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

Question

Determining the Propagation Delay of the CMOS Inverter

For the 0.25-μm process characterized by V_DD = 2.5 V, V_tn = −V_tp = 0.5 V, [latex]k_{n}^{′}[/latex] = 3.5 [latex]k_{p}^{′}[/latex] = 110 μA/V², find t_PLH, t_PHL, and t_P 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 V_DD = 2.0 V, by what factor does t_P change?

Accepted Answer

(a) Using the average current approach, we determine from Eq. (14.51),

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

and using Eq. (14.50),

[latex]t_{PHL} = \frac{α_{n} C}{k_{n}^{′}(W/L)_{n} V_{DD}} [/latex] (14.50)

[latex]t_{PHL} = \frac{1.7 × 10 × 10^{−15}}{110 × 10^{−6} × 1.5 × 2.5} = 41.2 ps[/latex]

Since |V_tp| = V_tn,

α_p = α_n = 1.7

and we can determine t_PLH from Eq. (14.52) as

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

[latex]t_{PHL} = \frac{1.7 × 10 × 10^{−15}}{(110/3.5) × 10^{−6} × 3 × 2.5} = 72.1 ps[/latex]

The propagation delay can now be found as

[latex]t_{P} = \frac{1}{2} (t_{PHL} + t_{PLH})[/latex]

[latex]= \frac{1}{2} (41.2 + 72.1) = 56.7 ps[/latex]

(b) Using the equivalent-resistance approach, we first find R_N from Eq. (14.56) as

[latex]R_{N} = \frac{12.5}{(W/L)_{n}} kΩ[/latex] (14.56)

[latex]R_{N} = \frac{12.5}{1.5} = 8.33 kΩ[/latex]

and then use Eq. (14.54) to determine t_PHL,

t_PHL= 0.69 R_NC (14.54)

t_PHL = 0.69 × 8.33 × 10³ × 10 × 10⁻¹⁵ = 57.5 ps

Similarly we use Eq. (14.57) to determine R_P,

[latex]R_{P} = \frac{30}{(W/L)_{p}} kΩ[/latex] (14.57)

[latex]R_{P} = \frac{30}{3} = 10 kΩ[/latex]

and Eq. (14.55) to determine t_PLH,

t_PLH = 0.69 R_PC (14.55)

t_PLH = 0.69 × 10 × 10³ × 10 × 10⁻¹⁵ = 69 ps

Thus, while the value obtained for t_PHL is higher than that found using average currents, the value for t_PLH is about the same. Finally, t_P can be found as

[latex]t_{P} = \frac{1}{2} (57.5 + 69) = 63.2 ps[/latex]

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 V_DD = 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 R_N and R_P.) Using Eq. (14.51), we write

[latex]α_{n} = \frac{2}{\frac{7}{4} - \frac{3 × 0.5}{2} + \left(\frac{0.5}{2}\right)^{2}} = 1.9[/latex]

The value of t_PHL can now be found by using Eq. (14.50):

[latex]t_{PHL} = \frac{1.9 × 10 × 10^{−15}}{110 × 10^{−6} × 1.5 × 2} = 57.6 ps[/latex]

Similarly, the value of α_p = α_n = 1.9 can be substituted in Eq. (14.52) to obtain,

[latex]t_{PLH} = \frac{1.9 × 10 × 10^{−15}}{(110/3.5) × 10^{−6} × 3 × 2}. = 100.8 ps[/latex]

and t_P can be calculated as

[latex]t_{P} = \frac{1}{2} (57.6 + 100.8) = 79.8 ps[/latex]

Thus, as expected, reducing V_DD has resulted in increased propagation delay.

Question 14.6: Determining the Propagation Delay of the CMOS Inverter For t...

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