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.
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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.
Step 1: Apply Virtual Short Concept. Since the op-amp is operating with negative feedback, we use the virtual short condition. Thus, \[ V_+ = V_-. \] Given $V_+ = 3\,\text{V}$, therefore: \[ V_- = 3\,\text{V}. \] Hence the emitter node is at 3 V. Step 2: Calculate Emitter Current. Emitter is connected to ground through 1 k$\Omega$. Voltage across 1 k$\Omega$ resistor is 3 V. Therefore, \[ I_E = \frac{3}{1k} = 3\,\text{mA}. \] Step 3: Relation between Base and Emitter Current. For a BJT, \[ I_E = I_C + I_B. \] If transistor $\beta$ is very large (ideal assumption in such problems), then: \[ I_B \approx 0. \] Thus, \[ I_C \approx I_E = 3\,\text{mA}. \] Step 4: Output Current of Op-Amp. The op-amp output supplies only the base current $I_B$. Since $\beta$ is assumed very large: \[ I_B \approx 0. \] Hence, \[ I_o \approx 0\,\text{mA}. \] Step 5: Verification of Active Region. Collector current is 3 mA. Voltage drop across 2 k$\Omega$: \[ V = (3\,\text{mA})(2k) = 6\,\text{V}. \] Thus collector voltage is: \[ V_C = 12 - 6 = 6\,\text{V}. \] Since $V_C > V_E (3\,\text{V})$, transistor is in active region. Therefore, final answer is: \[ I_o \approx 0\,\text{mA}. \]
