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Exercise 3.5.1 inverting amplifier
1. Derive the voltage gain $A_{\rm V}= {{U_{\rm O}}\over{U_{\rm I}}}$ for the inverting amplifier.
Use the procedure that was used for the non-inverting amplifier.
- What is required?
- Number of variables?
- Number of necessary equations?
- Set up the known equations
- Derivation of the voltage gain
Take into account that for the differential gain $A_\rm D$ of the ideal OPV applies: $A_\rm D \rightarrow \infty$. And the following also applies: $1/A_\rm D \rightarrow 0$
But the following doesn't always apply: ${{C}\over{U_x \cdot A_\rm D}} \rightarrow 0$, for an unknown constant $C$ and a voltage $U_x$!
$A_{\rm V} = \frac{U_{\rm O}}{U_{\rm I}}$
- 5 voltages: $U_{\rm I}$, $U_{\rm 1}$, $U_{\rm D}$, $U_2$, $U_{\rm O}$
- 5 currents: $I_1$, $I_{\rm m}$, $I_{\rm p}$, $I_2$, $I_{\rm o}$
- --> 10 variables
9, since one equation is to be determined
- Fundamental equation: (1) $U_{\rm O} = A_{\rm D} \cdot U_{\rm D}$
- Golden rules:
- $R_{\rm D} \rightarrow \infty$, and thus (2) + (3) $I_{\rm m} = I_{\rm p} = 0$
- $R_{\rm O} = 0$
- (4) $A_{\rm D} \rightarrow \infty$ and with (1) $U_{\rm D} = \frac{U_{\rm O}}{A_{\rm D}} \rightarrow 0$
- Mesh equations
- Mesh 1: (5) $-U_{\rm I} + U_1 - U_{\rm D} = 0$
- Mesh 2: (6) $U_{\rm D} + U_2 + U_{\rm O} = 0$
- Node equation: (7) $I_1 - I_2 + 0 = 0$
- Resistors:
- (8) $R_1 = \frac{U_1}{I_1}$
- (9) $R_2 = \frac{U_2}{I_2}$
\begin{align*}
A_V &= \frac{U_{\rm O}}{U_{\rm I}} \quad | \quad \text{using (5) and (6)} \\
A_V &= \frac{- U_{\rm D} - U_2}{U_1 - U_{\rm D}} \quad | \quad \text{using (8) and (9)} \\
A_V &= \frac{- U_{\rm D} - R_2 \cdot I_2}{R_1 \cdot I_1 - U_{\rm D}} \quad | \quad \text{using (1) }\\
A_V &= \frac{- R_2 \cdot I_2}{R_1 \cdot I_1} \quad | \quad \text{using (7)} \\
A_V &= -\frac{R_2 }{R_1 }
\end{align*}
2. Which type of amplifier circuit (inverting or non-inverting amplifier) has the lower input resistance? Why?