1. The low-Frequency Response To determine the low-frequency gain or transfer function of the common-source amplifier, we show in Fig.5(a) the circuit with the dc sources eliminated (current source I open­ circuited and voltage source VDD short-circuited). We shall perform the small-signal analysis directly on this circuit. However, we will ignore r o This is done in order to keep the analysis simple and thus focus attention on significant issues. The effect of r o on the low-frequency operation of this amplifier is minor, as can be verified by a SPICE simulation (Section 4.12). The analysis begins at the signal generator by finding the fraction of Vsig that appears at the transistor gate. Which can be written (**) ------(*)

2. Fig.5(a)Analysis of the CS amplifier to determine its low-frequency transfer function.For simplicity, ro is neglected. The equation(**) is recognized as the transfer function of a STC network of the high-pass type with a break or comer frequency ( ω 0 = 1 /C C1 (RG + Rsig). Thus the effect of the coupling capacitor C C1 is to introduce a high-pass STC response with a break frequency that we shall denote ω p1 - - - -(a)

3. Continuing with the analysis, we next determine the drain current Id by dividing Vg by the Total impedance in the source circuit which is [ ( 1 /g m) + ( 1 /s Cs)] to obtain: which can be written in the alternate form We observe that Cs introduces a frequency-dependent factor, which is also of the STC high-pass type. Thus the amplifier acquires another break frequency, To complete analysis, we find Vo by first using the current-divider rule to determine the fraction of Id that flows through R L ------(***) - - -(b)

4. and then multiplying 10 by RL to obtain from which we see that Cc 2 introduces a third STC high-pass factor, giving the amplifier a third break frequency at The overall low-frequency transfer function of the amplifier can be found by combining Equation (**);(***)&(****) and replacing the break frequencies by their symbols from equation(a);(b)&(c) ---(*****) ----(c) ------Eqn(6)

5. The low-frequency magnitude response can be obtained from Eq.(6) by replacing s by Jω and finding I Vo /Vsig I.In many cases, however, one of the three break frequencies can be Much higher than the other two, say by a factor greater than 4.In such case, it is this highest frequency break point that will determine the lower 3-dB frequency fL, and we do not have to do any additional hand analysis. For instance, because the expression for ω P2 includes g m (Eq.b), ω P2 is usually higher than ω P1 and ω P3.If ω P2 is sufficiently separated from ω P1 and ω P3, then which means that in such case, the bypass capacitor determines the low end of the midband Fig.5(b) shows a sketch of the low-frequency gain of a CS amplifier in which the Three break frequencies are sufficiently separated so that their effects appear distinct. Observe that at each break frequency, the slope of the asymptotes to the gain function increases by 20 dB/decade.And hence the break frequency associated with each of thethree capacitors. Theprocedure is simple: 1. Reduce Vsig to zero. 2. Consider each capacitor separately; that is, assume that the other two capacitors are acting as perfect short circuits. 3. For each capacitor, find the total resistance seen between its terminals. This is the resistance that determines the time constant associated with this capacitor.

6. Selecting Values for the Coupling and Bypass Capacitors : We now address the design issue of selecting appropriate values for Cc1, Cs, and Cc 2. The design objective is to place the lower 3-dB frequency fL at a specified value while minimizing the capacitor values. Fig.5(b)Sketch of the low-frequency magnitude response of a CS amplifier for which the three breakfrequencies are sufficiently separated for their effects to appear distinct.

7. Find the midband gain AM and the upper 3-dB frequency fH of a CS amplifier fed with a signal source having an internal resistance Rsig = 100 kΩ. The amplifier has RG = 4.7MΩ, R D = R L =15k Ω, g m = 1mA/V, r O = 150 kΩ, Cgs = 1pF,and Cgd=0.4 pF. Example1: Solution: Where Thus,

8. We wish to select appropriate values for the coupling capacitors Cc1 and Cc 2 and the bypass capacitor Ccs for the CS amplifier whose high-frequency response was analyzed in Example 1.The amplifier has R G = 4.7MΩ,R D =R L 15KΩ, Rsig = 100KΩ,and g m = 1mA/V. It is required to have fL at 100 Hz and that the nearest break frequency be at least a decade lower. Example2: Solution: We select Cs so that which yields and Which results in Thus,