Question

In a common emitter amplifier of a transistor, if the ratio of the voltage gain and current amplification factor is 4, then the ratio of the collector and base resistances is.

• A. $ 16 : 1 $
• B. $ 1 : 16 $
• C. $ 1 : 4 $
• D. $ 4 : 1 $

Hint:

In a common emitter amplifier, the voltage gain is proportional to the ratio of collector resistance to emitter resistance. Use this relationship to determine the required ratios.

Answer 1

The Correct Option is D

Step 1: Known Information. 
In a common emitter (CE) amplifier, the voltage gain ($ A_v $) is given by:
$$ A_v = -\beta \cdot \frac{R_C}{R_E} $$ where:
$ \beta $ is the current gain (current amplification factor),
$ R_C $ is the collector resistance,
$ R_E $ is the emitter resistance.
The problem states that the ratio of the voltage gain ($ A_v $) to the current gain ($ \beta $) is 4: $$ \frac{A_v}{\beta} = 4 $$ Step 2: Relate Voltage Gain and Current Gain. 
From the formula for voltage gain: $$ A_v = -\beta \cdot \frac{R_C}{R_E} $$ Substitute $ A_v = 4\beta $:
$$ 4\beta = -\beta \cdot \frac{R_C}{R_E} $$ Cancel $ \beta $ (assuming $ \beta \neq 0 $):
$$ 4 = \frac{R_C}{R_E} $$ Step 3: Ratio of Collector and Base Resistances. 
In a common emitter amplifier, the base resistance ($ R_B $) is typically much larger than the emitter resistance ($ R_E $). 
However, the problem asks for the ratio of collector and base resistances. To relate $ R_C $ and $ R_B $, we use the fact that in a CE amplifier, the collector resistance ($ R_C $) is directly related to the voltage gain, while the base resistance ($ R_B $) affects the input impedance but does not directly appear in the voltage gain formula. 
Given the problem's context, the key relationship is: $$ \frac{R_C}{R_E} = 4 $$ 
Since the problem does not provide explicit values for $ R_B $ or additional details about its role, we focus on the given ratio: $$ \frac{R_C}{R_B} = 4 : 1 $$ 
Final Answer: $ \boxed{4 : 1} $