29.1.1. Numerical Examples (Part B)

Yes, welcome back to our discussion Numerical Examples of CE Amplifier. And, we are discussing about CE amplifier with fixed-bias. So, what we said is that based on the value of R and R. We obtain the base current I_B = 20 µA and then for the value of β = 100 the I_C = 2 mA.

In this chunk, we are looking at the calculation of the base current (I_B) and collector current (I_C) for a common emitter (CE) amplifier with fixed bias. The base current is determined based on resistors (R and R) in the circuit configuration. Using the given transconductance parameter (β), we can relate the base current to the collector current. Here, I_B is calculated as 20 µA, which is small and indicates that the transistor is being driven with minimal input current. With β set at 100, this results in a collector current of 2 mA (I_C). The relationship between these three currents is a foundational concept in transistor operation, illustrating how a small input can control a larger output.

Think of a garden hose and a water sprinkler. When you slightly turn on the water (base current), it may seem insignificant, but it can lead to a strong spray from the sprinkler (collector current). In a similar way, a small I_B can control a larger I_C in our amplifier.

So, to get the expression the value of the gain of this circuit as well as input resistance and output resistance, we need to find a small signal parameter of the transistor. Namely, the important parameters are g_m, which is I_C / V_T. So, we can say that this is A/V. ... So, from that we can say that the circuit gain voltage gain A defined as v_out = -g_m × R_C.

In this section, we derive the expression for voltage gain (A) for the CE amplifier. The transconductance (g_m) of the transistor is calculated as I_C divided by the thermal voltage (V_T). This factor helps in understanding how the output voltage relates to the input voltage. The output voltage (v_out) is expressed in terms of g_m and the load resistance (R_C). The negative sign indicates a phase inversion, which is a critical characteristic of CE amplifiers; the output swings in the opposite direction to the input signal. The final equation shows how we can calculate the amplifier's gain based on the known parameters in the circuit.

Imagine a seesaw at a playground. If one child (input signal) pushes down on their side, the other side (output voltage) rises, illustrating the inversion. In our circuit, a small signal variation at the input leads to a larger variation at the output, while the phase between them is also reversed.

So, the input resistance of this circuit it is R_B coming in parallel with r_pi, so we can say that R_B = R in parallel with r_pi. ... The output resistance R_O is nothing but this R_C.

This chunk explains the determination of input and output resistance for the CE amplifier. The input resistance (R_B) is calculated as the parallel combination of the base resistor (R) and the small-signal emitter resistance (r_pi). Since R is typically much larger than r_pi, we can often simplify the calculation to focus mainly on r_pi. On the other hand, the output resistance (R_O) is taken directly as the collector resistance (R_C). These resistances play crucial roles in how the amplifier interacts with other circuit elements and affect overall circuit performance.

Consider a water faucet where the input is the opening of the tap (input resistance) and the pipe leading away is the output (output resistance). If the faucet is wide open (high input resistance), water flows freely; if the pipe is too narrow (low output resistance), it restricts the water flow out of the faucet. These principles apply similarly to the electrical properties of an amplifier.

Next to the amplifier gain I should say that output swing. Output swing means the output signal amplitude, either you may say peak to peak or amplitude which is quote and unquote distortion free... So, we can say that 5.4 minus this voltage. So, the negative side or we can say that V_out amplitude or magnitude it is equals to 5.4 minus this part which is 0.3. ... Power dissipation it will be V_CC multiplied by these two DC power, I_B + I_C.

This chunk covers two important parameters: output swing and power dissipation. The output swing refers to the maximum peak-to-peak amplitude of the signal at the output, which must remain distortion-free. The voltage swings are determined by the thresholds set by the DC biasing of the circuit. In this case, the output can swing from positive peaks above 5.4V and negative swings down to about -5.1V. Power dissipation is also discussed, calculated as the product of the supply voltage (V_CC) and the sum of the collector and base currents. Understanding power dissipation is crucial as it helps prevent overheating and ensures the longevity of the transistor.

Think of a seesaw again; its movement range (output swing) must not exceed the kids’ weight limits on either side to prevent tipping over (distortion). As for power dissipation, consider a car engine; if it generates too much heat (a result of power dissipation), it requires cooling for efficient operation. Similarly, we need to manage power dissipation in our circuits.

While we talk about the cutoff frequency. That is the other thing we must say that, so far we are assuming that the signal frequency and the value of the capacitors and then associated resistance... this output resistance in combination with maybe the load capacitance... forming one R-C circuit.

This chunk discusses the concept of cutoff frequencies, specifically focusing on how they affect amplifier performance. The concept is that as frequency changes, the behavior of capacitors and resistors in the circuit changes due to their reactive properties. At low frequencies, capacitors can restrict the signal flow, while at high frequencies, the output resistance and load capacitance can impede performance as well. Both lower and upper cutoff frequencies are defined, which sets the range of frequencies over which the amplifier can perform optimally. The bandwidth is thus determined by the difference between these cutoff frequencies.

Imagine a tunnel where certain frequencies correspond to different sizes of vehicles. Small cars can pass through a narrow entry (low frequencies) but might not fare well when it turns into a wider one at the end (high frequencies) due to road limitations. Similarly, our amplifier has a specific range of frequencies where it performs well (bandwidth) before losing its effectiveness.

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