EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 1 UNIT – I: IC FABRICATION AND CIRCUIT CONFIGURATION FOR LINEAR IC PART -A (2 Marks) 1. State the advantages of Integrated circuits over discrete components[AUC NOV 2013] Reduced size. Low power consumption. High speed of operation. 2. Define offset voltage of an operational amplifier[AUC NOV 2013] A small voltage applied to the input terminals to make the output voltage as zerowhen the two input terminals are grounded is called input offset voltage. 3. What are the two requirements to be met for a good current source? [AUC MAY 2012] The output current Io should not be dependent on β. The output resistance of the current source should be very high. 4. List the various methods of realizing high input resistance in a differential amplifier. [AUC MAY 2012] Use of darlington pair. Use of FET. Use of swamping resistance. 5. What are the applications of current source.[AUC MAY 2011] Transistor current sources are widely used in analog ICs both as biasingelements and as load devices for amplifier stages. 6. Define slew rate. [AUC MAY 2011] The slew rate is defined as the maximum rate of change of output voltage causedby a step input voltage.An ideal slew rate is infinite which means that op-amp’s outputvoltage should change instantaneously in response to input step voltage. 7. What is the need for frequency compensation in practical op-amps? [AUC MAY 2011] Frequency compensation is needed when large bandwidth and lower closed loopgain is desired. Compensating networks are used to control the phase shift and henceto improve the stability. 8. What is the need for constant current source in differential amplifier? [AUC MAY 2011] superior insensitivity of circuit performance to power supply variations and temperature. more economical than resistors in terms of die area required to provide bias currents of small value. When used as load element, the high incremental resistance of current source results in high voltage gain at low supply voltages. 9. Justify the statement: CMRR value of an ideal op-amp should be infinite. [ AUC MAY 2011] CMRR ρ = A d / A c For an op-amp A d is large and A c is zero . Hence CMRR is infinite.
43
Embed
UNIT I: IC FABRICATION AND CIRCUIT CONFIGURATION FOR …mahalakshmiengineeringcollege.com/pdf/ece/IVsem/EC2254/UNIT 1.pdf · EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 1
UNIT – I: IC FABRICATION AND CIRCUIT CONFIGURATION FOR LINEAR IC
PART -A (2 Marks)
1. State the advantages of Integrated circuits over discrete components[AUC NOV 2013]
Reduced size.
Low power consumption.
High speed of operation.
2. Define offset voltage of an operational amplifier[AUC NOV 2013]
A small voltage applied to the input terminals to make the output voltage as zerowhen the two input terminals are grounded is called input offset voltage.
3. What are the two requirements to be met for a good current source? [AUC MAY 2012]
The output current Io should not be dependent on β.
The output resistance of the current source should be very high.
4. List the various methods of realizing high input resistance in a differential amplifier. [AUC MAY 2012]
Use of darlington pair.
Use of FET.
Use of swamping resistance.
5. What are the applications of current source.[AUC MAY 2011]
Transistor current sources are widely used in analog ICs both as biasingelements and as load devices for amplifier stages.
6. Define slew rate. [AUC MAY 2011]
The slew rate is defined as the maximum rate of change of output voltage causedby a step input voltage.An ideal slew rate is infinite which means that op-amp’s outputvoltage should change instantaneously in response to input step voltage.
7. What is the need for frequency compensation in practical op-amps? [AUC MAY 2011]
Frequency compensation is needed when large bandwidth and lower closed loopgain is desired. Compensating networks are used to control the phase shift and henceto improve the stability.
8. What is the need for constant current source in differential amplifier? [AUC MAY 2011]
superior insensitivity of circuit performance to power supply variations and temperature.
more economical than resistors in terms of die area required to provide bias currents of small value.
When used as load element, the high incremental resistance of current source results in high voltage gain at low supply voltages.
9. Justify the statement: CMRR value of an ideal op-amp should be infinite. [ AUC MAY 2011]
CMRR ρ = Ad / Ac For an op-amp Ad is large and Ac is zero . Hence CMRR is infinite.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 2
10. Define CMRR. [AUC April 2004,MAY2010]
The relative sensitivity of an op-amp to a difference signal as compared to acommon –mode signal is called the common –mode rejection ratio. It is expressedin decibels. CMRR= Ad/Ac
11. What is the reason for drop in gain at high frequencies in op-amp?[AUC MAY2010]
The open-loop gain of op-amp decreases at higher frequencies due to thepresence of parasitic capacitance. The closed-loop gain increases at higher frequenciesand leads to instability.
12. Draw a neat sketch showing the frequency response of μA741 OP AMP.[AUC Nov 2004]
13. Write down the characteristics of ideal operational amplifier.[AUC May 2005,Nov 2006]
Open loop voltage gain Zout 0
Output impedance Zin Infinity
Input offset current Ios 0
Bandwidth BW 0
CMRR ρ Infinity
Slew rate S Infinity
PSRR PSRR 0
14. Draw the equivalent circuit of an ideal amplifier.[AUC June 2006]
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 3
15. An opamp circuit shown in figure has differential gain Ad=5mv.Calculate V0. [AUC Nov 2006]
Vo = -5v2 + 5 V1= 5(V1-V2)
16. Compare the ideal and practical characteristics of op-amp.[Nov 2006]
17. What does the term linear circuits generally convey?[Nov 2007]
In linear circuits the output electrical signals vary in proportion to the applied input signal.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 4
18. Give the circuit schematic of simple current mirror circuit.[AUC Nov 2007]
19. Define unity gain bandwidth of an Op-Amp.[AUC April 2008]
Unity gain bandwidth is defined as the bandwidth of an Op-Amp when the voltage gain is unity.
20. Define slew rate. What causes it ?[AUC May 2008,May2009]
The slew rate is defined as the maximum rate of change of output voltage caused by a
step inputvoltage.An ideal slew rate is infinite which means that op-amp’s output voltage
should changeinstantaneously in response to input step voltage.
21. State the applications of band gap reference circuit[AUC Nov 2008]
Voltage regulator
ADC and DAC circuits
V-F and F-V converters
22. What is a current mirror circuit? What are its advantages?[AUC Nov 2008]
The circuit in which the output current is forced to equal the input current is called as current mirror circuit.
Advantages:
Requires less components than constant current bias circuits.
Circuit is simple and easy to fabricate.
Using matching transistors, thermal stability is achieved.
23. Define current mirror.[AUC May 2009]
The circuit in which the output current is forced to equal the input current is called as current mirror circuit.
0
R2
1k
Q1
Q2N3721
0
Q2
Q2N3721
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 5
24. List the basic blocks of an op-amp. [AUC Nov2009]
Input stage
Intermediate stage
Level shifting stage
Output stage
25. How can you increase the input resistance of an op-amp.[AUC Nov2009]
Use of darlington pair.
Use of FET.
Use of swamping resistance.
PART-B
1. Explain the construction of a monolithic bipolar transistor [AUC NOV 2013] Step -1 Wafer preparation
P-type silicon wafer is prepared.
The wafers are usually of 10cm diameter and 0.4 mm thickness.
The resistivity of the wafer is 10 Ω -cm Step -2 Epitaxial Growth
An n type epitaxial film is grown over p- type substrate.
All active and passive components are fabricated wiyhin this layer.
The resistivity of n – type epitaxial layer is 0.1 to 0.5 Ω -cm
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 6
Step -3 Oxidation
A Sio2 layer of thickness of the order of 0.02 to 2 µm is grown on the n-epitaxial layer.
Step -4 Isolation Diffusion
The integrated circuits contains many transistors .
The collectors of all transistors are electrically connected together.
Hence , it is necessary to isolate one from another.
This is done by forming two windows in the Sio2 by means of photolithography and etching.
The P+ material is diffused through this windows until it reaches the substrate.
This process establishes an isolation island.
Step -5 Base Diffusion
The wafer is again covered by an Sio2 layer.
The base mask is used to form windows into which the p-type bases are diffused.
The base must be aligned so that it does not touch the isolator.
Step -6 Emitter Diffusion
A Sio2 layer is grown over to cover the wafer surface and an emitter mask is used to form the wndows.
The masking and etching process removes the Sio2 .
A heavily doped N-region is diffused through the windows.
The N+ region improves current gain and minimizes emitter resistance.
The base under the emitter diffuses into other region.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 7
Step -7 Contact mask
The contact mask cuts holes in the sides to allow electrical connection of the wire.
Step -8 Metallization The metallization is used to interconnect components and to provide external
access to the integrated circuits. Step -9 Passivation The entire surface is covered by a passivation layer for long term stability.
2. i)With neat diagrams ,explain the operation of voltage reference circuit using temperature compensation , voltage reference circuit using avalanche diode reference.(16) [AUC MAY 2012]
Voltage Reference circuit using temperature compensation scheme: The voltage reference circuit using basic temperature compensation scheme is shown below.
This design utilizes the close thermal coupling.
A constant current I is supplied to the avalanche diode DB and it provides a bias voltage of VB to the base of Q1.
The temperature dependence of the VBE drop across Q1 and those across D1 and D2 results in respective temperature coefficients.
Hence, with the use of resistors R1 and R2 with tapping across them at point N compensates for the temperature drifts in the base-emitter loop of Q1 .
This results in generating a voltage reference VR with normally zero temperature coefficient.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 8
Applying KCL at node N, we get
=
Assuming VBE(Q1)= VBE(D1)=VBE(D2)=VBE
Then Eq can be expressed as
(V B -2V BE - V R)/ R1 = (V R -V BE) / R2
Therefore, the voltage level VR is given by
V R =R2V B +V BE (R1 - 2R2) / R1 + R2
Differentiating VB and VBE in eq(2) partially with respect to temperature, we get
0 =R2 / (R1 + R2) V B / T +( R1 - 2R2 / R1 + R2) V BE / T V B / T R2 / (R1 + R2) = (R1 - 2R2 / R1 + R2 ) V BE / T
That is,
R1-2R2 / R2 =
Therefore, it can be inferred that eq(3) is to be satisfied for obtaining zero temperature coefficient.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 9
Voltage Reference circuit using Avalanche Diode Reference:
A voltage reference can be implemented using the breakdown phenomenon condition of a
heavily doped PN junction.
The zener breakdown is the main mechanism for junctions, which breakdown at a voltage of 5V
or less. For integrated transistors , the base-emitter breakdown voltage falls in the range of 6 to
8V.
Therefore, the breakdown in the junctions of the integrated transistor is primarily due to
avalanche multiplication.
The base bias for transistor Q1 is provided through register R1 and it also provides the dc
current needed to bias DB, D1and D2 .
The voltage at the base of Q1 is equal to the zener voltage VB added with two diode drops due
to D1 and D2 .
The voltage across R2 is equal to the voltage at the base of Q1 less the sum of the base –
mitter voltages of Q1 and Q2 .
Hence, the voltage across R2 is approximately equal to that across DB = VB . Since Q2 and Q3
act as a current mirror circuit, current I0 equals the current through R2 .
Therefore, I 0 = V B / R2
It shows that, the output current I0 has low temperature coefficient, if the temperature coefficient
of R2 is low, such as that produced by a diffused resistor in IC fabrication.
The zero temperature coefficient for output current can be achieved, if diodes are added in
series with R2 , so that they can compensate for the temperature variation of R2 and VB .
The temperature compensated avalanche diode reference source circuit is shown in figure.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 10
The transistor Q4 and Q5 form an active load current mirror circuit. The base voltage of Q1 is the voltage VB across zener DB . Then, VB = (VBE * n) +VBE across Q1 + VBE across Q2 + drop across R2 . Here, n is the number of diodes. It can be expressed as V B = Io R2 +( n+2) VBE Differentiating for VB , I0 , R2 and VBE partially, with respect to temperature T, we get
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 11
Therefore, zero temperature coefficient of I0 can be obtained, if the following condition is satisfied, That is,
3. a) Explain the working of a current source with a circuit diagram.[AUC Apr 2005] CURRENT MIRROR CIRCUIT:
DEFINITION: The circuit in which the output current is forced to equal the input current
is called as current mirror circuit.
Block diagram :
Analysis :
Circuit consists of two transistors q3 & q4.
The base emitter voltages and base currents of two transistors are same.
Vbe3=vbe4, ib3=ib4.
Similarly ic3=ic4.
CURRENT MIRROR
I source V0
I sink
Isource =Isink
0
R2
1k
Q1
Q2N3721
0
Q2
Q2N3721
i2 ib3
ic3=i2
Vbe3
Vbe4
-vee
ib4
To diff.
amp
Ic4
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 12
Applying KCL at node a:
I2=ic4+i
Applying KCL at node b:
I2=ib3+ib4
I=2ib3=2ib4
I2=ic4+2ib4
I2 =ic3+2ib3
We have ib3 =ic3 /β
I2=ic3 + 2ic3/ β
I2 =ic3(1+2/ β) = ic3+ic3(2/ β)
Β >> 2/ β
Hence i2 ~ ic3
The collector current of q3 is nearly equal to current i2 .hence current mirror circuit provides
constant current bias to the differential amplifier.
Advantages:
Requires less components than constant current bias circuits.
Circuit is simple and easy to fabricate.
Using matching transistors, thermal stability is achieved.
Uses:
Used in differential and operational amplifiers.
4. Explain the operation of a basic differential amplifier. [AUC Apr 2005,2006] (or)
Draw the circuit diagram of a symmetrical emitter coupled difference amplifier and show that a very high CMRR will result if the diff.amp is supplied by a constant current bias.[AUC Nov 2004]
Differential amplifier: The function of a differential amplifier is to amplify the difference between two signals. The need for differential amplifier arises in many physical measurements where response from dc to many MHz of frequency is required. This forms the basic input stage of an integrated amplifier.
EC2254 –LINEAR INTEGRATED CIRCUITS – II/IV SEM ECE - L.M.I.LEO JOSEPH ASST.PROF/ECE PAGE 13
The basic differential amplifier has the following important properties of 1. Excellent stability 2. High versatility and 3. High immunity to interference signals The differential amplifier as a building block of the op-amp has the advantages of 1. Lower cost. 2. easier fabrication as IC component. 3. closely matched components.
The two transistors Q! and Q2 have exactly matched characteristics.
The two collector resistances Rc1 and Rc2 are equal.
The differential amplifier can be obtained by using two emitter biased circuits .
This is done by coupling emitter of two transistors together.
Rc1 = Rc2 = Rc and RE1 =RE2 = RE .
The transistor Q1 is biased by supply voltage VS1 and Q2 by VS2. Operation The operation of differential amplifier is explained in two modes.