21.6.7. Numerical Example

So, let us consider these device parameters and then the dc voltage and the, this dc voltage and the V and the R. And, what we have to do here it is we need to find what will be this v . So, we need to find what will be the v in terms of this v or v ? So, this is same as v also.

In this numerical example, we will calculate the output voltage (v_out) using given parameters of a MOSFET circuit. The process begins by identifying the parameters such as dc voltage (V), threshold voltage (V_th), and resistance (R). We will express the output voltage in terms of the input voltage (v_in) and other parameters.

Think of the entire circuit as a water system where dc voltage acts as the main water line pressure, V_th is the minimum pressure needed to start the water flow (just like a faucet), and R acts as the resistance in the pipes affecting how fast the water can flow out.

Now, let me consider what are the parameters are given to us it is a say this is 10 V, let you consider this is maybe 2 V and let you consider V = 1 V for simplicity of calculation. And, then W say 10 µ, L = say maybe 1 µ and let you consider K the transconductance parameter K = say 200 µA/V2.

Next, we define the specific values for the parameters. For instance, we assume a dc voltage of 10V, a threshold voltage of 2V, and the transconductance parameter K of 200 µA/V². These values are vital for calculating the operating point of the MOSFET, as they dictate how the transistor behaves under given conditions.

This is similar to setting the specifications for a new engine in a car. By deciding the engine's horsepower, you determine how fast and efficient the car will perform, just like choosing parameters for the MOSFET impacts its behavior in a circuit.

First of all, the small signal parameters the depends on the operating point. So, first thing is that we need to find the operating point and to do that what you do let you consider the dc part and let me find the I . So, I = ( ) and then ( ). So, for the time being let you consider λ is very small so, ≈ 0. So, what we are getting here it is × ( ) this part it is approximately 1. So, that gives us 1 mA.

We begin by determining the operating point of the MOSFET. This is crucial because the small signal parameters depend on it. If we denote the current through the MOSFET as I_ds, we can compute it based on the given parameters. For this case, we assume the output conductance lambda (λ) to be approximately zero. This simplifies the calculations, allowing us to find that I_dc = 1 mA.

Imagine adjusting the temperature setting in a thermostat. The accurate temperature reading determines how effectively the heating system (just like our current) will operate. Similarly, determining the correct operating point is essential for the transistor's performance.

So, the dc current flowing here it is 1 mA so, the drop across this resistance it is 4 V. So, from that we can see that the V . So, that gives us V it is 10 V ‒ 4 × 1. So, that is the 6 V, V the V is 6 V.

After calculating I_dc, we can find the voltage drop across the resistor (R). Given the current of 1 mA and R = 4k ohms, the voltage drop equals 4V. Therefore, the output voltage V_out can be calculated by subtracting this voltage drop from the dc voltage V, resulting in V_out = 10V - 4V = 6V.

This step is like analyzing water flowing through a pipe with a filter that adds resistance. If the original water pressure is 10 units and the filter reduces it by 4 units, we would have effectively 6 units of pressure at the end of the pipe, similar to how V_out reflects the remaining voltage.

Now, we do have we can verify whether the device it is in saturation region or not. So, we do have 2 V here we do have 6 V here; obviously, the devices having pinch of happening at the draining. So, the device it is in saturation.

To ensure the calculations are valid, we need to verify if the MOSFET is in saturation. A MOSFET is considered to be in saturation when the drain-source voltage (V_ds) is greater than the difference between the gate-source voltage (V_gs) and the threshold voltage (V_th). Here, we observe V_gs = 6V and V_th = 2V, confirming that the device does enter saturation, validating our assumptions.

It's like checking whether a car has enough power to reach a hilltop. If the engine can overcome the slope (like a MOSFET over V_th), the car remains at an optimal speed. Similarly, checking saturation helps us understand if the transistor can efficiently transmit the input signal without falling out of its operational limits.

Now, then we can find the value of the small signal parameters. So, since we are considering λ = 0. So, we can see that g it is a small I mean it is 0 and then g . So g it is having different expression. So, either we can write in terms of I and V ‒ V say, for example, if we consider . So, that gives us ( ) .

We proceed to calculate the small signal parameters needed for further analysis. Since we have assumed λ = 0, the output conductance (g_d) is zero. The transconductance (g_m) is calculated based on the equation involving the drain current (I_ds) and the gate-source voltage (V_gs). This helps us in deriving the relationship between current and voltage fluctuations in the small signal model.

This is akin to calculating the efficiency of a car’s performance based on its horsepower and weight. Just as one needs to balance these factors to assess the vehicle's capability, we balance g_m and other parameters to understand the MOSFET's functionality under small signal conditions.

So, from this one what we can find is that this is it becomes ‒ R , that is 4 K and then g D m it is 2 m so, that gives us the gain of 8.

Finally, we substitute the values of R_D (4k) and g_m (2 mA/V) into our gain equation which simplifies to a gain of -R_D * g_m. This results in a voltage gain of 8. This step is critical since it shows how efficiently the circuit amplifies the input signal based on the earlier calculations.

Think of this as calculating the speed of a train based on its power and the weight it is carrying. If our calculations show a high speed (gain of 8), it indicates that the train (circuit) is properly optimized for performance, efficiently moving at that speed across the tracks (signal transmission).

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