Question

For the given circuit, the value of base current $I_b$ of the npn transistor is ........... mA. ($\beta$ is the current gain and assume the Op-Amp as ideal.) (Specify your answer in mA up to two digits after decimal.) 

Hint:

For ideal Op-Amps, the two input terminals always stay at the same voltage and input current is zero.

Answer 1

Correct Answer: 0.09

Step 1: Understand Op-amp behavior.
The Op-Amp keeps the inverting and non-inverting terminals at the same voltage (ideal Op-Amp). Both nodes thus stay at $+5\,\text{V}$.

Step 2: Compute current through 1k resistor.
The emitter resistor is 1k and bottom node is grounded, so emitter current flows through 1k from +10V:
$I_e = \frac{10 - 5}{1k} = 5\,\text{mA}$.

Step 3: Relate emitter and base currents.
For transistor: $I_e = I_b + I_c$ and $I_c = \beta I_b$.
Thus, $I_e = (\beta + 1)I_b = 51 I_b$.

Step 4: Compute base current.
$I_b = \frac{5\,\text{mA}}{51} \approx 0.098\,\text{mA} \approx 0.10\,\text{mA}$.