GATE 2025 PH – Question 60

Electronics · p-n diodes, bipolar junction transistors, field effect transistors · 2 marks · Numerical answer

In the transistor circuit shown, $V_{BE}=0.7$ V and $\beta_{DC}=400$. The base current in microampere (rounded off to one decimal place) is _____

NPN transistor with 100 kΩ base resistor, 1 kΩ collector and emitter resistors, and 10 V supply.

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Correct answer: 18 to 19

Explanation

**Base-emitter loop** for the transistor in the active region ($V_{BE}=0.7$ V, $\beta_{DC}=400$):

The emitter current is $I_E=(\beta+1)I_B=401\,I_B$. Kirchhoff's voltage law around the base loop (10 V supply, 100 kΩ base resistor, 0.7 V base-emitter drop, 1 kΩ emitter resistor):
$$10=100\,000\,I_B+0.7+1000\,(401\,I_B).$$

**Solve:**
$$9.3=100\,000I_B+401\,000I_B=501\,000\,I_B\;\Rightarrow\;I_B=\frac{9.3}{501\,000}=1.856\times10^{-5}\text{ A}.$$

The base current is **18.6 μA**.

Official GATE 2025 answer key: https://gate2026.iitg.ac.in/doc/download/2025_Key/PH_Keys.pdf