33.5.9. Numerical Problem

We are considering the same circuit here and also we have been given these parameters that K×W / L = 1 mA/V; suppose it is given to us. And, then threshold voltage of the transistor is say 1 V. And, on the other hand the DC parameters are rather bias circuit information it is given as that V_dd is say 12 V and then R_1 is 9 k and R_2 is 3 k.

In this chunk, we lay out the parameters for our numerical problem involving a circuit. First, we have a transistor with a certain gain (K×W / L) of 1 mA/V, which indicates how it behaves electrically. Additionally, the threshold voltage (V_th) of 1 V is critical since it helps us understand when the transistor will start conducting current. DC biasing is also important, with the supply voltage (V_dd) set to 12 V, and two resistances (R_1 and R_2) set to 9 kΩ and 3 kΩ respectively. These values are foundational as they will influence the circuit's operating point and performance.

Think of it like setting up a recipe. Just like you have specific quantities of ingredients (the parameters), they need to be just right before you can bake a cake (the circuit functioning properly). If you misjudge an ingredient, the cake might not rise as it should.

First thing is the DC operating point and to do that we need to get the DC voltage at the gate or gate to source. So, if you consider this R_1 and R_2 value and then this is V_dd = 12 V, what we are getting is V_G = V_dd × (R_2 / (R_1+R_2)). So, that gives us 12 × (3 / (9+3)) = 3 V.

Here, we apply a voltage divider rule to find the gate-to-source voltage (V_G). We calculate V_G using the formula derived from the resistive divider formed by R_1 and R_2, with the total supply voltage V_dd. By substituting the values of R_1 and R_2, we discover that V_G is 3 V. This voltage will influence the operation of the transistor since it determines if the transistor will be in the 'on' or 'off' state depending on the threshold voltage.

Imagine you are pouring juice into two glasses of different sizes. The amount of juice (V_dd) is the same, but if one glass (R_1) is wider than the other (R_2), the amount of juice each glass receives will be different. This calculation helps us understand how much voltage (V_G) we have 'poured' into our circuit.

In fact, this is same as V_G = 3 V here. Now, using this information I can say that I_DS = (K×W/L / 2) × (V_G - V_th)² = (1 mA/V / 2) × (3 V - 1 V)² = 2 mA of quiescent current I_DS.

We move on to calculate the quiescent current (I_DS) through the transistor. Using the given formula, which includes the gain factor (K×W/L) and the difference between the gate-to-source voltage (V_G) and the threshold voltage (V_th), we can plug in the values we've calculated. This gives us a quiescent current of 2 mA, indicating the steady-state current flowing through the transistor when no input signal is applied.

Think of the transistor as a water valve. The quiescent current is like the flow of water when the valve is slightly open. Even though there is no extra force pushing it (input signal), a steady flow is maintained based on how much you've opened the valve (V_G vs. V_th).

Also, if this R_D = 3 k ohm then V_D = V_dd - R_D × I_DS = 12 V - 3 kΩ × 2 mA = 6 V. So, the DC voltage at the output is V_D = 6 V.

Next, we determine the output voltage (V_D) across the drain resistor (R_D) based on the quiescent current we previously calculated. By substituting our known values into the simple voltage formula, we find that V_D equals 6 V. This output voltage is critical because it defines the operating point of the amplifier at the output.

Imagine you are balancing weights on a seesaw. The supply voltage (V_dd) is like the total weight on one side. By accounting for the weight of the resistor (R_D × I_DS), you can find the effective weight on the other side of the seesaw, which represents your output voltage (V_D).

Now, the value of g_m can be calculated as g_m = (K×W/L) × (V_G - V_th) = 1 mA/V × (3 V - 1 V) = 2 mA/V. Now, the gain voltage A is given as A = -g_m × R_D = - (2 mA/V) × (3 kΩ) = -6.

In this chunk, we find the transconductance (g_m), which indicates how effectively the transistor converts input voltage changes to output current changes. We then compute the voltage gain (A) using the previously derived formula. The negative sign signifies the phase inversion typical in common-source amplifiers, resulting in a gain of -6.

Think of g_m as a translator. If the translator (the transistor) is really good at converting words from one language (input voltage) to another (output current), it can effectively amplify the message (gain). The negative sign is like a misunderstanding in translation, where what was said is exactly the opposite of what was intended.

The minimum possible drain voltage we can say that (3 V - 1 V) leads to the valley being 2 V. The maximum swing from 6 V to reach to this 2 V is 6 V - 2 V = 4 V, so, the output swing is ± 4 V.

The final step is to compute the output voltage swing. The minimum output voltage is determined by the threshold voltage shift based on the input signal. The output swing of ± 4 V signifies that the output can vary between 2 V and up to 6 V without distortion, ensuring that the circuit remains within operational limits during fluctuations.

Imagine a swing set in a playground. The highest point (maximum voltage) is 6 V, while the lowest point (minimum voltage) is limited by the starting point of the swing (2 V). The total distance the swing travels back and forth (swing amplitude) is equivalent to the output swing of the circuit.

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