Holooly Plus Logo

Question 6.14: Implement the design in Example 6.13 by exploiting a properl......

Implement the design in Example 6.13 by exploiting a properly rescaled (with respect to the device periphery W) and integrated nanometer gate length device with W=0.25\text{ mm} and small-signal parameters C_{G S}=1.47 \times 10^{-14}\text{ F} , C_{G D}=0.1 \times 10^{-14}\text{ F} , C_{D S}=0.2 \times 10^{-13}\text{ F} , \text{g}_m=200\text{ mS} , \tau=0.05 \times 10^{-12}\text{ s} , R_I=1.6\;  \Omega ,R_{D S}=956 \; \Omega ; L_G=7 \times 10^{-12}\text{ H} , L_D=1 \times 10^{-12}\text{ H} , L_S=1 \times 10^{-12}\text{ H} , R_G=1.46\; \Omega ,R_S=1.55\; \Omega, R_D=1.6\; \Omega .

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

To obtain a 220 mS transconductance we need a rescaling ratio α :

\alpha=\frac{W^{\prime}}{W}=\frac{220}{200}=1.1,

where W is the original gate periphery, W^{\prime} the modified one, W^{\prime}=0.25 \cdot 1.1 = 0.275 mm. The new parameters (series elements scale according to 1/W, parallel elements according to W, \tau^{\prime}=\tau=0.05 \times 10^{-12}\text{ s} is unchanged) are listed in Table 6.8.
The (intrinsic and ideal) cutoff and maximum oscillation frequency of the rescaled device are:

\begin{aligned} f_T & =\frac{\text{g}_m^{\prime}}{2 \pi C_{G S}^{\prime}}=\frac{0.220}{2 \pi \cdot 1.62 \times 10^{-13}}=216\text{ GHz} \\ f_{\max } & =\frac{f_T}{2} \sqrt{\frac{R_{D S}^{\prime}}{R_I^{\prime}+R_G^{\prime}+R_S^{\prime}+2 \pi f_T R_{D S}^{\prime} R_G^{\prime} C_{G D}^{\prime}}}=1.3\text{ THz} . \end{aligned}

Notice that the theoretical value of f_{\max } is somewhat unrealistically high, since parasitics are only partially accounted for in (5.13).

f_{\max }=\frac{f_T}{2} \sqrt{\frac{R_{D S}}{R_I+R_G+R_S+2 \pi f_T R_{D S} R_G C_{G D}}}, \hspace{30 pt} \text{(5.13)}

However, the figures of merit obtained point out that the device is a high performance one.
With the parallel feedback of 550 Ω derived in Example 6.13, we obtain the results in Fig. 6.49 (stability coefficient) and in Fig. 6.50 (scattering parameters). The RL feedback results are obtained with an additional feedback peaking inductance L_p=2\text{ nH}. We notice the following points.
The device is unstable without feedback, and becomes stable with a resistive feedback. This points out the stabilizing effect of the resistive feedback. The use of a RL feedback, however, makes the device again potentially unstable at very high frequency (notice that this is still well below the device f_{\max } ), thus requiring additional out-of-band stabilization. In practice also a low-frequency additional stabilization is needed since a DC blocking capacitor must be inserted in the feedback loop, that opens the loop at low frequency thus canceling the stabilizing effect of feedback.
The values of S_{21} and S_{21 f} are lower than the theoretical ones (22 and 10, respectively) as evaluated on the basis of the ideal intrinsic equivalent circuit, see Example 6.13. This is due to the effect of parasitics (in particular the source resistance, that lowers the effective transconductance) and of the output resistance R_{D S}. The introduction of feedback clearly reduces the gain while increasing the amplifier bandwidth, as expected. Without feedback the 3 dB bandwidth is slightly below 20 GHz while it becomes above 35 GHz with resistive feedback and above 50 GHz with RL feedback. Notice that the amount of gain that can be obtained (on the basis of the scattering parameters) could in fact be improved by optimizing the closing resistance, chosen here as 50 Ω (reactive terminations are virtually impossible to implement in a very broadband structure). The peaking inductor leads to a significant amount of broadbanding, while decreasing at high frequency the input matching. Finally, the device matching is improved vs. the original (no feedback) condition, and, as expected, the feedback device is less unidirectional than the original one.

Table 6.8 Rescaled elements of the small-signal equivalent circuit in Example 6.14.
C_{G S}^{\prime}=\alpha C_{G S}=1.62 \times 10^{-13}\text{ F} C_{G D}^{\prime}=\alpha C_{G D}=1.1 \times 10^{-15}\text{ F}
C_{D S}^{\prime}=\alpha C_{D S}=2.2 \times 10^{-14}\text{ F} \text{g}_m^{\prime}=\alpha \text{g}_m=220\text{ mS}
R_I^{\prime}=\alpha^{-1} R_I=1.45 \; \Omega R_{D S}^{\prime}=\alpha^{-1} R_{D S}=869 \; \Omega
L_G^{\prime}=\alpha^{-1} L_G=6.36 \times 10^{-12}\text{ H} L_D^{\prime}=\alpha^{-1} L_D=9.09 \times 10^{-13}\text{ H}
L_S^{\prime}=\alpha^{-1} L_S=9.09 \times 10^{-13}\text{ H} R_G^{\prime}=\alpha^{-1} R_G=1.33 \; \Omega
R_S^{\prime}=\alpha^{-1} R_S=1.41 \; \Omega R_D^{\prime}=\alpha^{-1} R_D=1.45 \; \Omega
1102891 Figure 6.49
1102891 Figure 6.50

Related Answered Questions

Question: 6.7

Verified Answer:

Consider the simplified equivalent circuit of a fi...
Question: 6.6

Verified Answer:

Consider a two-port with the scattering matrix: [l...
Question: 6.2

Verified Answer:

Reworking (6.22) G_{ \text{op} }=\left|S_{...
Question: 6.1

Verified Answer:

The maximum of P_L vs. R_L[/...