Question

In a common emitter (CE) amplifier having a voltage gain \(\mathrm{G}\), the transistor used has transconductance \(0.03 \mathrm{mho}\) and current gain 25\. If the above transistor is replaced with another one with transconductance \(0.02 \mathrm{mh}\) and current gain 20 , the voltage gain will be : (1) \(1.5 \mathrm{G}\) (2) \(\frac{1}{3} \mathrm{G}\) (3) \(\frac{5}{4} \mathrm{G}\) (4) \(\frac{2}{3} \mathrm{G}\)

Short answer

The voltage gain will be \(\frac{2}{3} \text{G} \). Therefore, the correct answer is (4).

Step-by-step solution

Understanding Voltage Gain in CE Amplifier

In a common emitter (CE) amplifier, the voltage gain \(\text{G}\) is given by the product of the transconductance \(\text{g}_m\) and the load resistance \(R_L\). Mathematically, it can be expressed as \(\text{G} = \text{g}_m \times R_L\). However, for calculation simplicity, the effect of current gain \(\beta\) is also considered, influencing the overall gain.

Initial Transistor Analysis

Initially, the transistor has \(\text{g}_m = 0.03 \,\text{mho}\) and \(\beta = 25\). The voltage gain with this transistor is \(\text{G} = 0.03 \times \beta \times R_L = 0.03 \times 25 \times R_L\).

Replacement Transistor Analysis

The replacement transistor has \(\text{g}_m = 0.02 \,\text{mho}\) and \(\beta = 20\). The new voltage gain will be \(\text{G}' = 0.02 \times \beta \times R_L = 0.02 \times 20 \times R_L\).

Comparing Gains

The original gain is \(\text{G} = 0.03 \times 25 \times R_L\) and the new gain is \(\text{G}' = 0.02 \times 20 \times R_L\). Simplifying these expressions gives \(\text{G} = 0.75 \times R_L\) and \(\text{G}' = 0.40 \times R_L\).

Ratio of New Gain to Original Gain

The ratio of the new gain \(\text{G}'\) to the original gain \(\text{G}\) is \(\frac{\text{G}'}{\text{G}} = \frac{0.40 \times R_L}{0.75 \times R_L} = \frac{0.40}{0.75} = \frac{4}{7.5} = \frac{2}{3}\).

Key concepts

transconductance

Transconductance, symbolized as \( g_m \), is a fundamental parameter in the analysis of many types of amplifiers, particularly the common emitter (CE) amplifier. Transconductance measures the relationship between the current flowing through the output and the input voltage. In simpler terms, it tells us how much the output current will change for a given change in input voltage.
Mathematically, transconductance is given by:
\( g_m = \frac{I_c}{V_{BE}} \),
where \( I_c \) is the collector current and \( V_{BE} \) is the base-emitter voltage. Transconductance units are 'mho' or 'siemens' (a measure of conductance).

Understanding transconductance helps in determining the voltage gain in amplifiers. For instance, in the CE amplifier, it directly influences the voltage gain. The higher the transconductance, the higher the potential voltage gain, making it a crucial parameter in amplifier design and analysis.

current gain

Current gain, often denoted by \( \beta \) or h\textsubscript{FE}, is another key parameter in common emitter (CE) amplifiers. It represents the ratio of output current (collector current \( I_c \)) to input current (base current \( I_b \)).
Mathematically, it is expressed as:
\( \beta = \frac{I_c}{I_b} \).

In a CE amplifier, current gain plays a vital role in determining the overall voltage gain. The current gain tells us how effectively the transistor can amplify the input signal's current. Higher current gain means that a small input current can control a larger output current, contributing to a higher voltage gain when combined with transconductance and load resistance. Understanding the interplay between transconductance and current gain is essential for achieving the desired amplifier performance and functionality.

common emitter amplifier

The common emitter (CE) amplifier is one of the basic transistor amplifier configurations widely used in electronic circuits. In this arrangement, the emitter terminal of the transistor is common to both the input and output circuits.

Key features and advantages of the CE amplifier include:

• High Voltage Gain: Due to the multiplication of transconductance and load resistance, CE amplifiers provide significant voltage amplification.
• Phase Inversion: The output voltage signal is inverted compared to the input signal. This means a positive change in input results in a negative change in output and vice versa.
• Good Current Amplification: The CE amplifier also enhances the current, combining this with voltage gain to increase overall power gain.

The voltage gain (\text{G}) of a CE amplifier can be expressed as:
\( G = g_m \times R_L \),
where \( g_m \) is the transconductance and \( R_L \) is the load resistance. The overall gain is also influenced by the transistor's current gain (\beta). This interaction helps in understanding how changes in transistor parameters affect the amplifier's performance.