Question 14.6: Determining the Propagation Delay of the CMOS Inverter For t...
Determining the Propagation Delay of the CMOS Inverter
For the 0.25-μm process characterized by VDD = 2.5 V, Vtn = −Vtp = 0.5 V, k_{n}^{′} = 3.5 k_{p}^{′} = 110 μA/V², find tPLH, tPHL, and tP for an inverter for which (W/L)n = 1.5 and (W/L)p = 3, and for C = 10 fF. Use both the approach based on average currents and that based on equivalent resistances, and compare the results obtained. If to save on power dissipation, the inverter is operated at VDD = 2.0 V, by what factor does tP change?
(a) Using the average current approach, we determine from Eq. (14.51),
α_{n} = 2 / \left[\frac{7}{4} – \frac{3 V_{tn}}{V_{DD}} + \left(\frac{V_{tn}}{V_{DD}}\right)^{2} \right] (14.51)
α_{n} = \frac{2}{\frac{7}{4} – \frac{3 × 0.5}{2.5} + \left(\frac{0.5}{2.5}\right)^{2}} = 1.7
and using Eq. (14.50),
t_{PHL} = \frac{α_{n} C}{k_{n}^{′}(W/L)_{n} V_{DD}} (14.50)
t_{PHL} = \frac{1.7 × 10 × 10^{−15}}{110 × 10^{−6} × 1.5 × 2.5} = 41.2 ps
Since |Vtp| = Vtn,
αp = αn = 1.7
and we can determine tPLH from Eq. (14.52) as
t_{PHL} = \frac{α_{p} }{k_{p}^{′}(W/L)_{p} V_{DD}} (14.52)
t_{PHL} = \frac{1.7 × 10 × 10^{−15}}{(110/3.5) × 10^{−6} × 3 × 2.5} = 72.1 ps
The propagation delay can now be found as
t_{P} = \frac{1}{2} (t_{PHL} + t_{PLH})
= \frac{1}{2} (41.2 + 72.1) = 56.7 ps
(b) Using the equivalent-resistance approach, we first find RN from Eq. (14.56) as
R_{N} = \frac{12.5}{(W/L)_{n}} kΩ (14.56)
R_{N} = \frac{12.5}{1.5} = 8.33 kΩ
and then use Eq. (14.54) to determine tPHL,
tPHL = 0.69 RNC (14.54)
tPHL = 0.69 × 8.33 × 103 × 10 × 10−15 = 57.5 ps
Similarly we use Eq. (14.57) to determine RP,
R_{P} = \frac{30}{(W/L)_{p}} kΩ (14.57)
R_{P} = \frac{30}{3} = 10 kΩ
and Eq. (14.55) to determine tPLH,
tPLH = 0.69 RPC (14.55)
tPLH = 0.69 × 10 × 103 × 10 × 10−15 = 69 ps
Thus, while the value obtained for tPHL is higher than that found using average currents, the value for tPLH is about the same. Finally, tP can be found as
t_{P} = \frac{1}{2} (57.5 + 69) = 63.2 ps
which is a little higher than the value found using average currents.
To find the change in propagation delays obtained when the inverter is operated at VDD = 2.0 V, we have to use the method of average currents. (The dependence on the power-supply voltage is absorbed in the empirical values of RN and RP.) Using Eq. (14.51), we write
α_{n} = \frac{2}{\frac{7}{4} – \frac{3 × 0.5}{2} + \left(\frac{0.5}{2}\right)^{2}} = 1.9
The value of tPHL can now be found by using Eq. (14.50):
t_{PHL} = \frac{1.9 × 10 × 10^{−15}}{110 × 10^{−6} × 1.5 × 2} = 57.6 ps
Similarly, the value of αp = αn = 1.9 can be substituted in Eq. (14.52) to obtain,
t_{PLH} = \frac{1.9 × 10 × 10^{−15}}{(110/3.5) × 10^{−6} × 3 × 2}. = 100.8 ps
and tP can be calculated as
t_{P} = \frac{1}{2} (57.6 + 100.8) = 79.8 ps
Thus, as expected, reducing VDD has resulted in increased propagation delay.