Determining the Effective Load Capacitance C and the Propagation Delay Consider a CMOS inverter fabricated in a 0.25-μm process for which Cox = 6 fF/μm², μnCox = 110 μA/V²,

Question

Determining the Effective Load Capacitance C and the Propagation Delay

Consider a CMOS inverter fabricated in a 0.25-μm process for which C_ox = 6 fF/μm², μ_nC_ox = 110 μA/V², μ_pC_ox = 30 μA/V², V_tn = –V_tp = 0.5 V, and V_DD = 2.5 V. The W/L ratio of Q_N is 0.375 μm/0.25 μm, and that for Q_P is 1.125 μm/0.25 μm. The gate–source and gate–drain overlap capacitances are specified to be 0.3 fF/μm of gate width. Further, the effective (large-signal) values of drain–body capacitances are C_dbn = 1 fF and C_dbp = 1 fF. The wiring capacitance C_w = 0.2 fF. Find t_PHL, t_PLH , and t_P when the inverter is driving an identical inverter.

Accepted Answer

First, we determine the value of the equivalent capacitance C using Eqs. (14.58) and (14.59),

C_g3 + C_g4 = (WL)₃C_ox + (WL)₄C_ox + C_gsov3 + C_gdov3 + C_gsov4 + C_gdov4 (14.58)

C = 2 C_gd1 + 2 C_gd2 +C_db1 + C_db2 + C_g3 + C_g4 + C_w (14.59)

where

C_gd1 = 0.3 × W_n = 0.3 × 0.375 = 0.1125 fF

C_gd2 = 0.3 × W_p = 0.3 × 1.125 = 0.3375 fF

C_db1 = 1 fF

C_db2 = 1 fF

C_g3 = 0.375 × 0.25 × 6 + 2 × 0.3 × 0.375 = 0.7875 fF

C_g4 = 1.125 × 0.25 × 6 + 2 × 0.3 × 1.125 = 2.3625 fF

C_w = 0.2 fF

Thus,

C = 2 × 0.1125 + 2 × 0.3375 + 1 + 1 + 0.7875 + 2.3625 + 0.2 = 6.25 fF

Next we use Eqs. (14.51) and (14.52) to determine t_PHL,

[latex]α_{n} = 2 / \left[\frac{7}{4} - \frac{3 V_{tn}}{V_{DD}} + \left(\frac{V_{tn}}{V_{DD}}\right)^{2} \right] [/latex] (14.51)

[latex]t_{PHL} = \frac{α_{p} }{k_{p}^{′}(W/L)_{p} V_{DD}}[/latex] (14.52)

[latex]α_{n} = \frac{2}{\frac{7}{4} - \frac{3 × 0.5}{2.5} + \left(\frac{0.5}{2.5}\right)^{2}} = 1.7[/latex]

[latex]t_{PHL} = \frac{1.7 × 6.25 × 10^{−15}}{110 × 10^{−6} ×(0.375/0.25) × 2.5} = 25.8 ps[/latex]

Similarly, we use Eqs. (14.53) and (14.54) to determine t_PHL ,

[latex]α_{p} = 2 / \left[\frac{7}{4} - \frac{3 |V_{tp}|}{V_{DD}} + \left|\frac{V_{tp}}{V_{DD}}\right|^{2} \right][/latex] (14.53)

t_PHL= 0.69 R_NC (14.54)

α_p = 1.7

[latex]t_{PLH} = \frac{1.7 × 6.25 × 10^{−15}}{30 × 10^{−6} ×(1.125/0.25) × 2.5} = 31.5 ps[/latex]

Finally, we determine t_P as

[latex]t_{P} = \frac{1}{2} (25.8 + 31.5) = 28.7 ps[/latex]

Question 14.7: Determining the Effective Load Capacitance C and the Propaga...

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