68.1.9. Small Signal Gain and Resistance

So, from that we can calculate the small signal parameters of the transistors namely in a g say g of transistor-1 it is thermal equivalent voltage we can consider that is 26 m mV. So, this is . So, that is ℧. So, likewise g it is also = ℧, then r = β of transistor-1 divided by g of the transistor.

In this chunk, we are calculating the small signal parameters for two transistors used in the circuit. Each transistor has a transconductance 'g', which represents its ability to control the output current based on the input voltage. The thermal equivalent voltage is mentioned as 26 mV, which is a standard value for many transistors at room temperature. Additionally, the internal resistance, denoted as 'r', is determined by dividing the transistor's beta (β, also known as current gain) by the transconductance (g). This calculation helps in understanding how the transistor behaves when small signal inputs are applied.

Think of the transistor as a valve that controls water flow. The flow rate is analogous to the current flowing through the transistor. The g and r parameters tell us how sensitive the valve is to adjustments in pressure (or voltage), similar to how much the valve opens or closes when you adjust a handle.

So, to get the DC voltage at the output node say V what we can do? we can compare β I × ( ) of transistor-1 = β I of transistor-2 × ( ) of transistor-2.

Here, we are determining the output DC voltage (V_OUT) by equating the collector current equations of both transistors. The equations factor in the beta (β) of each transistor, which affects the collector current based on the input. By ensuring that the product of the beta and the input current for both transistors is equal, we can derive the output voltage. This helps in ensuring balanced operation of the transistors within the amplifier circuit.

Imagine two water tanks connected by pipes. The height of the water in each tank is crucial for maintaining the system's balance. If one tank (transistor) has a higher water level (more current), it will affect the flow to the other tank. Here, we ensure both tanks are at equal levels (currents) so that the system operates smoothly.

So, to find the output resistance what we have to do? We can stimulate this circuit by a signal called v and then we can observe the corresponding current here.

This chunk discusses how to compute the output resistance of the circuit. To do this, we apply a test signal and measure the current that flows as a result. The output resistance is determined by the ratio of the output signal voltage to the output current. It's important to consider that while testing, all other voltage sources in the circuit should be at zero to ensure accurate measurement.

Think of output resistance like the resistance of a drinking straw to flow. If you use your finger to block the top of the straw and then blow into it (applying a signal), you can measure how hard you have to blow (voltage) to achieve a certain flow of air (current). By measuring this, you can determine how much resistance the straw is adding to your effort.

So, while you are talking about the input capacitance we have to consider the C and so this is C and then this is C.

In this part, we are analyzing the input capacitance of the circuit, which affects how quickly the amplifier can respond to changes in input signals. The capacitance values for different components in the circuit are considered, and their combined effects help determine the overall input capacitance. This is important for setting the upper cutoff frequency of the amplifier, thereby influencing its bandwidth.

Input capacitance can be thought of like a sponge that absorbs water. If the sponge is soaked (charged) with water (input signal), it takes time to dry up (respond to new input changes). The combined capacity of all sponges (capacitors) will determine how fast or slow the system can respond to new water (signal).

So, in summary what we can say that the cutoff frequency it is getting reduced gain got increased and output resistance also got increased.

In this final chunk, a summary is provided regarding the performance of the circuit, focusing on how gains and bandwidth (cutoff frequencies) are affected by using active loads. It is noted that while the gain increases, the cutoff frequency tends to reduce due to higher capacitance and added electrical load. Understanding these trade-offs is crucial for the design and optimization of amplifiers.

Consider a high-performance sports car (the amplifier with active load) versus a regular sedan (passive load). The sports car accelerates faster (higher gain) but has less torque (reduced bandwidth), just like the amplifier performs exceptionally well in gain but has other limitations, which could affect its overall performance in certain scenarios.

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