EECE 310 Homework 9, Fall 2011
American University of Beirut · Electronics · 1 page
Stuck on a question? A tutor can walk you through it.
More EECE 310 files
- Homework Spice3 Solution · Summer 2012Homework · with solutionView
- Homework Spice2 Solution · Summer 2012Homework · with solutionView
- Homework Spice1 Solution · Summer 2012Homework · with solutionView
- Homework 1 Solution · Fall 2011Homework · with solutionView
- Homework 2 · Fall 2011HomeworkView
- Homework 1 · Fall 2011HomeworkView
- Homework 1 · Fall 2011HomeworkView
- Homework 1 Solution · Fall 2011Homework · with solutionView
Show the text of this homework
American University of Beirut
DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING
EECE 310 – Electronics
Fall 2011 – 2012
Due Wednesday December 21, 2011, at 9:00 am
Homework 10
1. Consider a CMOS inverter with the following circuit parameters:
The N-channel MOSFET has L = 0.18 m, W = 0.4 m, Vt = 0.6 V, and k’ = 330 A/V2.
The P-channel MOSFET has L = 0.18 m, W = 0.8 m, Vt = –0.7 V, and k’ = 170 A/V2.
The supply voltage is VDD = 1.8 V.
a) Find the values of VOL and VOH for this inverter.
b) Find the value of VOL when a 10 k resistor is connected from the output to VDD.
The input is at VDD. What is the total circuit power dissipation in this case?
c) Find the value of VOH when a 10 k resistor is connected from the output to
ground. The input is at 0 V. What is the total circuit power dissipation in this
case?
d) A 50 fF capacitor is connected from the output of the inverter to ground. Find the
dynamic power dissipation of the inverter is the input voltage is a square wave
with a cycle time of 2 ns.
In the following two problems, assume that for a BJT in the active region, |VBE| = 0.7 V.
2. The base current of the transistor shown in Fig. 1 is 30 A. The BJT has = 120.
a) Assume that the BJT is active. Find IE and the required value of RB.
b) Find VCE and verify that the BJT is operating in the active region.
c) If the emitter of the BJT is grounded, would the transistor remain in the active region?
Use the value of RB calculated in part (a).
3. The PNP transistor shown in Fig. 2 has = 0.989. Find RE and RC to get IE = 0.15 mA
and VEC = 3.3 V.
Figure 1 Figure 2