An ideal OmAmp is an OpAmp with input resistance $R_{in}=\infty$, output resistance $R_{out}=0$, and gain $A=\infty$. In practice, real OpAmps have relatively high input resistance (usually over $1 M\Omega$), low output resitance (below $100 \Omega$) and very high gain ($over $10^6$) and can be often approximated with an ideal OpAmp.

OpAmps are unually represented in simplified form, as shown in Fig. 1. Notice that, just like in the case of non-ideal OpAmp, the OpAmp is a 4-terminal device and, according to KCL the algebraic sum of the currents flowing out through the 4-terminals is equal to 0. Notice again that if we do not show the ground electrode (see the right diagram in Fig. 1), we are still assuming that OpAmmp contains a 4th (ground electrode); for this reason the algebraic sum of the currents flowing out through the 3-terminals of the OpAmp shown in the right diagram in Fig. 1 is not necessarily 0. This does not mean that KCL is not satisfied but it only says that the OpAmp contains a 4th electrode (the electrode connected to the ground) which is not shown on the diagram.

A few standard configurations with ideal OpAmps are presented below.

Since OpAmps have 4 terminals (excluding the lines that power the dependent voltage source) they can be described electrically using three equations (notice that a resistor, which is a 2-terminal device, is described using one equation-Ohm's law; a transistor, which is a 3- terminal deivice, is described using 2 equations, etc.). Using the notations shown in Fig. 2, we have:

• Infite input resistance which implies the folowing two equations

$$\begin{equation}I_{+} = I_{-} = 0\end{equation}$$

• Infite gain which implies that

$$\begin{equation}V_{+} = V_{-}\end{equation}$$ Notice that, when we solve probably with ideal OpAmps we do not impose any conditions on the output current or potential as they will be computed when we solve the entire system of nodal analysis (assuming the diagram is correct). A few examples with ideal OpAmps are presented below.

A standard configuration for OpAmps is the inverting amplifier shown in Fig. 3. The nodal equtions for nodes $v_1$ and $v_2$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}\frac{v_1-v_2}{R_2} + \frac{v_1-v_3}{R_1}=0\end{equation}$$ Since $v_1=0$ and $v_3=V_{out}$ we obtain that $$\begin{equation}\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1} \end{equation}$$ We notice that in the case of the inverter configuration the total gain depends only on the value of resistors $R_1$ adn $R_2$ and $V_{out}$ in inverted with respect to the input signal.

Another standard configuration for OpAmps is the non-inverting amplifier (follower) shown in Fig. 4. The nodal equtions for nodes $v_1$ and $v_3$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}\frac{v_3-v_2}{R_2} + \frac{v_3}{R_1}=0\end{equation}$$ Since $v_1=v_3$ and $v_2=V_{out}$ we obtain that $$\begin{equation}\frac{V_{out}}{V_{in}}=1+\frac{R_2}{R_1}\end{equation}$$ We notice that in the case of the follower configuration the total gain depends only on the value of resistors $R_1$ adn $R_2$ and $V_{out}$ is directly proportioal with the input signal.

A particular case of the fallower configuration is the unit-gain amplifier shown in Fig. 5. In this case $$\begin{equation}V_{out}=V_{in}\end{equation}$$ An important benefit of unit gain amplifiers is that the input impedance is very large (infinity in the case of ideal OpAmps), because they do not draw any current from the input source.

Notice that we could also build a "negative" unit gain amplifier using the inverter configuration.