54.1. Common Base and Common Gate Amplifiers (Contd.): Numerical Examples (Part D)

So, to start with let me consider that output swing it is see if this is 12 V and the requirement maybe a ± 4 V; which means that the requirement is 8 V P-P. So, the first step it is that the voltage drop across this resistance, it should be more than 4 V and the so that will ensure the +ve swing of the output voltage it is at least it is 4 V. On the other hand ‒ve side if you see the gate voltage of this MOS transistor should be sufficiently low. So, if the output voltage it is changing from it is quiescent voltage by an amount of say 4 V towards the ‒ve side then we have to ensure that the device it is in saturation region.

In this part, we discuss the necessary voltage drop needed to achieve the desired output swing of an amplifier. If we require an output swing of ±4 V from a 12 V supply, then the total output swing is 8 V. To ensure the positive swing reaches at least 4 V, the voltage drop across the resistance must exceed 4 V. Conversely, we must also account for the negative swing, ensuring that the gate voltage remains low enough for the transistor to remain in saturation.

Imagine a seesaw at a playground. If one side of the seesaw wants to go up (the positive swing), it needs enough force pushing down on the other side (the voltage drop across the resistance). If the weight isn’t enough, it won't go up fully. Similarly, to ensure our amplifier works properly, we must strategically manage both sides of the 'seesaw'—the positive and negative swings.

Which means that the DC voltage here at the output it should be such that DC voltage here and DC voltage at the gate it should be such that the V_GD should be at least 3 V. Why 3 V, because the V_th should have 1 V.

The relationship between DC output voltage and gate voltage must satisfy certain conditions to ensure the MOS device operates in the desired region. The critical point is that the difference between the gate and drain voltages (V_GD) needs to be at least 3 V to keep the transistor in saturation. Here, V_th is the threshold voltage, which is generally around 1 V, indicating the minimum gate-to-source voltage required for the device to turn on.

Think of a water fountain that only starts working when a certain water pressure is achieved. If the pressure (or voltage in our case) at the gate is too low, the fountain (MOS transistor) won't operate. We need to ensure that there's enough pressure difference (3 V in this case) for the fountain to function properly.

Let you consider V_DC voltage it is say we consider a drop of 5 V so, 12 ‒ 5 so, that is 7 V. And so, likewise here also we can keep a margin of 1 V. So, we can say that the gate voltage it can be the 7 V ‒ 4 so that is a 3 V.

To select resistor values for managing output DC voltages, we start by calculating the required drop across the resistors from the supply voltage. For a 12 V supply and a specified drop of 5 V across the load, we find that the gate voltage can remain at 3 V. This helps ensure that the device operates correctly while still allowing room for fluctuations in voltage.

Imagine budgeting for a monthly bill. If your total income is like the supply voltage (12 V), and your fixed expenses are represented by the required voltage drop (5 V), you need to ensure you have enough left (7 V) for other unexpected expenses (like our margin of voltage).

And then next thing is that the you must be having the requirement of the input impedance and the input impedance we know that the it is expression it is and suppose this is given to us that this is a 250 Ω, which means that g_m of the transistor it is we are looking for more right.

The next key element involves determining the input impedance, which is specified as 250 Ω. This requires calculating transconductance (g_m), which represents how effectively a transistor can control its output current based on a change in input voltage. A higher transconductance indicates that the transistor is more effective at controlling the output based on the input signal.

Think of a valve controlling water flow. The input impedance is like the valve's ability to respond to pressure changes in the water supply. If the valve (transistor) can adjust well to the pressure changes (input signals), it will maintain a steady flow (output current).

So, what we have obtained here it is let me summarize, output DC voltage is 7 V, we are converging to 4 V instead of 3 V, I_DS is 4 mA ... the resistance here it is 250 Ω. So, we conclude that the design successfully meets the input impedance and output swing requirements.

In summary, we have calculated all component values required to meet design specifications. The output DC voltage is confirmed to be 7V, with the adjusted gate voltage properly set to ensure functioning within operational limits. The final calculated components resulted in maintaining the expected input impedance at 250 Ω, enabling satisfactory overall performance for the common gate amplifier configuration.

After following a recipe to bake a cake, you check that you have the right ingredients and measurements. Just like ensuring that your cake turns out right, we’ve made sure that all our amplifier values are correct so it functions as intended.

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