Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Quizon “Active and Passive Components of IC – 2”.

1. The number of leads in schottky barrier diode are
A. Four
B. Three
C. Two
D. Six

Answer: C
Clarification: Schottky barrier diode has two contact leads namely,
1. Schottky Barrier contact
2. Ohmic contact.

2. In Schottky barrier diode, which contact has similar characteristics to that of an ordinary PN diode?
A. Ohmic contact
B. Schottky barrier contact
C. Both ohmic and Schottky barrier contact
D. None of the mentioned

Answer: B
Clarification: A metal semiconductor is formed when aluminium is deposited directly upon n-type silicon. Its characteristics is found to be same as in an ordinary PN junction diode (Although the physical mechanism is different).

3. Find the symbol for Schottky barrier diode from the given circuit diagram?

Answer: A
Clarification: The mentioned symbol is the symbol for metal semiconductor diode or Schottky diode.

4. How the ohmic contact is formed in metal semiconductor diode? (AL-Aluminium)
A. n⁺ diffusion in p-region near AL lead
B. p⁺ diffusion in p-region near AL lead
C. n⁺ diffusion in n-region near AL lead
D. p⁺ diffusion in n-region near AL lead

Answer: C
Clarification: Aluminium is p-type impurity in silicon. So, when it is used to make a contact with n-type silicon, its essential contact is ohmic and no PN-junction is formed. Therefore, the contact is done by making n⁺ diffusion in the region near the surface where aluminium is deposited.

5. The flow of current in Schottky barrier diode is due to
A. Majority and Minority carriers
B. Majority carriers
C. Minority carriers
D. None of the mentioned

Answer: B
Clarification: When the diode is forward biased, electron flow from semiconductor to metal (where electrons are abundant). Hence, the majority carrier ‘electrons’ carry current in Schottky diode.

6. Find the application areas, where Schottky diode can be used?
A. Radio frequency
B. Power rectifier
C. Clamping diode
D. All of the mentioned

Answer: D
Clarification: Schottky diode can be used for ideal clamping or as detector in high frequency microwave ICs. Therefore, it is used for all these applications.

7. Which of the following resistor is not used as an integrated resistor?
A. Poly gate resistor
B. Pinched resistor
C. Epitaxial resistor
D. Thin film resistor

Answer: A
Clarification: Except poly gate resistor, other resistors are integrated resistor.

8. Which of the following is not true about diffused resistor?
A. Limitation due to small range of resistance
B. Resistance depends upon surface geometry
C. Resistance depends on diffusion characteristic of material
D. Diffused resistors are non-economical

Answer: D
Clarification: In diffused resistor method, the resistors are very economical as no extra fabrication steps are required.

9. Determine the formula for sheet resistance (R_s).
A. R×L×W
B. R×(L×ρ)/W
C. R×(W/L)
D. R×(W×ρ)/L

Answer: C
Clarification: The formula for sheet resistance of a material of surface dimension L and W is
R_s = R×(W/L).

10. Consider a 52cm×52cm material of uniform resistivity 100Ωm and thickness 3cm. Find the area and resistance of this sheet of material.
A. 16 m², 1.923 Ω/square
B. 8112 cm², 1.733 Ω/square
C. 156 cm², 33.33 Ω/square
D. 901 cm², 3.333 Ω/square

Answer: C
Clarification: Area = L × t= 52cm×3cm=156 cm²
=> Sheet resistance R_S = (ρ×L)/(L×t) = ρ/t = 100 Ω m/3m =33.33 Ω/square.

11. If a 25Ω diffused resistor is to be designed for an emitter resistor, determine the pattern in which it is fabricated?
A. 20mil long by 5mil wide
B. 25mil long by 1mil wide
C. 5mil long by 1mil wide
D. 16mil long by 4mil wide

Answer: B
Clarification: The sheet resistance of n-type diffused resistor is 5Ω/square.
=> L/W=R/R_S= 25Ω/5Ω = 5/1
=> 5mil long by 1mil wide.

12. The number of square contained in the integrated resistor by diffused resistor method depend on ratio of
A. ρ/t
B. ρ×L/W
C. W/L×t
D. L/W

Answer: D
Clarification: The number of square contained in the resistor depends on the surface geometry. Which is given by, the ratio L/W is called the aspect ratio of surface geometry.

13. Match the sheet resistance value for the following region in diffused resistor

1. Epitaxial Collector region	i. 200Ω/square
2. p-type base region	ii. 1 to 10kΩ/square
3. n-type emitter region	iii. 5Ω/square

A. 1-I, 2-ii, 3-iii
B. 1-ii, 2-I, 3-iii
C. 1-iii, 2-I, 3-ii
D. 1-iii, 2-ii, 3-i

Answer: B
Clarification: The mentioned values are sheet resistance values for respective diffused resistor.

Linear Integrated Circuit Multiple Choice Questions on “Input Offset Voltage – 1”.

1. What makes the output voltage equals to zero in practical op-amp?
A. Input offset voltage
B. Output offset voltage
C. Offset minimizing voltage
D. Error voltage

Answer: A
Clarification: Input offset voltage is the differential input voltage that exists between two input terminals of an op-amp without any external input and force the output voltage to zero.

2. What happens due to mismatch between two input terminals in an op-amp?
A. Input offset voltage
B. Output offset voltage
C. Bothe the input and output offset voltage
D. None of the mentioned

Answer: B
Clarification: The input offset voltage in op-amp force the output voltage to zero due to the mismatch between two input terminal, there will be voltage produced at the output and this voltage is called output offset voltage.

3. Define polarity of the output offset voltage in a practical op-amp?
A. Positive polarity
B. Negative polarity
C. Positive or negative polarity
D. None of the mentioned

Answer: C
Clarification: The output offset voltage is a DC voltage, it may be positive or negative in polarity depending on whether the potential difference between two input terminal is positive or negative.

4. The input offset voltage of 741 op-amp has an absolute maximum value of 6mv, which means
A. Minimum difference between input terminals in 741 op-amp can be large as 6mv DC
B. Minimum difference between input terminals in 741 op-amp can be large as 6mv AC
C. Maximum difference between input terminals in 741 op-amp can be large as 6mv DC
D. Maximum difference between input terminals in 741 op-amp can be large as 6mv AC

Answer: C
Clarification: Given, the absolute maximum value for a 741 is V_io= 6mv. Therefore, voltage at the non-inverting input terminal may differ from that at the inverting input terminal by as much as 6mv dc. Also the output offset voltage is a DC voltage and it cannot be AC voltage.

5. If three different 741 op-amps are taken and the corresponding output offset voltage for each of them is measured. The output voltage in these three op-amps have
A. Same amplitude and polarity
B. Different amplitude and polarity
C. Same amplitude and different polarity
D. Different amplitude and same polarity

Answer: B
Clarification: Even though the op-amps are of the same type, the output voltage in these three op-amps are not of the same amplitude and polarity, because of mass production.

6. To reduce the output offset voltage V_ooT to zero
A. Input offset voltage compensating network is added at the inverting input terminal
B. Input offset voltage compensating network is added at the non-inverting input terminal
C. Input offset voltage compensating network is added at the output terminal
D. None of the mentioned

Answer: D
Clarification: To reduce the V_ooT to zero, the external circuit is added at the input terminal of the op-amp that will give the flexibility of obtaining input offset voltage of proper amplitude and polarity. The input terminal can be inverting or non-inverting.

7. Which of the following op-amp does not need compensating network?
A. 777
B. 741
C. 748
D. All of the mentioned

Answer: D
Clarification: The compensating network is not needed for these op-amps because, they have offset null pins.

9. What will the condition of op-amp, before applying any external input
A. Compensated
B. Biased
C. Balanced
D. Zero

Answer: C
Clarification: Before applying external input to the op-amp, the output offset voltage should be reduced to zero with the help of an offset voltage compensating network. At this condition, the op-amp is said to be balanced or nulled.

Linear Integrated Circuit Multiple Choice Questions on “DC and AC Amplifiers”.

1. An inverting amplifier that amplifies dc input level is called
A. DC and AC amplifier
B. AC amplifier
C. DC amplifier
D. None of the mentioned

Answer: C
Clarification: In a DC amplifier, the output signal changes it response to changes in its dc input level. A DC amplifier can be inverting, non-inverting or differential.

2. Find the DC differential amplifier with offset null circuitry?

Answer: C
Clarification: The mentioned circuit is the DC differential amplifier, since DC voltage is applied to the differential inputs of the amplifier.

3. Why DC amplifier uses offset null circuitry?
A. To reduce the distortion in output
B. To improve the accuracy of amplifier
C. To get large output gain
D. All of the mentioned

Answer: B
Clarification: A DC amplifier uses offset null circuitry to reduce the output offset voltage to zero, that is , to improve the accuracy of DC amplifier.

4. Why are coupling capacitor used in audio receiver system?
A. All of the mentioned
B. Thermal drift
C. Component tolerance
D. Variation in signal

Answer: A
Clarification: An audio receiver system consists of a number of stages because of thermal drift, components tolerance and variation, which introduces a dc level. To prevent the amplification of such dc level, the coupling capacitors are used.

5. Find the AC inverting amplifier from the circuits given below?

Answer: A
Clarification: AC amplifier consists of external offset voltage compensating network. Therefore, the external offset voltage is connected to the non-inverting input terminal of the amplifier.

6. Determine the bandwidth of the AC inverting amplifier for high cut-off frequency of 15 Hz?

A. 4.321Hz
B. 8.356Hz
C. 7.056Hz
D. 2.334Hz

Answer: B
Clarification: Input resistance R_IF=R₁=100Ω (for ideal inverting amplifier).
=> The source resistance R_IN=R_O=5kΩ.
Therefore, Low frequency cut-off, f_L=1/2πC_i(R_IF+R_O) = 1/ [2π×4.7µF×(5kΩ+100Ω)] = 6.64Hz. and the bandwidth is calculated as, BW= f_H-f_L = 15Hz-6.64Hz = 8.356Hz.

7. When does the offset voltage compensating network must be used in inverting configuration?
A. When the input is AC voltage
B. When the input is DC voltage
C. When the input is either AC or DC voltage
D. None of the mentioned

Answer: B
Clarification: To reduce the output offset voltage to zero, the offset minimizing resistor is used to minimize the effect of input bias currents on the output offset voltage. However, when the inputs are DC voltages, the offset compensating network must be used.

8. State the application in which summing, scaling or averaging amplifiers are used?
A. Multiplexers
B. Counters
C. Audio mixers
D. All of the mentioned

Answer: C
Clarification: Summing, scaling or averaging amplifiers are commonly used in audio mixers, in which a number of inputs are added up to produce a desired output.

Linear Integrated Circuit Multiple Choice Questions on “Integrator – 1″.

1. The circuit in which the output voltage waveform is the integral of the input voltage waveform is called
A. Integrator
B. Differentiator
C. Phase shift oscillator
D. Square wave generator
Answer: A
Clarification: Integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

2. Find the output voltage of the integrator
A. V_o = (1/R×C_F)×^t∫₀ V_indt+C
B. V_o = (R/C_F)×^t∫₀ V_indt+C
C. V_o = (C_F/R)×^t∫₀ V_indt+C
D. V_o = (R×C_F)×^t∫₀ V_indt+C
Answer: A
Clarification: The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RC_F.
V_o = (1/R×C_F)×^t∫₀ V_indt+C
Where C-> Integration constant and C_F-> Feedback capacitor.

3. Why an integrator cannot be made using low pass RC circuit?
A. It require large value of R and small value of C
B. It require large value of C and small value of R
C. It require large value of R and C
D. It require small value of R and C
Answer: C
Clarification: A simple low pass RC circuit can work as an integrator when time constant is very large, which require large value of R and C. Due to practical limitations , the R and C cannot be made infinitely large.

4. How a perfect integration is achieved in op-amp?
A. Infinite gain
B. Low input impedance
C. Low output impedance
D. High CMRR
Answer: A
Clarification: In an op-amp integrator the effective input capacitance becomes C_F×(1-A_v). Where A_v is the gain of op-amp. The gain is infinite for ideal op-amp. So, effective time constant of the op-amp integrator becomes very large which results in perfect integration.

5. The op-amp operating in open loop result in output of the amplifier to saturate at a voltage
A. Close to op-amp positive power supply
B. Close to op-amp negative power supply
C. Close to op-amp positive or negative power supply
D. None of the mentioned
Answer: C
Clarification: In practice, the output of op-amp never becomes infinite rather the output of the op-amp saturate at a voltage close to op-amp positive or negative power supply depending on the polarity of the input dc signal.

6. The frequency at which gain is 0db for integrator is
A. f=1/(2πR_FC_F)
B. f=1/(2πR₁C_F)
C. f=1/(2πR₁R₁)
D. f=(1/2π)×(R_F/R₁)
Answer: B
Clarification: The frequency at which the gain of the integrator becomes zero is f=1/(2πR₁C_F).

7. Why practical integrator is called as lossy integrator?
A. Dissipation power
B. Provide stabilization
C. Changes input
D. None of the mentioned
Answer: D
Clarification: To avoid saturation problems, the feedback capacitor is shunted by a feedback resistance(R_F). The parallel combination of R_F and C_F behave like a practical capacitor which dissipates power. For this reason, practical integrator is called as a lossy integrator.

8. Determine the lower frequency limit of integration for the circuit given below.

A. 43.43kHz
B. 4.82kHz
C. 429.9kHz
D. 4.6MHz
Answer: B
Clarification: The lower frequency limit of integration, f= 1/(2πR_FC_F) = 1/(2π×1kΩ×33nF) = 4.82kHz.

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Linear Integrated Circuit Multiple Choice Questions on “Comparator”.

1. Depending on the value of input and reference voltage a comparator can be named as
A. Voltage follower
B. Digital to analog converter
C. Schmitt trigger
D. Voltage level detector
Answer: D
Clarification: A comparator is some time called as voltage level detector because, for a desired value of reference voltage, the voltage level of the input can be detected.

2. Why clamp diodes are used in comparator?
A. To reduce output offset voltage
B. To increase gain of op-amp
C. To reduce input offset current
D. To protect op-amp from damage
Answer: D
Clarification: The diodes protect the op-amp from damage due to excessive input voltage. Because of these diodes the difference input voltage of the op-amp is clamped to 0.7v or -0.7 v, hence these diodes are clamp diodes.

3. Find the non-inverting comparator

Answer: A
Clarification: In a non-inverting comparator a fixed reference voltage V_ref of 1v is applied to positive inverting input terminal and the other time vary in signal voltage is applied to non-inverting input terminal of the op-amp.

4. How the op-amp comparator should be choosen to get higher speed of operation?
A. Large gain
B. High slew rate
C. Wider bandwidth
D. None of the mentioned
Answer: C
Clarification: The bandwidth of the op-amp comparator must be wider so that the output of comparator can switch rapidly between saturation levels. Also, the op-amp responds instantly to any change in condition at the input.

5. How to obtain high rate of accuracy in comparator?
A. Input offset
B. High voltage gain
C. High CMRR
D. All of the mentioned
Answer: D
Clarification: High voltage gain causes comparator output voltage to switch between saturation levels. High CMRR rejects noise at input terminal and input offset (voltage & current) help to keep changes in temperature variation very slight.

6. How to keep the output voltage swing of the op-amp comparator within specific limits?
A. External resistors or diodes are used
B. External zeners or diodes are used
C. External capacitors or diodes are used
D. External inductors or diodes are used
Answer: B
Clarification: To keep the output voltage swing within specific limit, op-amps are used with external wired components such as zeners or diodes. In the resulting circuit, the outputs are limited to predetermined values.

7. Zero crossing detectors is also called as
A. Square to sine wave generator
B. Sine to square wave generator
C. Sine to triangular wave generator
D. All of the mentioned
Answer: B
Clarification: In zero crossing detectors, the output waveform is always a square wave for the applied sinusoidal input signal.

8. What is the drawback in zero crossing detectors?
A. Low frequency signal and noise at output terminal
B. High frequency signal and noise at input terminal
C. Low frequency signal and noise at input terminal
D. High frequency signal and noise at output terminal
Answer: C
Clarification: Due to low frequency signal, the output voltage may not switch quickly from one saturation voltage to other. The presence of noise can fluctuate the output between two saturation voltages.

9. State a method to overcome the drawback of zero crossing detectors?
A. Increasing input voltage
B. Use of positive feedback
C. Connect a compensating network
D. None of the mentioned
Answer: B
Clarification: The drawback of zero crossing detectors can be in cured with the use of regenerative or positive feedback that causes the output to change faster and eliminate any false output transition due to noise signals at the input.

10. Name the comparator that helps to find unknown input.
A. Time marker generator
B. Zero crossing detectors
C. Phase meter
D. Window detector
Answer: D
Clarification: Sometimes it is necessary to find the instant at which an unknown input is between two threshold levels. This can be achieved by a circuit called window detector.

11. Find the instance at which the input can be fed to the op-amp in a three level comparator with LED indicator.

A. When Green LED glow
B. When Yellow LED glow
C. When Red LED glow
D. All of the mentioned
Answer: A
Clarification: The input can be fed to the op-amp when the green LED glows, which is considered to be safe input that is when the input voltage is between 3v and 6v.

12. Find the output voltage at the point V₂ from the given circuit.


Answer: B
Clarification: The output of the zero crossing detector is differentiated by an RC circuit (RC>>1). So, the voltage at V₂ is a series of positive and negative pulses.

13. Mention the application areas of time marker generator can be used
A. Monoshots
B. SCR
C. Sweep voltage of CRT
D. All of the mentioned
Answer: D
Clarification: A diode connected at the output of time marker generator circuit converts the sinusoidal signal into a train of positive pulses. So, these pulses are used in triggering the monoshot, SCR, sweep voltage of CRT, etc.

14. Which among the following is used to increase phase angle between different voltages?
A. Phase detector
B. Window detector
C. Zero crossing detector
D. None of the mentioned
Answer: A
Clarification: Phase angle between different voltages can be measured using phase detector circuit. The corresponding voltage to be measured is converted into spikes and the time interval between the pulse spikes is measured, which is proportional to the phase difference.

15. For the comparator shown below, determine the transfer curves if an ideal op-amp with V_Z1= V_Z2=9v.


Answer: A
Clarification: The open loop voltage gain of an ideal op-amp A_Ol=∞, even a small positive or negative voltage at the input drives the output to ±V_sat. So, the output voltage V_O = ±( V₂ +V_sat)
Therefore, V_O = ±(V_Z+V_Sat) =± (9+0.7) = ±9.7 v.

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Linear Integrated Circuit Multiple Choice Questions on “IC Voltage Regulator”.

1. Which is not considered as a linear voltage regulator?
A. Fixed output voltage regulator
B. Adjustable output voltage regulator
C. Switching regulator
D. Special regulator
Answer: C
Clarification: In linear regulator’s the impedance of active element may be continuously varied to supply a desired current to the load. But in the switching regulator, a switch is turned on and off.

2. What is the dropout voltage in a three terminal IC regulator?
A. |V_in| ≥ |V_o|+2v
B. |V_in| o|-2v
C. |V _in| = |V_o|
D. |V_in| ≤ |V_o|
Answer: A
Clarification: The unregulated input voltage must be atleast 2V more than the regulated output voltage. For example, if V_o=5V, then V_in=7V.

3. To get a maximum output current, IC regulation are provided with
A. Radiation source
B. Heat sink
C. Peak detector
D. None of the mentioned
Answer: B
Clarification: The load current may vary from 0 to rated maximum output current. To maintain this condition, the IC regulator is usually provided with a heat sink; otherwise it may not provide the rated maximum output current.

4. For the given circuit, let V_EB(ON)=1v, ß= 15 and I_O=2mA. Calculate the load current

A. I_L = 23.45A
B. I_L = 46.32A
C. I_L = 56.87A
D. I_L = 30.75A
Answer: D
Clarification: The equation for load current, I_L = [(ß+1)I_O]-[ß×(V_EB(ON)/R₁)]=[(15+1)×2]–[15×(1v/12 Ω)] =32-1.25 =30.75A.

5. Which type of regulator is considered more efficient?
A. All of the mentioned
B. Special regulator
C. Fixed output regulator
D. Switching regulator
Answer: D
Clarification: The switching element dissipates negligible power in either on or off state. Therefore, the switching regulator is more efficient than the linear regulators.

6. State the reason for thermal shutdown of IC regulator?
A. Spikes in temperature
B. Decrease in temperature
C. Fluctuation in temperature
D. Increase in temperature
Answer: D
Clarification: The IC regulator has a temperature sensor (built-in) which turn off the IC, when it becomes too hot (usually 125^oC-150^oC.. The output current will drop and remains there until the IC has cooled significantly.

7. Find the difference between output current having a load of 100Ω and 120Ω for 7805 IC regulator. Consider the following specification: Voltage across the load = 5v; Voltage across the internal resistor= 350mv.
A. 8.4mA
B. 7mA
C. 9mA
D. 3.4mA
Answer: A
Clarification: Given the voltage across the internal resistor to be 350mv, which is less than 0.7v. Hence the transistor in 7805 is off.
When load = 100Ω, I_L= I_O= I_i= 5v/100 Ω = 50mA
When load=120Ω, I_O= 5v/120 Ω = 41.6mA.
So, the difference between the output voltage = 50-41.6mA = 8.4mA.

8. The change in output voltage for the corresponding change in load current in a 7805 IC regulator is defined as
A. All of the mentioned
B. Line regulation
C. Load regulation
D. Input regulation
Answer: C
Clarification: Load regulation is defined as the change in output voltage for a change in load current and is also expressed in millivolts or as a percentage of output voltage.

9. An IC 7840 regulator has an output current =180mA and internal resistor =10Ω. Find the collector current in the output using the transistor specification: ß=15 and V_EB(ON) =1.5v.
A. 270mA
B. 450mA
C. 100mA
D. 50mA
Answer: B
Clarification: The collector current from transistor, I_C= ßI_B
Where, I_B= I_O-(V_EB(ON)/R₁) = 180mA-(1.5v-10Ω) = 0.03A.
Therefore, I_C= 15×0.03 = 0.45A = 450mA.

10. How the average temperature coefficient of output voltage expressed in fixed voltage regulator?
A. miilivolts/^oC
B. miilivolts^oC
C. None of the mentioned
D. ^oC/ miilivolts
Answer: A
Clarification: The temperature stability or average temperature coefficient of output voltage, is the change in the output voltage per unit change in temperature and expressed in miilivolts/^oC.

11. In the circuit given below, let V_EB(ON)=0.8v and ß=16. Calculate the output current coming from 7805 IC and collector current coming from transistor Q₁ for a load of 5Ω.

A. I_O =111mA, I_C= 808mA
B. I_O =111mA, I_C= 829mA
C. I_O =111mA, I_C= 881mA
D. I_O =111mA, I_C= 889mA
Answer: D
Clarification: When load = 5Ω, I_L= 5v/5Ω =1A. The voltage across R₁ is 7Ω × 1A=7v. Since, I_L is more than 100mA, the transistor Q₁ turns on and supplies the extra current required.
Therefore, I_L =(ß+1)I_O-[ß×(V_EB(ON)/R₁)
I_O = [I_L/(ß+1)]+ [ß×(V_EB(ON)/R₁) = [1/(16+1)]+[16×(0.8/2Ω)] ≅111mA.
=> I_C=I_L-I_O=1A-111mA =889mA.

12. Calculate the output voltage for LM314 regulator. The current I_ADJ is very small in the order of 100µA. (Assume V_REF=1.25v)

A. 17.17v
B. 34.25v
C. 89.34v
D. 23.12v
Answer: A
Clarification: The output voltage, V_O =V_REF[1=(R₂/R₁)]+(I_ADJ×R₂)=1.25V_in× [1+(3kΩ/240Ω)] +( 100µA×3kΩ )= 16.875 +0.3.
=> V_O=17.17v.

13. Compute the input voltage of 7805c voltage regulator with a current source that will deliver a 0.725A current to 65Ω, 10w load. (Assume reference voltage =5v)
A. V_in = 84v
B. V_in = 34v
C. V_in = 54v
D. V_in = 64v
Answer: C
Clarification: V_O=V_REF+V_L =V_REF+(I_L×R_L) = 5v+(0.725A×65Ω) = 52.125v
=> Input voltage, V_in = V_O + dropout voltage = 52.125v+2v.
=> V_in ≅54v.

14. Which of the following is not a characteristic of adjustable voltage regulators?
A. Non-versatile
B. Better performance
C. Increased reliability
D. None of the mentioned
Answer: A
Clarification: Adjustable voltage regulators are versatile; it has improved over-load protection allowing greater output current over operating temperature range.

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