$\frac{- \beta R_C}{r_e+(1+\beta)R_E}$
The voltage gain of a common emitter amplifier with emitter resistance (R_E) is a crucial parameter in electronic circuit design. This gain indicates how much the amplifier will amplify an input signal. Let us solve this by understanding the circuit dynamics:
The voltage gain (A_v) for a common emitter amplifier with an emitter resistance can be derived from the small-signal model:
The small signal voltage gain formula is given by:
A_v = \frac{V_{out}}{V_{in}} = -\frac{\beta \cdot R_C}{R_E + r_e}
where:
In practical circuits with emitter degeneration (a resistor R_E is added to stabilize the gain), the effective emitter resistance seen at the base is (1 + \beta)R_E. Therefore, the correct formula for the small-signal voltage gain incorporating R_E becomes:
A_v = \frac{- \beta R_C}{r_e + (1 + \beta)R_E}
Thus, among the provided options, the correct formulation for the voltage gain of a common emitter amplifier with R_E is:
\frac{- \beta R_C}{r_e + (1 + \beta)R_E}
This expression accounts for the degenerative feedback brought by R_E, which stabilizes the gain but reduces its magnitude.
Find $I_C$, $V_{BE}$=0.7V.

Match the following :
| List – I (Biasing of BJT) | List – II (Functions) |
| a. E-B junction forward bias and C-B junction reverse bias | i. Very low gain amplifier |
| b. Both E-B and C-B junctions forward bias | ii. Saturation condition |
| c. E-B junction reverse bias and C-B junction forward bias | iii. High gain amplifier |
| d. Both E-B and C-B junctions reverse bias | iv. Cut-off condition |
Codes :
The doping concentration of n-p-n transistor are
i. 5 × 1018 / cm3
ii. 1017 / cm3
iii. 2 × 107 / cm3
Identify the regions in the above order i, ii and iii.
For the BJT to be biased in its linear or active operating region :
(a) the base-emitter Junction must be forward biased
(b) the base-collector Junction must be reverse biased
(c) the base-emitter Junction must be reverse biased
(d) the base-collector Junction must be forward biased
Assertion (A) : Two P-N diodes connected back to back cannot be used as a transistor.
Reason (R) : Fabrication of a transistor requires controlled doping of Emitter, base and collector regions.
Select your answer using the codes given below :