Estimate the −3-dB bandwidth of a resistively-degenerated common-source stage. Assume λ = γ = 0.

Question

Accepted Answer

Shown in Fig. 6.53(a), the small-signal model is of third order,[latex]^8[/latex] providing little intuition. The zero-value time constant method can give a rough estimate of the circuit’s bandwidth, thereby revealing the contribution of each capacitor.

We begin with the time constant associated with [latex]C_{GS}[/latex] and set [latex]C_{G D } and C_L[/latex] to zero. As depicted in Fig. 6.53(b), the resistance seen by [latex]C_{GS} is V_X /I_X[/latex] . We denote this resistance by [latex]R_{CGS}[/latex]. Since [latex]V_1 = V_X[/latex] and the current flowing through [latex]R_S[/latex] is equal to [latex]g_m V_1 − I_X = g_m V_X − I_X[/latex] , we write a KVL as follows:

[latex]I_X R_G = V_X + (g_m V_X − I_X )R_S[/latex] (6.123)

obtaining

[latex]R_{CGS} =\frac{ R_G + R_S} {1 + g_m R_S}[/latex] (6.124)

For the resistance seen by [latex]C_{G D}[/latex], we have from Example 6.21

[latex]R_{CGD} = R_D +\left(\frac{ g_m R_D}{1 + g_m R_S}+ 1\right)R_G[/latex] (6.125)

Finally, the resistance seen by [latex]C_L[/latex] is simply equal to [latex]R_D[/latex]. It follows that the −3-dB bandwidth is given by

[latex]ω^{−1}_{−3d B} = \frac{R_G + R_S}{1 + g_m R_S}C_{GS} +\left[R_D +\left(\frac{g_m R_D}{1 + g_m R_S}+ 1\right) R_G \right] C_{G D} + R_DC_L [/latex] (6.126)

With no degeneration, this result reduces to Eq. (6.35). With a finite RS, the effect of [latex]C_{GS} and R_G[/latex] is reduced by a factor of [latex]1 + g_m R_S,[/latex] albeit at the cost of voltage gain.

[latex]ω_{p1} = \frac{1}{R_S(1 + g_m R_D)C_{G D} + R_SC_{GS} + R_D(C_{G D} + C_{DB})}[/latex] (6.35)

[latex]^8[/latex] This can be seen by observing that it is possible to impose three independent initial conditions across the three capacitors
without violating KVL.

Question 6.22: Estimate the −3-dB bandwidth of a resistively-degenerated co...

Related Answered Questions

Repeat the above example for a common-gate stage containing a gate resistance of RG and a source resistance of RS. ...

Verified Answer:

A CG stage contains a resistance RG in series with the gate [Fig. 6.50(a)]. If only CG D is considered, determine the pole frequency. ...

Verified Answer:

Repeat the above example while including both CF and a capacitor, CB, from node B to ground. ...

Verified Answer:

Using the EET, find the transfer function of the circuit in Fig. 6.47(a) ...

Verified Answer:

Consider the cascode stage shown in Fig. 6.35(a), where the load resistor is replaced by an ideal current source. Neglecting the capacitances associated with M1, representing Vin and M1 by a Norton equivalent as in Fig. 6.35(b), and assuming γ = 0, compute the transfer function. ...

Verified Answer:

Does cascoding increase the GBW product? Assume that the output pole is dominant. ...

Verified Answer:

Not all fully differential circuits are free from mirror poles. Figure 6.42(a) illustrates an example where current mirrors M3–M5 and M4–M6 “fold” the signal current. Estimate the low-frequency gain and the transfer function of this circuit. ...

Verified Answer:

For the common-gate stage shown in Fig. 6.32(a), calculate the transfer function and the input impedance, Zin. Explain why Zin becomes independent of CL as this capacitance increases. ...

Verified Answer:

Can we construct a (two-terminal) inductor from a source follower? ...

Verified Answer:

Neglecting channel-length modulation and body effect, calculate the transfer function of the circuit shown in Fig. 6.27(a). ...

Verified Answer: