Chapter Goal. Zulfiqar Ali

Transcription

1 Chapter Goal Understand behaior and characteristics of ideal differential and op amps. Demonstrate circuit analysis techniques for ideal op amps. Characterize inerting, non-inerting, summing and instrumentation amplifiers, oltage follower and integrator. Learns factors inoled in circuit design using op amps. Proide and introduction to actie filters Explore applications of op amps in nonlinear circuits, such as precision rectifiers. Proide examples of multiibrator circuits employing positie feedback. Demonstrate use of ac analysis capability of SPICE.

2 Differential Basic Model epresented by: A= open-circuit oltage gain id = (+--) = differential input signal oltage id = amplifier input resistance o = amplifier output resistance Signal deeloped at amplifier output is in phase with the oltage applied at + input (non-inerting) terminal and 80 0 out of phase with that applied at - input (inerting) terminal.

3 Differential Amplifier Model: With Source and Load L = load resistance S = Theenin equialent resistance of signal source s = Theenin equialent oltage of signal source id A = = s o s id + id S and * o = id + L o + id S L = A id L o + L Op amp circuits are mostly dc-coupled amplifiers. Signals o and s may hae a dc component representing a dc shift of the input away from Q-point. Op-amp amplifies both dc and ac components.

4 Differential Amplifier Model: With Source and Load (Example) Problem: Calculate oltage gain Gien Data: A=00, id =00kΩ, o = 00Ω, S =0kΩ, L =000Ω Analysis: A = o = id + + L s id S o L 00kΩ 000Ω = 00 0kΩ+ 00kΩ 00Ω+ 000Ω = 8.6= 38.3dB Ideal amplifier s output depends only on input oltage difference and not on source and load resistances.this can be achieed by using fully mismatched resistance condition ( id >> S or infinite id and o << L or zero o ). or A = open-loop gain (maximum oltage gain aailable from the deice)

5 Ideal Operational Amplifier Ideal op amp is a special case of ideal differential amplifier with infinite gain, infinite id and zero o. = o id A and lim = 0 id A If A is infinite, id is zero for any finite output oltage. Infinite input resistance id forces input currents i + and i - to be zero. Ideal op amp has following assumptions: Infinite common-mode rejection, power supply rejection, openloop bandwidth, output oltage range, output current capability and slew rate Zero output resistance, input-bias currents and offset current, input-offset oltage.

6 Inerting Amplifier: Configuration Positie input is grounded. Feedback network, resistors and connected between inerting input and signal source and amplifier output node respectiely.

7 Inerting Amplifier:Voltage Gain s is i o = 0 But is=i and -=0 (since id=+-- =0) i s s = A = o s = Negatie oltage gain implies 800 phase shift between dc/sinusoidal input and output signals. Gain greater than if > Gain less than if > Inerting input of op amp is at ground potential (not connected directly to ground) and is said to be at irtual ground.

8 Inerting Amplifier: Input and Output esistances = s = in i s = s = in i s out is found by applying a test current (or oltage) source to amplifier output and determining the oltage(or current) and turning off all independent sources. Hence, s = 0 x x = = i + i But i=i i ( + ) Since - = 0, i=0 and x = 0 irrespectie of the alue of ix. out =0

9 Inerting Amplifier: Example Problem:Design an inerting amplifier Gien Data: A =0 db, in =0kΩ, Assumptions: Ideal op amp Analysis: Input resistance is controlled by and oltage gain is set by /. A and A =-00 A minus sign is added since the amplifier is inerting. = = 0kΩ in =0 40dB/0dB = 00 A = = 00 = MΩ

10 Non-inerting Amplifier: Configuration Input signal is applied to the non-inerting input terminal. Portion of the output signal is fed back to the negatie input terminal. Analysis is done by relating oltage at to input oltage s and output oltage o.

11 Non-inerting Amplifier: Voltage Gain, Input esistance and Output esistance Since i=0 But id = 0 = and o s = + id + o = s + A o = = = + s = s = in i + s = Since i+=0 out is found by applying a test current source to amplifier output and setting s = 0 and is identical to the output resistance of inerting amplifier i.e. out =0

12 Non-inerting Amplifier: Example Problem:Determine the characteristics of gien non-inerting amplifier Gien Data: = 3kΩ, =43kΩ, s =+0. V Assumptions: Ideal op amp Analysis: 43kΩ A = + = + 3kΩ o = A s = (5.3)(0.V) =.53V Since i=0 io = o.53v = = 33.3μA + 43kΩ+ 3kΩ

13 Unity-gain Buffer A special case of non-inerting amplifier, also called oltage follower with infinite and zero. Hence A =. Proides excellent impedance-leel transformation while maintaining signal oltage leel. Ideal oltage buffer does not require any input current and can drie any desired load resistance without loss of signal oltage. Unity-gain buffer is used in may sensor and data acquisition systems.

14 Summing Amplifier Since i=0, i3=i+i 3 o = 3 Since negatie amplifier input is at irtual ground, i = i = i 3 = o 3 Scale factors for the inputs can be independently adjusted by proper choice of and. Any number of inputs can be connected to summing junction through extra resistors. This is an example of a simple digital-to-analog conerter.

15 Difference Amplifier o = - i = - i + - = ( -) = also + = + - Also called a differential subtractor, amplifies difference between input signals. in is series combination of and because i+ is zero. For =0, in=, as the circuit reduces to an inerting amplifier. For general case, i is a function of both and.

16 Difference Amplifier: Example Problem:Determine Vo, V+, V-, Io, I, I, I3. Gien Data: = 0kW, =00kW, V=5 V, V=3 V Assumptions: Ideal op amp. Hence, V-= V+ and I-= I+= 0. Analysis:Using dc alues, 00kΩ V + = V - = V = 3V =.73V + 0kΩ+ 00kΩ I Vo = I V V = - 5V-.73V = = 7μA 0kΩ = V I I = 5V (7μA)(0k Ω ) = 0.0V Io = I = 7μA

17 Integrator d o = s dτ C o (t)= C t s (τ) dτ + o (0) 0 o (0)=V c (0) Feedback resistor in the inerting amplifier is replaced by capacitor C. The circuit uses frequencydependent feedback. i s = s ic = C d o dt Voltage at the circuit s output at time t is gien by the initial capacitor oltage integral of the input signal from start of integration interal, here, t=0. Integration of an input step signal results in a ramp at the output.

18 Differentiator i = o i s = C d s dt Since i= is Input resistor in the inerting amplifier is replaced by capacitor C. Deriatie operation emphasizes high-frequency components of input signal, hence is less often used than the integrator. o = C d s dt Output is scaled ersion of deriatie of input oltage.