Question 14.5: Calculating the Propagation Delay of a Simple Inverter Retur...
Calculating the Propagation Delay of a Simple Inverter
Return to the inverter of Fig. 14.17(a) and consider the case where a capacitor C is connected between the output node and ground. If at t = 0, vI goes low, and assuming that the switch opens instantaneously, find the time for vO to reach \frac{1}{2} (V_{OH} + V_{OL}). This is the low-to-high propagation time, tPLH . Calculate the value of tPLH for the case R = 25 kΩ and C = 10 fF.
Before the switch opens, vO = VOL. When the switch opens at t = 0, the circuit takes the form shown in Fig. 14.28(a). Since the voltage across the capacitor cannot change instantaneously, at t = 0+ the output will still be VOL. Then the capacitor charges through R, and vO rises exponentially toward VDD. The output waveform will be as shown in Fig. 14.28(b), and its equation can be obtained by substituting in Eq. (14.45):
y(t) = Y_{∞} – (Y_{∞} – Y_{0+}) e^{-t/τ} (14.45)
vO(∞) = VOH = VDD and vO(0+) = VOL. Thus,
v_{O}(t) = V_{OH} – (V_{OH} – V_{OL}) e^{-t/τ}
where τ = CR. To find tPLH , we substitute
v_{O}(t_{PLH}) = \frac{1}{2} (V_{OH} + V_{OL})
Thus,
\frac{1}{2} (V_{OH} + V_{OL}) = V_{OH} – (V_{OH} – V_{OL}) e^{-t_{PLH}/τ}
which results in
tPLH = τ ln 2 = 0.69 τ
Note that this expression is independent of the values of VOL and VOH. For the numerical values given,
tPLH = 0.69 RC
= 0.69 × 25 × 103 × 10 × 10−15
= 173 ps
