View
242
Download
1
Category
Preview:
Citation preview
ElectronicsEECE2412 — Spring 2013Exam #2 with Solutions
Prof. Charles A. DiMarzioDepartment of Electrical and Computer Engineering
Northeastern UniversityFile:11785/exam2
23 April 2013
1 BJT Amplifier
The following figure shows a BJT amplifier, which we are going to analyze.The transistor has β = 160, and VA = 100 Volts. RS is part of the source.The input and output voltages are at the indicated locations, vin and vout,respectively.
1
1 BJT AMPLIFIER 1.1 Amplifier Type
1.1 Amplifier Type
What type of amplifier is it? Common Collector or Emitter Follower
What is the typical situation in which this amplifier is used?
Used when unity voltage gain is needed
and source alone cannot supply current.
1.2 Small–Signal Parameters
What are the small–signal parameters?
DiMarzio 11785/exam2, Apr 2013 Page 2DiMarzio 11785/exam2, Apr 2013 Page 2DiMarzio 11785/exam2, Apr 2013 Page 2
1 BJT AMPLIFIER 1.3 AC Circuit
gm = IDCC /VT =
1 mA
25 mV= 0.04 A/V
rπ = β/gm =160
0.04 A/V= 4000 Ω
re = IDCE /VT =
rπβ + 1
= 24.8 Ω
R0 =VA
IDCC
=100 Volts
10−3 Amperes= 105 Ω
gm = 0.04 A/V
rπ = 4 kΩ
re = 24.8 Ω
R0 = 100 kΩ
1.3 AC Circuit
Draw the AC circuit, using a T model for the transistor.
DiMarzio 11785/exam2, Apr 2013 Page 3DiMarzio 11785/exam2, Apr 2013 Page 3DiMarzio 11785/exam2, Apr 2013 Page 3
1 BJT AMPLIFIER 1.4 Amplifier Characteristics
AC circuit in the following figure:
1.4 Amplifier Characteristics
What are the AC characteristics of this amplifier?
Voltage Gain: Look at the voltage divider. Neglect
R0 in comparison to R:.
AV =Vout
Vin=
5000 ω
5025 ω
Current Gain: Note that the dependent source gm (vin − vout)
can also be written as iacc = βiacb . The current going to
re and thus mostly to RL is the sum of i(ac)b and i
(ac)c or
β + 1 times the input current.
Output resistance:
vout:oc ≈ vin
DiMarzio 11785/exam2, Apr 2013 Page 4DiMarzio 11785/exam2, Apr 2013 Page 4DiMarzio 11785/exam2, Apr 2013 Page 4
1 BJT AMPLIFIER 1.4 Amplifier Characteristics
iout:sc ≈ vin/25 Ω
Rout =vout:oc2iout:sc
5 Ω
Voltage Gain (With Load) 0.995
Current Gain 161
Output Resistance 25 Ω
DiMarzio 11785/exam2, Apr 2013 Page 5DiMarzio 11785/exam2, Apr 2013 Page 5DiMarzio 11785/exam2, Apr 2013 Page 5
2 DISCRETE FET AMPLIFIER
2 Discrete FET Amplifier
The following figure shows an amplifier using a discrete FET and a resistor.The transistor has a drain current of 250 µAmp when VGS = 5 Volts, andthe threshold voltage is zero. The Early Voltage is VA = 70 Volts.
What type of amplifier is it? Common Drain
What is the typical situation in which this amplifier is used?
Unity Voltage gain is desired
and source cannot provide current
DiMarzio 11785/exam2, Apr 2013 Page 6DiMarzio 11785/exam2, Apr 2013 Page 6DiMarzio 11785/exam2, Apr 2013 Page 6
2 DISCRETE FET AMPLIFIER 2.1 Circuit Design
2.1 Circuit Design
We want a drain current of 250 µAmperes in this circuit. What resistancedo we need for RS?
I(DC)D = 250 µAmperes → V
(DC)GS = 5 Volts
V(DC)G = 0 → V
(DC)S = −5 Volts
RS =10 Volts
250× 10−6 Amperes= 4× 104 Ω
RS =40 kΩ
2.2 Small–Signal Parameters
What are the small–signal parameters?
gm =2I
(DC)D
V(DC)GS
=250× 10−6 Amperes
5 Volts= 10−4 A/V
R0 =VA
I(DC)D
=70 Volts
250 µA= 2.8× 10−5 Ω
gm = 100 µA/V
R0 = 280 kΩ
2.3 AC Circuit
Draw the AC circuit, using a Pi model for the transistor.
DiMarzio 11785/exam2, Apr 2013 Page 7DiMarzio 11785/exam2, Apr 2013 Page 7DiMarzio 11785/exam2, Apr 2013 Page 7
2 DISCRETE FET AMPLIFIER 2.4 Amplifier Characteristics
AC circuit is shown below
2.4 Amplifier Characteristics
What are the AC characteristics of this amplifier?
v(AC)out:oc = gm
(v(AC)in − v
(AC)out:oc
)(R0 ‖ Rs)
v(AC)out:oc [1 + gm (R0 ‖ Rs)] = gmv
(AC)in (R0 ‖ Rs)
v(AC)out:oc =
gm (R0 ‖ Rs)
1 + gm (R0 ‖ Rs)v(AC)in
Neglect R0 in comparison to RS.
v(AC)out:oc =
10−4 A/V × 4× 104 Ω
1 + 10−4 A/V × 4× 104 Ωv(AC)in = 0.80v
(AC)in
DiMarzio 11785/exam2, Apr 2013 Page 8DiMarzio 11785/exam2, Apr 2013 Page 8DiMarzio 11785/exam2, Apr 2013 Page 8
2 DISCRETE FET AMPLIFIER 2.4 Amplifier Characteristics
Input resistance is infinite because of the oxide between
the gate and the channel.
Output resistance is RS ‖ R0 ≈ RS.
Open–Circuit Gain 0.80
Input Resistance Infinite
Output Resistance 40 kΩ
DiMarzio 11785/exam2, Apr 2013 Page 9DiMarzio 11785/exam2, Apr 2013 Page 9DiMarzio 11785/exam2, Apr 2013 Page 9
3 CMOS AMPLIFIER
3 CMOS Amplifier
The figure below shows a CMOS amplifier. The transistors are all the same,and have a drain current of 250 µAmp when VGS = 5 Volts, and the thresholdvoltage is zero (This is the same as in Problem 2). However, in this case,VA = 100 Volts.
3.1 Amplifier Type
What type of amplifier is it? Common Drain
What is the typical situation in which this amplifier is used?
Unity Voltage gain is desired
and source cannot provide current.
DiMarzio 11785/exam2, Apr 2013 Page 10DiMarzio 11785/exam2, Apr 2013 Page 10DiMarzio 11785/exam2, Apr 2013 Page 10
3 CMOS AMPLIFIER 3.2 Transistor Characteristics
3.2 Transistor Characteristics
What is the constant, µnCoxW/L for this transistor?
In saturation,
iD = µnCoxW
L
(vGS − vthr)2
2
µnCoxW
L=
2iD
(vGS − vthr)2 =
2× 250× 10−6 A
(5 V2)= 2×10−5 A/V2.
20 µA/V2
3.3 Small–Signal Parameters
What are the small–signal parameters?
I(DC)D = 0.1 mA
gm =I(DC)D
V(DC)GS
=
√2I
(DC)D µnCox
W
L= 6.3× 10−5 AV
R0 =VA
I(DC)D
=100 V
10−4 A= 106 Ω
gm = 63 µA/V
R0 = 1 MΩ
DiMarzio 11785/exam2, Apr 2013 Page 11DiMarzio 11785/exam2, Apr 2013 Page 11DiMarzio 11785/exam2, Apr 2013 Page 11
3 CMOS AMPLIFIER 3.4 AC Circuit
3.4 AC Circuit
Draw the AC circuit, using a Pi model for the transistor.
3.5 Amplifier Characteristics
What are the AC characteristics of this amplifier?
vout:oc =R0
2gm (vin − vout:oc)
vout:oc
(1 +
R0
2gm
)=
R0
2gmvin
vout:oc =R02
1 + R02 gm
gmvin =32
33vin
Again, the input resistance is infinite because the input
is on the gate.
Open–Circuit Gain 0.97
Input Resistance Infinite
DiMarzio 11785/exam2, Apr 2013 Page 12DiMarzio 11785/exam2, Apr 2013 Page 12DiMarzio 11785/exam2, Apr 2013 Page 12
4 CMOS LOGIC
4 CMOS Logic
The following figure shows a CMOS digital logic circuit. As usual, +5 Voltsrepresents “true,” and zero represents “false.” There are two inputs, V A andV B, and one output, Vout at the voltage probe shown in the figure.
4.1 Truth Table
Construct the truth table for this circuit, for all possible combinations ofinputs V A and V B being “true” or “false.”
A B M1 M2 M3 M4 Out
H H O O X X L
H L X O O X L
L H O X X O L
L L X X O O H
What is the function of this circuit?
NOR
DiMarzio 11785/exam2, Apr 2013 Page 13DiMarzio 11785/exam2, Apr 2013 Page 13DiMarzio 11785/exam2, Apr 2013 Page 13
4 CMOS LOGIC 4.2 Output Voltages
4.2 Output Voltages
The following figure shows output voltage as a function of input voltage, V A.In the figure, the bottom curve (along the Vout = 0 axis) is for V B = 5, thenext curve is for V B = 2.5 Volts and the top curve is for V B = 0 Volts.Explain these curves in terms of what you know about FETs.
VB = 5 V: Output Always Low
VB = 0 V: Output changes with VA.
Similar to VB = 0 V but M4 is only partially on.
4.3 Current
Draw a plot of the current, iD as a function of V A for V B = 5 Volts. Youdon’t need to provide a correct current axis, because I have not given you
DiMarzio 11785/exam2, Apr 2013 Page 14DiMarzio 11785/exam2, Apr 2013 Page 14DiMarzio 11785/exam2, Apr 2013 Page 14
4 CMOS LOGIC 4.4 Alternative Output
the transistor parameters.
4.4 Alternative Output
With the addition of two transistors at the output, you can convert this toeither an “AND” or “OR” gate. Which one? Draw the additional circuit.
Add an inverter.
“AND” or ”OR?” OR
DiMarzio 11785/exam2, Apr 2013 Page 15DiMarzio 11785/exam2, Apr 2013 Page 15DiMarzio 11785/exam2, Apr 2013 Page 15
Recommended