_{out}=-V_1\cdot\dfrac{A_{VOL}\cdot R_2}{R_2+R_1(1+A_{VOL})}}[/tex]
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d)
As with the expression, the output voltage is found by multiplying VX by -AVOL:
[tex]V_{out}=\dfrac{(-1)(-10000)}{10101}V\\\\\boxed{V_{out}=\dfrac{10000}{10101}V\approx 0.990001V}[/tex]
