# Part 11 Operational Enhancers and Applications.

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Part Goals. Comprehend the
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﻿Part 11 Operational Amplifiers and Applications

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Chapter Goals Understand the "enchantment" of negative input and the attributes of perfect operation amps. Comprehend the conditions for non-perfect operation amp conduct so they can be maintained a strategic distance from in circuit plan. Exhibit circuit investigation methods for perfect operation amps. Describe transforming, non-upsetting, summing and instrumentation enhancers, voltage adherent and first request channels. Take in the components required in circuit plan utilizing operation amps. Discover the increase attributes of fell speakers . Exceptional Applications: The transformed step DAC and progressive guess ADC

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Differential Amplifier Model: Basic Represented by: A = open-circuit voltage pick up v id = ( v + - v - ) = differential info signal voltage R id = enhancer input resistance R o = speaker yield resistance The sign created at the intensifier yield is in stage with the voltage connected at the + input (non-rearranging) terminal and 180° out of stage with that connected at the - input (upsetting) terminal.

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LM741 Operational Amplifier: Circuit Architecture Current Mirrors

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Ideal Operational Amplifier The "perfect" operation amp is an exceptional instance of the perfect differential speaker with vast addition, unbounded R id and zero R o . what\'s more, If An is unbounded, v id is zero for any limited yield voltage. Unbounded info resistance R id strengths input streams i + and i - to be zero. The perfect operation amp works with the accompanying presumptions: It has interminable normal mode dismissal, power supply dismissal, open-circle data transmission, yield voltage range, yield current capacity and slew rate It additionally has zero yield resistance, input-predisposition streams, input-balance current, and info counterbalance voltage.

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The Inverting Amplifier: Configuration The positive information is grounded. A "criticism system" made out of resistors R 1 and R 2 is associated between the upsetting info, signal source and intensifier yield hub, separately.

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Inverting Amplifier:Voltage Gain The negative voltage pick up suggests that there is a 180 0 stage shift between both dc and sinusoidal information and yield signals. The addition greatness can be more prominent than 1 if R 2 > R 1 The increase size can be under 1 if R 1 > R 2 The modifying contribution of the operation amp is at ground potential (in spite of the fact that it is not associated straightforwardly to ground) and is said to be at virtual ground . Be that as it may, i s = i 2 and v - = 0 (since v id = v + - v - = 0) and

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Inverting Amplifier: Input and Output Resistances R out is found by applying a test current (or voltage) source to the intensifier yield and deciding the voltage (or current) subsequent to killing every free source. Henceforth, v s = 0 But i 1 =i 2 Since v - = 0, i 1 =0. In this way v x = 0 regardless of the estimation of i x .

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Inverting Amplifier: Example Problem : Design a transforming intensifier Given Data : A v = 20 dB, R in = 20k W , Assumptions: Ideal operation amp Analysis : Input resistance is controlled by R 1 and voltage increase is set by R 2/R 1 . what\'s more, A v = - 100 A short sign is included since the speaker is modifying.

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The Non-upsetting Amplifier: Configuration The info sign is connected to the non-transforming input terminal. A segment of the yield sign is sustained back to the negative information terminal. Investigation is finished by relating the voltage at v 1 to info voltage v s and yield voltage v o .

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Non-altering Amplifier: Voltage Gain, Input Resistance and Output Resistance Since i - = 0 and But v id = 0 Since i + = 0 R out is found by applying a test current source to the enhancer yield in the wake of setting v s = 0. It is indistinguishable to the yield resistance of the reversing intensifier i.e. R out = 0.

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Non-upsetting Amplifier: Example Problem : Determine the yield voltage and current for the given non-altering intensifier. Given Data : R 1 = 3k W , R 2 = 43k W , v s = +0.1 V Assumptions: Ideal operation amp Analysis : Since i - = 0,

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Finite Open-circle Gain and Gain Error A b is called circle pick up. For A b >>1, This is the "perfect" voltage increase of the speaker . In the event that A b is not >>1, there will be "Increase Error". is known as the criticism element.

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Gain Error Gain Error is given by GE = (perfect increase) - (genuine addition) For the non-reversing enhancer, Gain mistake is likewise communicated as a partial or rate blunder.

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Gain Error: Example Problem : Find perfect and genuine pick up and pick up mistake in percent Given information : Closed-circle addition of 100,000, open-circle increase of 1,000,000. Approach : The intensifier is intended to give perfect addition and deviations from the perfect case must be resolved. Henceforth, . Note: R 1 and R 2 aren\'t intended to make up for the limited open-circle addition of the intensifier. Investigation :

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Output Voltage and Current Limits Practical operation amps have constrained yield voltage and current reaches. Voltage: Usually constrained to a couple of volts not as much as force supply range. Current: Limited by extra circuits (to point of confinement force dissemination or ensure against inadvertent shortcircuits). As far as possible is habitually determined as far as the base burden resistance that the intensifier can drive with a given yield voltage swing. Eg: For the reversing intensifier,

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Example PSpice Simulations of Non-modifying Amplifier Circuits

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The Unity-pick up Amplifier or "Support" This is an extraordinary instance of the non-rearranging enhancer, which is likewise called a voltage supporter, with vast R 1 and zero R 2 . Thus A v = 1. It gives a fantastic impedance-level change while keeping up the sign voltage level. The "perfect" cradle does not require any information current and can drive any craved burden resistance without loss of sign voltage. Such a support is utilized as a part of numerous sensor and information securing framework applications.

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The Summing Amplifier Since i - = 0, i 3 = i 1 + i 2 , Scale components for the 2 sources of info can be autonomously balanced by the best possible decision of R 2 and R 1 . Any number of sources of info can be associated with a summing intersection through additional resistors. This circuit can be utilized as a basic advanced to-simple converter. This will be shown in more detail, later. Since the negative speaker information is at virtual ground,

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The Difference Amplifier Since v - = v + This circuit is additionally called a differential enhancer, since it opens up the contrast between the information signals. R in2 is arrangement blend of R 1 and R 2 since i + is zero. For v 2 = 0, R in1 = R 1 , as the circuit diminishes to a reversing speaker . For general case, i 1 is a component of both v 1 and v 2 . For R 2 = R 1 Also,

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Difference Amplifier: Example Problem : Determine v o Given Data : R 1 = 10k W , R 2 =100k W , v 1 =5 V, v 2 =3 V Assumptions: Ideal operation amp. Thus, v - = v + and i - = i + = 0. Investigation: Using dc values, Here A dm is known as the "differential mode voltage increase" of the distinction speaker.

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Finite Common-Mode Rejection Ratio (CMRR) An (or A dm ) = differential-mode pick up A cm = normal mode pick up v id = differential-mode input voltage v ic = basic mode input voltage A genuine speaker reacts to flag basic to both data sources, called the basic mode input voltage ( v ic ) . When all is said in done, A perfect speaker has A cm = 0, however for a genuine intensifier,

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Finite Common-Mode Rejection Ratio: Example Problem : Find yield voltage mistake presented by limited CMRR. Given Data : A dm = 2500, CMRR = 80 dB, v 1 = 5.001 V, v 2 = 4.999 V Assumptions: Op amp is perfect, with the exception of CMRR. Here, a CMRR in dB of 80 dB compares to a CMRR of 10 4 . Investigation: The yield mistake presented by limited CMRR is 25% of the normal perfect yield.

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uA741 CMRR Test: Differential Gain

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Differential Gain A dm = 5 V/5 mV = 1000

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uA741 CMRR Test: Common Mode Gain

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Common Mode Gain A cm = 160 mV/5 V = .032

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CMRR Calculation for uA741

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Instrumentation Amplifier NOTE Combines 2 non-transforming enhancers with the distinction intensifier to give higher addition and higher information resistance. Perfect information resistance is boundless in light of the fact that info current to both operation amps is zero. The CMRR is resolved just by Op Amp 3.

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Instrumentation Amplifier: Example Problem : Determine V o Given Data : R 1 = 15 k W , R 2 = 150 k W , R 3 = 15 k W, R 4 = 30 k W V 1 = 2.5 V, V 2 = 2.25 V Assumptions: Ideal operation amp. Subsequently, v - = v + and i - = i + = 0. Investigation: Using dc values,

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The Active Low-pass Filter Use a phasor way to deal with increase examination of this modifying speaker. Let s = j w . f c is known as the high recurrence "cutoff" of the low-pass channel.

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Active Low-pass Filter (proceeded) At frequencies beneath f c ( f H in the figure), the speaker is an upsetting intensifier with increase set by the proportion of resistors R 2 and R 1 . At frequencies above f c , the enhancer reaction "moves off" at - 20dB/decade. Notice that cutoff recurrence and increase can be freely set. greatness stage

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Active Low-pass Filter: Example Problem : Design a dynamic low-pass channel Given Data : A v = 40 dB, R in = 5 k W , f H = 2 kHz Assumptions: Ideal operation amp, indicated pick up speaks to the fancied low-recurrence pick up. Investigation : Input resistance is controlled by R 1 and voltage increase is set by R 2/R 1 . The cutoff recurrence is then set by C. The nearest standard capacitor estimation of 160 pF brings cutoff recurrence down to 1.99 kHz. also, .:tslidesep

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