The steady state output (V_out), of the circuit shown below, will

The correct answer is: saturate to -V_EE

Step-by-Step Explanation:

1. Analyze the Circuit Diagram:
By inspecting the provided circuit diagram, we can identify the following components and connections:
- The circuit is configured as an inverting integrator using an operational amplifier (op-amp).
- A DC voltage source of value $0.1V$ is connected to the inverting input terminal (-) via the resistor $R_1$.
- A feedback capacitor $C_1$ is connected between the inverting input terminal and the output node $V_{\mathrm{out}}$.
- The non-inverting terminal (+) is connected to ground through a resistor $R_2$.
- The op-amp is powered by positive and negative bias supply voltages, labeled as $+V_{\mathrm{DD}}$ and $-V_{\mathrm{EE}}$ respectively.

2. Apply Op-Amp Analysis Rules:
Assuming an ideal op-amp operating in its active/linear region initially:
- No current enters the input terminals of the ideal op-amp. Consequently, there is no voltage drop across resistor $R_2$, which sets the voltage at the non-inverting terminal to:
$$V_{+}=0V$$
- By the virtual short concept, the voltage at the inverting terminal is equal to the non-inverting terminal:
$$V_{-}=V_{+}=0V$$
Therefore, the inverting terminal acts as a virtual ground.

3. Calculate the Input Current:
The current $I_{\mathrm{in}}$ flowing from the $0.1V$ DC source through the resistor $R_1$ to the virtual ground is:
$$I_{\mathrm{in}}=\frac{0.1-V_{-}}{R_1}=\frac{0.1-0}{R_1}=\frac{0.1}{R_1}$$
Since no current enters the inverting terminal of the op-amp, all of this current $I_{\mathrm{in}}$ must flow through the feedback capacitor $C_1$ towards the output:
$$I_C=I_{\mathrm{in}}=\frac{0.1}{R_1}$$

4. Determine the Output Voltage Behavior:
The current through the capacitor $C_1$ is given by the relation:
$$I_C=C_1\frac{d(V_{-}-V_{\mathrm{out}})}{dt}=-C_1\frac{d{V}_{\mathrm{out}}}{dt}$$
Equating $I_C$ to the constant input current:
$$-C_1\frac{d{V}_{\mathrm{out}}}{dt}=\frac{0.1}{R_1}\Rightarrow \frac{d{V}_{\mathrm{out}}}{dt}=-\frac{0.1}{R_1{C}_1}$$
Integrating with respect to time $t$ gives:
$$V_{\mathrm{out}}\left(t\right)=-\frac{0.1}{R_1{C}_1}t+V_{\mathrm{out}}\left(0\right)$$
This equation represents a continuously decreasing linear ramp voltage with a negative slope.

5. Steady-State Saturation:
As time $t$ approaches infinity ($t\to \infty$), the output voltage $V_{\mathrm{out}}\left(t\right)$ decreases without bound.
However, the output of any practical op-amp is limited by its power supply voltage rails. The negative limit of the output voltage range is defined by the lower supply voltage $-V_{\mathrm{EE}}$.
Thus, as the capacitor continues to charge up, the output voltage will eventually hit the negative limit and remain locked at that voltage.
Therefore, in the steady state, the output voltage will saturate to -V_EE.