Lecture – 33: Common Source Amplifier (Part B)

Today, we're going to discuss the voltage gain of the common source amplifier. Can anyone tell me what voltage gain is?

Is it the ratio of output voltage to input voltage?

Exactly! It’s essential for understanding amplifier performance. In a common source amplifier, the voltage gain can be expressed as A = -g_m * R_D, where g_m is the transconductance.

What is transconductance again?

Good question! Transconductance, g_m, defines the relationship between the gate-source voltage and the drain-source current. A higher g_m means a stronger response to input voltage changes.

Can we write it in a simpler way?

Sure! Just remember: gain 'A' is proportional to the product of g_m and R_D. If you think of 'Gain = Gain (g_m) x Resistance (R_D)', it can help you recall the relationship.

Does that mean a higher R_D will always give us better gain?

Not always, Student_4! While a larger R_D increases gain, it can also affect other parameters like output resistance. The key is to optimize both.

In summary, the voltage gain is a critical parameter evaluated as A = -g_m * R_D, linking input to output voltage effectively.

Now, let's discuss output resistance. Who can tell me why we care about output resistance in amplifiers?

I think it affects how much current can be supplied!

Absolutely! The output resistance, R_O, impacts how the amplifier interacts with its load. In the case of the common source amplifier, R_O is largely determined by R_D.

So if we connect different loads, how does that change our output?

Great question! When load resistance is connected, it divides the output voltage. The lower the R_O compared to this load, the less voltage you will observe across it. Always consider R_O during load calculations!

So, how do we calculate R_O practically?

From our circuit, once you set the input to zero and look into the output, R_O can typically be simplified to just R_D. Just remember, R_O impacts how well our circuit drives real-world loads.

To recap, R_O influences the performance of our amplifier in real scenarios and is calculated primarily as the drain resistance, R_D.

Now we will explore input resistance. Can anyone tell me what influences input resistance in a common source amplifier?

Isn’t it just how much resistance is connected at the input?

Exactly! Specifically, it's primarily the parallel combination of R_1 and R_2, which connects to the gate. Remember this: Input Resistance = R_1 || R_2.

And how do parasitic capacitances fit into this?

Excellent point! At high frequencies, parasitic capacitances like C_gs and C_gd come into play, affecting the overall input impedance.

So does more capacitance cause problems at high frequency?

Yes! Increased capacitance gives you Miller effects, making the input look like it has larger capacitances. It's a phenomenon that can severely limit bandwidth!

In summary, input resistance is defined as the parallel combination of resistors at the gate, heavily influenced by parasitic capacitances which challenge high-frequency operations.

Let's apply our knowledge to a numerical example. If we set a bias voltage of 12V with resistances of 9kΩ and 3kΩ, can anyone help me with the gain?

First, we need to find V_GS, right? That's V_dd * R_2 / (R_1 + R_2).

Correct! What do we get for V_GS?

That's 3V!

Exactly. Now, what is our quiescent current I_D?

Using the equation, I_D = K * (W/L) * (V_GS - V_th)^2. We get 2 mA.

Great! Now how do we calculate the voltage gain, A?

We can use A = -g_m * R_D. But first, we must find g_m!

Right! If K is 1 mA/V and V_th is 1V, can we compute g_m?

Yes! g_m = K * (W/L) * (V_GS - V_th) = 2 mA/V. Then A = -2 mA/V * 3 kΩ!

Excellent! So we find A to be -6. To round it out, higher gains indicate better amplifiers, but don’t forget the trade-offs with bandwidth.