29.4.1. Numerical Example

We obtain the base current $I_B = 20 \mu A$ and then for the value of $\beta = 100$ the $I_C = 2 mA$.

In this chunk, we start by determining the base current, denoted as $I_B$. Given the fixed bias configuration of the common emitter amplifier, we find that the base current is $20 \mu A$. Next, we calculate the collector current, $I_C$, which is derived from the relationship between the base current and beta ($\beta$), the transistor's current gain. With $\beta = 100$, the formula we use is $I_C = \beta \cdot I_B$, giving us $I_C = 100 \cdot 20 \mu A = 2 mA$.

Think of the base current as the amount of water flowing through a hose (base) and the collector current as the water that flows out of a sprinkler (collector). The sprinkler's output depends on how much water is coming from the hose, and here, the relationship is defined by the factor $\beta$.

We need to find small signal parameters of the transistor. Namely, the important parameters are $g_m$, which is $I_C/V_T$. We have discussed about this.

This chunk introduces small signal parameters that describe the transistor's behavior in a linear approximation regime. The transconductance $g_m$ is calculated as the ratio of collector current $I_C$ to the thermal voltage $V_T$. This relationship is crucial because it allows us to derive the gain of the amplifier. The thermal voltage $V_T$ is approximately 26 mV at room temperature. Knowing $I_C = 2mA$ and using $V_T \approx 0.026 V$, we get $g_m = \frac{2 mA}{26 mV}$.

Imagine a microphone converting sound (input) into an electrical signal (output). The efficiency of the microphone in converting sound into electrical means is like $g_m$; it indicates how well the device can transform one type of energy into another.

Let us draw the small signal equivalent circuit: R connected to V_CC, with the base$r_π$ and the emitter terminal connected to ground.

Here, we illustrate how to depict the small signal equivalent circuit, which simplifies the analysis of the amplifier. In this representation, $R$ is our load resistor connected to the supply voltage $V_{CC}$, while the base resistance $r_π$ connects the base terminal to ground. Understanding this equivalent circuit aids in visualizing how input signals affect the transistor's output under small signal conditions.

Consider a model of a city where different paths of water flow represent the connections in our circuit. Each component (like banks, tunnels) plays a critical role in determining how effectively resources are distributed – just like how our circuit operates.

The circuit gain $A_v = -g_m \times R_C$. So the voltage gain is shown to be approximately -208.

In this segment, we calculate the voltage gain of the amplifier. The formula used is $A_v = -g_m \times R_C$, where $R_C$ is the collector resistor. In our example, we find that the voltage gain is roughly $-208$ (the negative sign indicating phase inversion). This gain means that for every 1V of input, there will be an approximate 208V output, but inverted in phase.

Think of this amplification process like a loudspeaker: small signals (like a voice) inputted to the mic are amplified tremendously to produce a loud sound, with the speaker acting like the amplifier output.

The input resistance $R_{in}$ is $R_B // r_π$ and the output resistance $R_{out} = R_C$.

This section elaborates on calculating the input and output resistances of the circuit. The input resistance $R_{in}$ is derived from the parallel combination of the biasing resistor $R_B$ and the transistor's small signal base-emitter resistance $r_π$. Similarly, the output resistance $R_{out}$ is directly equivalent to the collector resistance $R_C$, which influences how much of the output voltage can be dropped across the load.

Imagine the circuit as a highway system. Input resistance can be viewed as how many cars can enter a highway lane (input), and the output resistance as the capacity of the highway (output) to handle traffic. Adjusting these allows for smoother flow depending on the situation.

If we draw the equivalent circuit, we find the input resistance is around 1.3 kΩ and the output resistance is 3.3 kΩ.

In conclusion, we summarize the key performance metrics of the amplifier. The input resistance found is approximately 1.3kΩ and output resistance is about 3.3kΩ. These values define how the amplifier interfaces with other circuit elements, impacting its overall performance. High input resistance is favorable for minimizing loading on previous stages, whereas appropriate output resistance ensures effective power delivery to the next stage.

Consider an amplifier as an intermediary in a conversation. The ability of the amplifier to listen without interrupting (high input resistance) while also making itself heard clearly (output resistance) ensures that the message (signal) flows effectively.

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