68.1.5. Output Voltage Calculation

So, end of it what we are getting here it is collector current of transistor-1 assuming the device it is in active region of operation it is it is β × I and I it is supply voltage minus V of transistor-1 divided by R, and that is 100 multiplied by. So, this is 12, this is 0.6 so, that is 11.4 and then R it is 570 k. So, that gives us this part it is 20 µA × 100. So, that gives us 2 mA.

In this chunk, we calculate the collector current for transistor-1. The formula used is based on the assumption that the transistor is operating in its active region. The collector current (I_C1) is determined by multiplying the transistor's beta (β) by the base current (I_B). The base current is derived from the supply voltage (12V), minus the V_BE(on) voltage of 0.6V, divided by the base resistor (R_B1) of 570 kΩ. This calculation leads us to find that the collector current is 2 mA.

Think of a water faucet (the transistor) where the base current is the amount of water you control with the faucet handle. The beta (β) acts like the efficiency of the faucet in letting water flow out; a higher beta means more water (current) comes out for less handle movement (base current). If you turn the faucet handle slightly, a large amount of water could flow out, just like how a small base current results in a larger collector current.

To get the DC voltage at the output node say V_OUT, what we can do? we can compare β I × (V_CE of transistor-1) = β I of transistor-2 × (V_CE of transistor-2). Now this part and this part we have seen here they are equal. So, now just by equating this factor what we can get here it is since the early voltage of the 2 transistors they are equal. So, that gives us V_CE1 = V_CE2 and also we know that V_CE1 + V_CE2 = V_CC. So, this is V_CE of transistor-2, this is V_CE of transistor-1 that is V_CC which is 12 V and from that we can say that both of them are equal and they are equal to 6 V.

This chunk explains how to derive the output voltage (V_OUT) from the DC voltage across the transistors. By equating the products of beta and the collector-emitter voltage (V_CE) for both transistors, we conclude that since the early voltages are equal, the collector-emitter voltages for both transistors must also be the same. Given that the total supply voltage (V_CC) is 12V, it follows that V_CE1 and V_CE2 each must be 6V. Therefore, the output voltage is 6V.

Imagine two identical water tanks (the transistors) connected in a series with a 12V supply (the total height of the water column). If both tanks are the same size, the water level (voltage) in each tank must be the same, which would naturally be half of the total height (6V in this case). Thus, each tank's output is identical, just as the transistor voltages match in this setup.

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

This section outlines how to determine small-signal parameters, which are crucial for analyzing the small-signal behavior of transistors. For transistor-1, the transconductance (g_m) can be identified using a standard thermal equivalent voltage of 26 mV. The relationship between beta (β) and g_m is employed to derive the input resistance (r_pi), where r_pi is equal to beta multiplied by a derived value related to g_m.

Think of a sensitive microphone (representing the transistor) that amplifies tiny sound waves (small signals) into something audible. The microphone's sensitivity (g_m) helps determine how well it can amplify these small sounds into larger signals, much like how the small-signal parameters allow us to understand how effectively a transistor will respond to small voltage changes.

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