Electronics EECE2412 | Spring 2013 Exam #2 with Solutions...Electronics EECE2412 | Spring 2013 Exam...

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 DiMarzio 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.


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


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


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


Unity Voltage gain is desired

and source cannot provide current


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



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.


2 DISCRETE FET AMPLIFIER 2.4 Amplifier Characteristics

AC circuit is shown below



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


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


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


Unity Voltage gain is desired

and source cannot provide current.


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



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


3 CMOS AMPLIFIER 3.4 AC Circuit

3.4 AC Circuit

Draw the AC circuit, using a Pi model for the transistor.



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


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


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


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


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