63.6.1. Output Voltage Gain Calculation

To start with let we yeah, let me use different color here. So, to start with we do have here it is supply voltage is 12 V and then we do have R which is 570 kΩ and then we do have the V_be is approximately 0.6 V drop. So, from that we can calculate what is the I_B1? In fact, I_B1 is 12 V supply minus V_be of 0.6 divided by R1, which is 570 kΩ, so that gives us 20 µA. And we do have β here which is 100, from that we can get the collector current I_C1 = 2 mA.

In this chunk, we are calculating the base current (I_B1) and collector current (I_C1) for the first transistor in the circuit. We start with a supply voltage (V_supply) of 12 volts. The base-emitter voltage (V_be) drop is around 0.6 volts, which is typical for BJTs. By subtracting V_be from the supply voltage and dividing by the biasing resistor (R1 = 570 kΩ), we can compute I_B1. This gives us a base current of 20 µA. Knowing the current gain (β) is 100 (which means the collector current is 100 times the base current), we can determine I_C1 to be 2 mA. This step is crucial as it sets the operating point for the transistor.

Think of the supply voltage as the water pressure in a pipe system. The V_be acts like a valve that controls how much water (current) can flow through. The resistor (R1) is like a narrow section of the pipe that limits the flow. Just as the pressure downstream after the valve depends on these factors, the currents in our transistor circuit depend on the input voltage, voltage drops, and resistance.

So, we can say that I_C is approximately equal to it is pure emitter current, which is equal to I_C1 and that is 2 mA. So, we can say that this current of 2 mA of current, it is also flowing through R3 and we do not have this current source; as we said that the bias current we are assuming here it is 0. So, the drop across this R3, it is 2.8 k × 2 mA, so that gives us 5.6 V; which implies that, the voltage at the collector of transistor-2, V_C = 12 ‒ 5.6, so that is 6.4 V.

This chunk refers to calculating the collector voltage (V_C) of the second transistor in the cascode amplifier configuration. As we've determined a collector current (I_C2) of 2 mA flowing through a resistor (R3) of 2.8 kΩ, we can now calculate the voltage drop across R3. The voltage drop across R3 can be found with Ohm's Law (V = I * R), yielding 5.6 V. By subtracting this drop from the supply voltage (12 V), we find the collector voltage (6.4 V). This plays a critical role in maintaining the transistors in their active operating region.

Imagine a two-story water tank system. The first tank is the input water supply (12 V), and the second tank's height corresponds to the collector voltage. As water flows out of the pipes (current), it leaves behind a lower water level (voltage drop). You can find the level in the second tank by subtracting the amount of water flowed out from the original height.

We can say that g_m1 corresponding . So, this = ℧. So, likewise we can get g_m2, it is also same ℧; because the collector currents they are the same, the output resistance r_o1 of the transistor Q1, it is coming from the early voltage here and then 2 mA of current. So, this gives us r_o = 50 k; in fact r_o is also 50 kΩ.

In this section, we identify the transconductance (g_m) for both transistors, which is significant in determining the amplifier's gain. When the collector currents are equal, the transconductance values are the same, allowing us to use simplified calculations for amplifying signals. The output resistances (r_o) are derived from the early voltage (which affects how the current varies with voltage) and current flowing through the transistors, yielding both r_o values as 50 kΩ. Knowing these parameters aids in calculating voltage gains more effectively.

Think of the transconductance as the effectiveness of a faucet in controlling the flow of water. If both faucets (transistors) are set to deliver the same maximum flow when turned fully open, then knowing their settings helps us predict how much water we can get from either faucet before affecting the overflow (gain) at a certain height (desired signal level).

So, the output voltage v_out will depend on the calculated currents and resistance values. Therefore, the gain A is defined as the output voltage divided by the input voltage.

This crucial section summarizes how we calculate the output voltage (v_out) in relation to the various currents and resistances identified previously. The voltage gain (A) of the amplifier is calculated by taking the ratio of output voltage to input voltage. Understanding this ratio will help us determine how well our amplifier converts input signals into larger output signals, which is essentially the purpose of amplification.

Imagine you're amplifying a voice in a hall using a microphone. The small sound (input voltage) enters the microphone and gets amplified into a louder sound (output voltage). The gain tells us how many times louder the output is compared to someone speaking directly without a microphone. If the output is significantly higher than the input, we say the microphone is performing well.

So, the input capacitance here it is 20 pF. Now this is of course, one of the important point that, the gain of this cascode amplifier you may recall if I consider R3; and then the overall gain, it was.

In this chunk, we discuss capacitance (C_in) and its role in determining the availability of bandwidth in the circuit. A lower input capacitance (in this case, 20 pF) allows the amplifier to achieve a higher bandwidth, which is crucial for processing a wide range of frequencies without distorting the signal. The selection and consideration of capacitors are pivotal in optimizing amplifier performance, especially for signals that change rapidly over time.

Think of a highway that allows vehicles to travel at high speed. If the highway is narrow (like high capacitance), fewer cars can pass simultaneously, reducing efficiency. A wider highway (lower capacitance) allows more cars to travel together without slowing down, enabling fast-moving traffic (high bandwidth) without congestion (signal degradation).

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