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

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,

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

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

$α_{n} = \frac{2}{\frac{7}{4}  –  \frac{3  ×  0.5}{2.5}  +  \left(\frac{0.5}{2.5}\right)^{2}} = 1.7$

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

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

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

t_PHL = 0.69 R_NC                            (14.54)

α_p = 1.7

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

Finally, we determine t_P as

$t_{P} = \frac{1}{2} (25.8 + 31.5)  = 28.7  ps$