Consider the circuit shown in the Figure with $V_i=3V$ and $V_{\mathrm{CC}}=12V$. Assume $V_{\mathrm{BE}}=0.7V$ and ${\beta }_{\mathrm{dc}}=99$ for the BJT. Which of the following options is the correct value of the current $I_o$ ?

The correct option is 30 µA.

1. Circuit Analysis and Image Observations:
From the provided circuit diagram, we analyze the connection of the operational amplifier (op-amp) and the NPN bipolar junction transistor (BJT):
- The non-inverting input terminal (labeled +) of the op-amp is connected to the input voltage $V_i$.
- The inverting input terminal (labeled -) is connected in a feedback loop directly to the emitter node of the BJT.
- The emitter of the BJT is connected to ground through a resistor of value $1\text{k}\Omega$.
- The collector of the BJT is connected to the supply voltage $V_{\mathrm{CC}}$ through a resistor of value $2\text{k}\Omega$.
- The output of the op-amp is connected to the base of the BJT, supplying the base current labeled as $I_o$.

2. Finding the Emitter Node Voltage:
Due to the negative feedback configuration, the op-amp operates under the virtual short-circuit condition, maintaining equal potential at its inputs:

$V_{-}=V_{+}=V_i=3\text{V}$

Since the emitter terminal of the transistor is tied to the inverting input ($V_{-}$), the emitter voltage is:

$V_E=3\text{V}$

3. Determining the Emitter Current:
Using Ohm's law, we compute the current flowing through the emitter resistor to ground:

$I_E=\frac{V_E}{R_E}=\frac{3\text{V}}{1\text{k}\Omega }=3\text{mA}$

4. Calculating the Base Current ($I_o$):
For a BJT operating in the active region, the relationship between emitter current ($I_E$) and base current ($I_B$ or $I_o$) is given by:

$I_E=(1+{\beta }_{\mathrm{dc}})I_o$

Rearranging the formula to solve for $I_o$:

$I_o=\frac{I_E}{1+{\beta }_{\mathrm{dc}}}$

Substituting the given parameters ($I_E=3\text{mA}$ and ${\beta }_{\mathrm{dc}}=99$):

$I_o=\frac{3\text{mA}}{1+99}=\frac{3\times {10}^{-3}\text{A}}{100}=3\times {10}^{-5}\text{A}=30\mu \text{A}$