MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
4th Edition
ISBN: 9781266368622
Author: NEAMEN
Publisher: MCG
bartleby

Videos

Question
Book Icon
Chapter 14, Problem 14.40P

a.

To determine

Effective resistance R1' and R2' .

a.

Expert Solution
Check Mark

Answer to Problem 14.40P

Value of effective resistance R1'=R2'=490Ω

Explanation of Solution

Given:

Bipolar diff-amp with active load and a pair of offset-null terminal is shown below.

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 14, Problem 14.40P , additional homework tip  1

Given parameters are:

  R1=R2=500ΩRX=50kΩVT=26mV

Wipe arm of potentiometer is exactly at center.

Calculation:

As wipe arm of potentiometer is exactly at center. So, R1andR2 will have resistance RX2 connected in parallel. Therefore,

  R1'=R1||( R X 2)R2'=R2||( R X 2)

As R1=R2

So, R1'=R2'=R1||(RX2)

  R1'=R2'=500||( 50× 10 3 2)R1'=R2'=500||25000R1'=R2'=500×25000500+25000R1'=R2'=490Ω

b.

To determine

Values of x and (1x) of potentiometer in order to compensate the mismatches of transistor.

b.

Expert Solution
Check Mark

Answer to Problem 14.40P

Value of x=0.183 and (1x)=0.817

Explanation of Solution

Given:

Bipolar diff-amp with active load and a pair of offset-null terminal is shown below.

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 14, Problem 14.40P , additional homework tip  2

Given parameters are:

  R1=R2=500ΩRX=50kΩVT=26mVIQ=250μAiC1=iC2=125μAIS3=2×1014AIS4=2.2×1014A

Calculation:

Now, KVL between terminals of Q3,Q4andV

  vBE3+iC1R1'=vBE4+iC2R2'......(1)

Here,

  R1' and R2' are effective resistances in emitters of Q3andQ4 .

Now, on including the effect of potentiometer RX gives,

  R1'=R1||xRX.....(2)

  R2'=R2||(1x)RX......(3)

The base emitter voltages are,

  vBE3=VTln(i C1I S3).......(4)

And

  vBE4=VTln(i C2I S4).......(5)

Substituting values from equation (4) and (5) in equation (1)

  VTln( i C1 I S3 )+iC1R1'=VTln( i C2 I S4 )+iC2R2'VT[ln( i C1 I S3 )ln( i C2 I S4 )]=iC2R2'iC1R1'26×103[ln( 125× 10 6 2× 10 14 )ln( 125× 10 6 2.2× 10 14 )]=125×106R2'125×106R1'26×103[ln(6.25× 10 9)ln(5.682× 10 9)]=125×106(R2'R1')26×103[ln( 6.25× 10 9 5.682× 10 9 )]=125×106(R2'R1')R2'R1'=( 26× 10 3 125× 10 6 )×ln(1.1)R2'R1'=208×0.09531R2'R1'=19.8

Now, substituting value from equation (2) and (3) in above equation,

  [R2||(1x)RX][R1||xRX]=19.8[ R 2×( 1x) R X R 2+( 1x) R X][ R 1×x R X R 1+x R X]=19.8[( 1x) R X R 2( R 1 +x R X )x R X R 1( R 2 +( 1x ) R X )( R 2 +( 1x ) R X )( R 1 +x R X )]=19.8[( 1x) R X R 2 R 1+x( 1x) R X 2 R 2=x R X R 1 R 2x( 1x) R X 2 R 1( R 2 +( 1x ) R X )( R 1 +x R X )]=19.8[( 12x) R X R 1 R 2 R 1 R 2+x R X R 2+( 1x) R X R 1+x( 1x) R X 2]=19.8

Substituting the values

   [ ( 12x )( 50× 10 3 )( 500 )( 500 ) ( 500 )( 500 )+x( 50× 10 3 )( 500 )+( 1x )( 50× 10 3 )( 500 )+x( 1x ) ( 50× 10 3 ) 2 ]=19.8

   [ ( 12x )( 1.25× 10 10 ) ( 25× 10 4 )+( 25× 10 6 )+x( 25× 10 8 ) x 2 ( 25× 10 8 ) ]=19.8

   [ 1.25× 10 10 2×1.25× 10 10 x 25.25× 10 6 +x( 25× 10 8 ) x 2 ( 25× 10 8 ) ]=19.8

   1.25× 10 10 2.5× 10 10 x=19.8[ 25.25× 10 6 +x( 25× 10 8 ) x 2 ( 25× 10 8 ) ]

   1.25× 10 10 2.5× 10 10 x=( 499.95× 10 6 )+x( 495× 10 8 ) x 2 ( 495× 10 8 )

   1.25× 10 10 ( 499.95× 10 6 )=2.5× 10 10 x+( 495× 10 8 )x x 2 ( 495× 10 8 )

   1.2× 10 10 =( 7.45× 10 10 )x( 4.95× 10 10 ) x 2

   4.95 x 2 7.45x+1.2=0

So,

  x=( 7.45)± ( 7.45 ) 2 ( 4×4.95×1.2 )2×4.95x=7.45± 31.74259.9x=7.45±5.6349.9x=7.45+5.6349.9or7.455.6349.9x=13.0849.9or1.8169.9x=1.322or0.183

Since, x should be less than 1. Therefore, x=0.813

So, to compensate the mismatches of transistor the value of x=0.183 and

  1x=10.1831x=0.817

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The following figure shows a Zener diode regulator designed to hold 5.0 V at the output. Assume the Zener resistance is zero and the Zener current ranges from 2 mA minimum (2x) to 30 mA maximum (IZM). What are the minimum and maximum input source voltages for these currents? Vs = ? R₁ W 560 02 Vz +5.0 V
Consider the Zener diode circuit shown in Figure. The Zener Breakdown Voltage (Vz) is 5.6 V at Iz= 0.1mA. Zener resistance is rz= 10 0. What is Vo without load (RL= ) ? R= 0.5 ko Vo R Vps= 10 V
A full-wave rectifier produces a 32V peak rectified voltage from a 50Hz ac source. Determine the filter capacitor needed to provide a DC output voltage 24V across a 500 ohm load.

Chapter 14 Solutions

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

Ch. 14 - Find the closedloop input resistance of a voltage...Ch. 14 - An opamp with openloop parameters of AOL=2105 and...Ch. 14 - A 0.5 V input step function is applied at t=0 to a...Ch. 14 - The slew rate of the 741 opamp is 0.63V/s ....Ch. 14 - Prob. 14.8TYUCh. 14 - Prob. 14.8EPCh. 14 - Consider the active load bipolar duffamp stage in...Ch. 14 - Prob. 14.10EPCh. 14 - Prob. 14.11EPCh. 14 - Prob. 14.12EPCh. 14 - For the opamp circuit shown in Figure 14.28, the...Ch. 14 - Prob. 14.9TYUCh. 14 - List and describe five practical opamp parameters...Ch. 14 - What is atypical value of openloop, lowfrequency...Ch. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Describe the gainbandwidth product property of a...Ch. 14 - Define slew rate and define fullpower bandwidth.Ch. 14 - Prob. 9RQCh. 14 - What is one cause of an offset voltage in the...Ch. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 14.1PCh. 14 - Consider the opamp described in Problem 14.1. In...Ch. 14 - Data in the following table were taken for several...Ch. 14 - Prob. 14.4PCh. 14 - Prob. 14.5PCh. 14 - Prob. 14.6PCh. 14 - Prob. 14.7PCh. 14 - Prob. 14.8PCh. 14 - An inverting amplifier is fabricated using 0.1...Ch. 14 - For the opamp used in the inverting amplifier...Ch. 14 - Prob. 14.11PCh. 14 - Consider the two inverting amplifiers in cascade...Ch. 14 - The noninverting amplifier in Figure P14.13 has an...Ch. 14 - For the opamp in the voltage follower circuit in...Ch. 14 - The summing amplifier in Figure P14.15 has an...Ch. 14 - Prob. 14.16PCh. 14 - Prob. 14.18PCh. 14 - Prob. 14.19PCh. 14 - Prob. 14.20PCh. 14 - Prob. 14.21PCh. 14 - Prob. 14.22PCh. 14 - Three inverting amplifiers, each with R2=150k and...Ch. 14 - Prob. 14.24PCh. 14 - Prob. 14.25PCh. 14 - Prob. 14.26PCh. 14 - Prob. 14.27PCh. 14 - Prob. D14.28PCh. 14 - Prob. 14.29PCh. 14 - Prob. 14.30PCh. 14 - Prob. 14.31PCh. 14 - Prob. 14.32PCh. 14 - Prob. 14.33PCh. 14 - Prob. 14.34PCh. 14 - Prob. 14.35PCh. 14 - Prob. 14.36PCh. 14 - Prob. 14.37PCh. 14 - In the circuit in Figure P14.38, the offset...Ch. 14 - Prob. 14.39PCh. 14 - Prob. 14.40PCh. 14 - Prob. 14.41PCh. 14 - Prob. 14.42PCh. 14 - Prob. 14.43PCh. 14 - Prob. 14.44PCh. 14 - Prob. 14.46PCh. 14 - Prob. D14.47PCh. 14 - Prob. 14.48PCh. 14 - Prob. 14.50PCh. 14 - Prob. 14.51PCh. 14 - Prob. D14.52PCh. 14 - Prob. D14.53PCh. 14 - Prob. 14.55PCh. 14 - Prob. 14.56PCh. 14 - Prob. 14.57PCh. 14 - The opamp in the difference amplifier...Ch. 14 - Prob. 14.61P
Knowledge Booster
Background pattern image
Electrical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Text book image
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Text book image
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Text book image
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Text book image
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,
Routh Hurwitz Stability Criterion Basic Worked Example; Author: The Complete Guide to Everything;https://www.youtube.com/watch?v=CzzsR5FT-8U;License: Standard Youtube License