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## Q. 16.1

Objective: Determine the transition point, minimum output voltage, maximum drain current, and maximum power dissipation of an NMOS inverter with resistor load.

Specifications: Consider the circuit in Figure 16.3(a) with parameters $V_{DD} = 2.5 V$ and $R_{D} = 20 kΩ$. The transistor parameters are $V_{T N} = 0.5 \ V \text{ and } K_n = 0.3 \ mA/V^{2}$

## Verified Solution

The input voltage at the transition point is found from Equation (16.9). We have
$(0.3)(25)(V_{I t} − 0.5)^{2} + (V_{I t} − 0.5) − 2.5 = 0$
which yields
$V_{I t} − 0.5 = 0.515 V$      or     $V_{I t} = 1.015 V$
The output voltage at the transition point is
$V_{O t} = V_{I t} − V_{T N} = 1.015 − 0.5 = 0.515 V$
When $v_{I}$ is high at $v_{I} = 2.5 V$, the output voltage is found from Equation (16.11). We find

$v_{O} = V_{DD} − K_{n} R_{D} [2(v_{I} − V_{T N})v_{O} − v^{2}_{O} ]$               (16.11)

$v_{O} = 2.5 − (0.3(25)[2(2.5 − 0.5)v_{O} − v^{2}_{O}]$
which yields the output low level as
$v_{O} = V_{O L} = 82.3 mV$
The maximum drain current in the inverter occurs when $v_{O} = V_{O L}$ and is found to be
$i_{D,max} = \frac{2.5 − 0.0823}{25} ⇒ 96.7 µA$
The maximum power dissipated in the inverter is
$P_{D,max} = i_{D,max} \cdot V_{DD} = (0.0967)(2.5) = 0.242 mW$
Comment: The level of $V_{O L}$ is less than the threshold voltage $V_{T N}$ ; therefore, if the output of this inverter is used to drive a similar inverter, the driver transistor of the load inverter would be cut off and its output would be high, which is the desired condition. We will compare the maximum drain currents and maximum  power dissipations of the three basic NMOS inverters