Question

In the given circuit, the non-inverting input of the op-amp is at 3 V. The op-amp drives the base of a transistor as shown. The emitter is connected to a 1 k$\Omega$ resistor to ground and the collector is connected to 12 V through a 2 k$\Omega$ resistor. Find the output current $I_o$ supplied by the op-amp. 
 

Hint:

In op-amp + BJT circuits, the op-amp usually supplies only the base current. If $\beta$ is large, base current is negligible compared to collector/emitter current.

Answer 1

Correct Answer: 0

Step 1: Understanding the Question
This question involves a circuit that combines an operational amplifier (op-amp) and a Bipolar Junction Transistor (BJT).
The goal is to find the current $I_o$ that flows from the output of the op-amp, which is connected to the base of the BJT.
Step 2: Key Formula or Approach 
For an ideal op-amp with negative feedback, the virtual short concept applies, which means the voltage at the inverting input terminal ($V_-$) is equal to the voltage at the non-inverting input terminal ($V_+$).
The key BJT current relationship is $I_E = (\beta + 1) I_B$, where $\beta$ is the current gain. For large $\beta$, we can approximate $I_E \approx I_C$ and $I_B \approx 0$.
Step 3: Detailed Explanation 
Applying the Virtual Short Concept: 
The op-amp is in a negative feedback configuration because its inverting input is connected to the BJT's emitter, which in turn is part of the output loop.
The non-inverting input ($V_+$) is connected to a 3 V source. Therefore:
\[ V_+ = 3\,\text{V} \] Due to the virtual short, the inverting input voltage is also 3 V:
\[ V_- = V_+ = 3\,\text{V} \] The inverting input terminal is directly connected to the emitter of the transistor. Hence, the emitter voltage ($V_E$) is forced to be 3 V.
\[ V_E = 3\,\text{V} \] Calculating Emitter and Base Currents: 
The emitter is connected to ground through a 1 k$\Omega$ resistor. Using Ohm's Law, we can calculate the emitter current ($I_E$):
\[ I_E = \frac{V_E}{R_E} = \frac{3\,\text{V}}{1\,\text{k}\Omega} = 3\,\text{mA} \] The output current of the op-amp, $I_o$, is the base current ($I_B$) of the BJT. The relationship between emitter and base current is:
\[ I_B = \frac{I_E}{\beta + 1} \] In typical problems of this type, the BJT is assumed to have a very large current gain ($\beta \rightarrow \infty$). Under this assumption, the base current becomes negligibly small.
\[ I_o = I_B \approx 0\,\text{mA} \] Verifying the Transistor's Operating Region: 
To ensure our assumptions are valid, we should check if the transistor is in the active region ($V_C > V_E$).
Assuming $\beta$ is large, the collector current $I_C \approx I_E = 3\,\text{mA}$.
The collector voltage ($V_C$) is:
\[ V_C = V_{CC} - I_C R_C = 12\,\text{V} - (3\,\text{mA} \times 2\,\text{k}\Omega) = 12\,\text{V} - 6\,\text{V} = 6\,\text{V} \] Since $V_C = 6\,\text{V}$ and $V_E = 3\,\text{V}$, we have $V_C > V_E$. This confirms that the BJT is operating in the forward-active region, so our analysis is correct.
Step 4: Final Answer 
The current supplied by the op-amp is the base current, which is approximately zero.
\[ I_o \approx 0\,\text{mA} \]