Foundations of Analog and Digital Circuits Mas - Wordpress ...

transitions from low to high):
vC = VTH + (VS − VTH)e
11.3 Power Dissipation in Logic Gates CHAPTER ELEVEN 607
−t
RTHCL .
By substituting t = 0, we can verify that the capacitor is initially charged to
VS when the MOSFET just turns on. Similarly, by substituting t =∞, we can
confirm that the final voltage on the capacitor is VTH.
The rest of the derivation for w1 follows the steps in Section 11.2. When
T1 ≫ RTHCL, we obtain the following simplified expression for w1:
V
w1 =
2 S
RL + RON
Energy Dissipated During Interval T2
T1 + V2 SR2 LCL . (11.23)
2(RL + RON) 2
Now, let us consider the second interval T2 in which the input signal is low
and the switch is off. During T2, the capacitor charges through the resistor RL.
The initial voltage on the capacitor is VTH.
As in the previous section, let us first determine vC. When the switch
is off, the circuit shown in Figure 11.12 applies. Since the initial voltage on
the capacitor is VTH and the final voltage is VS, we can write the following
expression for vC:
vC = VTH + (VS − VTH) 1 − e
−t
RLCL Notice that as t →∞, the capacitor voltage vC → VS. Similarly, for t = 0, the
capacitor voltage is VTH.
Following the derivation in Section 11.2, we can derive the following
expression for w2 when T2 ≫ RLCL:
w2 =
V2 SR2 LCL .
2(RL + RON) 2
Total Energy Dissipated
Combining the expressions for w1 and w2, we obtain total energy dissipated
by the inverter in a cycle:
w = w1 + w2 =
V 2 S
RL + RON
.
T1 + V2 SR2 LCL 2(RL + RON) 2 + V2 SR2 LCL .
2(RL + RON) 2
+
VS -
R L
v C
C L
FIGURE 11.12 Equivalent
circuit with the switch open for the
inverter.