GATE EE 2018 Set 1 — Question 51

The Thevenin resistance seen at terminals a-b is found with the independent dc sources set to zero (both the 15 V and the 10.7 V source are replaced by short circuits).Step 1: Bias point.
The base is held at $V_B = 10.7\text{ V}$ (the base current through $10\text{ k}\Omega$ is small) and $V_{BE} = 0.7\text{ V}$, so$$V_E \approx 10.7 - 0.7 = 10\text{ V}, \qquad I_E \approx \frac{10\text{ V}}{1\text{ k}\Omega} = 10\text{ mA}.$$The collector sits near 15 V, above the base, so the transistor is in the active region and $I_C = \beta I_B$ holds.Step 2: Resistance looking into the emitter.
Because $V_{BE}$ is a constant $0.7\text{ V}$, a change in emitter voltage produces the same change in base voltage (there is no incremental base-emitter resistance). Apply a test voltage $v$ at the emitter (with the sources shorted). The base voltage changes by $v$, so the base current changes by $v/R_B$ with $R_B = 10\text{ k}\Omega$, and the emitter current changes by $(\beta+1)$ times that:$$i = (\beta+1)\frac{v}{R_B} \;\Rightarrow\; R_{e}' = \frac{v}{i} = \frac{R_B}{\beta+1} = \frac{10\,000}{101} = 99.01\ \Omega.$$(The $10\ \Omega$ collector resistor does not affect this, since the collector current depends only on the base current in the active region.)Step 3: Combine with the emitter resistor.
At terminals a-b this resistance appears in parallel with the $1\text{ k}\Omega$ resistor:$$R_{Th} = 1000 \parallel 99.01 = \frac{1000 \times 99.01}{1000 + 99.01} = 90.09\ \Omega.$$Answer: $R_{Th} \approx 90.09\ \Omega$