EE 230 Lecture 2 Nonlinear Op Amp Applications Nonlinear analysis methods Comparators with Hysteresis
Quiz 5 Plot the transfer charactristics of the following circuit. Assume the op amp has =2 and SATL =-2. OUT IN 2
And the number is? 3 8 5 4 2? 6 9 7
Quiz 5 Plot the transfer charactristics of the following circuit. Assume the op amp has =2 and SATL =-2. OUT IN 2 Solution: + IN = 4 2 = + if > OUT + SATL if < 6 IN OUT if IN>6 = SATL if IN <6-2 SATL
eview from Last Lecture The Comparators (Assume = DD, SATL = SS ) OUT DD for IN>EF = for < SS IN EF OUT DD for IN<EF = for > SS IN EF Op Amps make good comparators when operated open-loop Some ICs are manufactured to serve only as comparators and often have better performance than op amps
eview from Last Lecture The Op Amp is Highly Nonlinear when Over-Driven The comparator circuit is also highly nonlinear Many useful applications of op amps when operating nonlinearly Many other nonlinear devices exist that are also very useful
eview from Last Lecture Nonlinear Circuits and Applications Definition: A circuit is nonlinear if one or more devices in the circuit do not operate linearly Superposition can not be used to analyze circuit Nonlinear circuit applications Will first consider applications where op amp operates nonlinearly Will then consider other nonlinear devices Will first discuss the concepts of nonlinear circuits and nonlinear circuit analysis techniques
eview from Last Lecture Methods of Analysis of Nonlinear Circuits Will consider three different analysis requirements and techniques for some particularly common classes of nonlinear circuits. Circuits with continuously differential devices Interested in obtaining transfer characteristics of these circuits or outputs for given input signals 2. Circuits with piecewise continuous devices interested in obtaining transfer characteristics of these circuits or outputs for a given input signals 3. Circuits with small-signal inputs that vary around some operating point Interested in obtaining relationship between small-signal inputs and the corresponding small-signal outputs. Will assume these circuits operate linearly in some suitably small region around the operating point Other types of nonlinearities may exist and other types of analysis may be required but we will not attempt to categorize these scenarios in this course
. Nonlinear circuits with continuously differential devices Use KL and KCL for analysis epresent nonlinear models for devices either mathematically or graphically Solve the resultant set of equations for the variables of interest Example: I Nonlinear Device I = 3e x ( - ) G = I IN I = 3e x ( IN- ) = I = +3e IN Solution relating to IN is highly nonlinear Circuit with Nonlinear Device =? Explicit expression for in terms of IN does not exist!
. Nonlinear circuits with continuously differential devices Example: =? = -3e IN Solution relating to IN is highly nonlinear Explicit expression for in terms of IN does not exist! Solution for even modestly more complicated circuits can be really messy Explicit expressions for IN or or both are often impossible to obtain Most useful nonlinear circuits will have reasonably simple final expressions for output variables of interest and a systematic procedure for analyzing these circuits
2. Circuits with piecewise continuous devices Example:. Guess region of operation 2. Solve resultant circuit using the previous method 3. erify region of operation is valid 4. epeat the previous 3 steps as often as necessary until region of operation is verified 0 < = 4 - > Y Y ( ) Y =?
2. Circuits with piecewise continuous devices Example:. Guess region of operation 2. Solve resultant circuit using the previous method 3. erify region of operation is valid 4. epeat the previous 3 steps as often as necessary until region of operation is verified 0 < = Y 4 ( - ) > egion egion 2 Y =? Guess egion Y =0 verify <? =8 erification fails (solution not valid) verify Guess egion 2 Y =4(8-)=28 >? =8 =8 > erification passes (solution valid) Y = 28
3. Circuits with small-signal inputs that vary around some operating point Determine the operating point (using method or 2 discussed above after all small signal independent inputs are set to 0) Develop small signal (linear) model for all devices in the region of interest (around the operating point or Q-point ) Create small signal equivalent circuit by replacing all devices with small-signal equivalent Solve the resultant small-signal (linear) circuit Can use KCL, KL, and other linear analysis tools such as superposition, voltage and current divider equations, Thevenin and Norton equivalence Determine boundary of region where small signal analysis is valid Example: I 0 <0 I = 2 v x>0 v x IN is a small signal, much smaller than 6
3. Circuits with small-signal inputs that vary around some operating point Example: I 0 I = 2 v < 0 x > 0 v x IN is a small signal, much smaller than 6 Will go through the mechanics of this process at this time but will develop and justify the steps later in the course! This is a very useful process that is used widely in the electronics field and in many other fields as well Students will not be expected to do this type of analysis until the process is formally developed
3. Circuits with small-signal inputs that vary around some operating point Determine the operating point (using method or 2 discussed above after all small signal independent inputs are set to 0) Develop small signal (linear) model for all devices in the region of interest (around the operating point or Q-point ) Create small signal equivalent circuit by replacing all devices with small-signal equivalent Solve the resultant small-signal (linear) circuit Can use KCL, KL, and other linear analysis tools such as superposition, voltage and current divider equations, Thevenin and Norton equivalence Determine boundary of region where small signal analysis is valid Example: 0 I = 2 < 0 v > 0 x Will linearize the circuit at the operating point (Q-point ( Q, I Q )) Linear region shown
3. Circuits with small-signal inputs that vary around some operating point Determine the operating point (using method or 2 discussed above after all small signal independent inputs are set to 0) Develop small signal (linear) model for all devices in the region of interest (around the operating point or Q-point ) Create small signal equivalent circuit by replacing all devices with small-signal equivalent Solve the resultant small-signal (linear) circuit Can use KCL, KL, and other linear analysis tools such as superposition, voltage and current divider equations, Thevenin and Norton equivalence Determine boundary of region where small signal analysis is valid Example: G EQ I = (,I ) Q Q 0 I = 2 v < 0 x > 0 Solution of Small-Signal Linear Circuit OUT = + G EQ IN Small-signal equivalent circuit Overall output if required OUT = OUTQ +OUT = Q + + G EQ IN
3. Circuits with small-signal inputs that vary around some operating point Example: 0 I = 2 v < 0 x > 0 egion egion 2 Circuit with small-signal sources zeroed out Small signal equivalent circuit 0 I = 2 v < 0 x > 0 Guess egion 2 (must verify x >0) 6 = I + 2 I = =2 + - 6=0 2 with =, obtain the solution veriify >0 ( Q,I Q )=(2,4) G G EQ I = (,I ) Q Q 2 = = 2 = 4 EQ (2,4) (,I ) Q = = = 0.2 + G + 4 Q OUT IN IN IN EQ
3. Circuits with small-signal inputs that vary around some operating point Example: egion 0 <0 I = 2 v x>0 egion 2 OUT OUT = 0.2 IN = Q + IN= 2+ 0.2 + G EQ IN If IN =0. sinωt, OUT =.02sin ωt OUT = 2+.02sinωt
Nonlinear Circuits Will now investigate several nonlinear circuits
Before trashing the bad circuit which we saw was unstable, lets see if it has any useful properties! If ideal op amps both have gain A FB =+ 2 Usually the good circuit Usually the bad circuit This circuit is unstable!
Consider this circuit 2 This circuit is an unstable noninverting amplifier and is usually not useful as an amplifier But what does the circuit really do?
Consider this circuit 2 θ = + 2 DIFF> = SATL OUT A0 IN < DIFF< A SATL DIFF< Assume in egion (must verify) = OUT alid for θ- IN> A θ > 0 IN 0 0 SATL A 0 A A 0 egion egion 2 egion 3 θ SATL
Consider this circuit 2 θ = + 2 DIFF> = SATL OUT A0 IN < DIFF< A SATL DIFF< 0 0 SATL A 0 A A 0 egion egion 2 egion 3 Assume in egion 3 (must verify) = OUT SATL alid for θout - IN< A SATL SATL 0 θ < θ SATL θ IN SATL Note two-valued for a range of IN
Consider this circuit 2 θ = + 2 DIFF> = SATL OUT A0 IN < DIFF< A SATL DIFF< 0 0 SATL A 0 A A 0 egion egion 2 egion 3 Assume in egion 2 (must verify) DIFF= θout - IN alid for A = A OUT 0 DIFF <θ - < SATL OUT IN 0 A0 (must use exact value for OUT to avoid degenerate conditions) θ < < θ SATL IN OUT= - θ-a0 OUT= θ IN IN θ SATL θ SATL Note tripple-valued for a range of IN
Comparator with Hysteresis θ = + 2 2 +
Comparator with Hysteresis θ = + 2 θ θ SATL
Comparator with Hysteresis If unstable region is entered from the left θ = + 2 θ θ SATL
Comparator with Hysteresis If unstable region is entered from the right θ = + 2 θ θ SATL
Comparator with Hysteresis θ = + 2 If unstable region is entered from the left or right θ θ θ SATL θ SATL
Comparator with Hysteresis θ = + 2 θ θ SATL
Comparator with Hysteresis θ = + 2 θ Width of Hysteresis egion θ ( ) SATL θ SATL
Comparator with Hysteresis θ θ = + 2 θ SATL Width of Hysteresis egion ( ) θ SATL Bistable if IN is in the hysteresis window Often θ is very small Widely used in control applications Serves as a memory if IN =0
Comparator as Boolean Memory DD LD LOW θ HOLD OUT θ SATL SS LD HIGH 2 θ = + 2 Not cost-competitive with other memory structures May be useful, though, in limited applications
Movement of Hysteresis Loop? CENT
Modifications of Comparator with Hysteresis OUT Hysteresis egion θ IN θ = + 2 θ SATL SATL IN OUT OUT ( -θ) +θ EF 2 EF θ = + 2 OUT ( -θ) +θ EF SATL SATL OUT IN Hysteresis egion 2 IN θ = 2 Note this is the basic inverting amplifier with op amp terminals interchanged Hysteresis egion -θ SATL -θ SATL IN Many other ways to control position and size of hysteresis window
Movement of Hysteresis Loop 2 θ= + 2 -θ= + 2 Consider adding a dc voltage Edges of hysteresis loop determined by condition where + = - For this circuit that is where = IN It follows from the 2-input voltage divider equation that =θ + -θ ( ) OUT Substituting the second equation into the first, the edges of the hysteresis loop can be obtained by solving for the two possible values of OUT, and SATL
Movement of Hysteresis Loop 2 θ= + 2 -θ= + 2 DD SATL SS Shifted Inverting Comparator with Hysteresis =θ + -θ ( ) IN OUT OUT =θ + ( -θ) HYH =θ + ( -θ) HYL SATL HYL HYH IN CENT SATL
Movement of Hysteresis Loop SATL SS DD 2 θ= + 2 -θ= + 2 =θ + ( -θ) HYH =θ + ( -θ) HYL SATL HYL W HYH = - W HYH HYL =θ - ( ) W SATL CENT + HYH HYL = CENT 2 θ( + ) SATL = + ( -θ) CENT 2
Movement of Hysteresis Loop θ= + 2 Shifted Inverting Comparator with Hysteresis HYL CENT W HYH =θ( - ) W SATL ( ) θ + 2 ( ) SATL = + -θ CENT If = DD, STAL = SS =- DD =2θ W CENT DD = -θ ( ) Shift can be to left or right depending upon sign of
Inversion of Hysteresis Loop? Inverting Comparator with Hysteresis Noninverting Comparator with Hysteresis Strategies Precede or follow inverting structure with an inverting amplifier Modify input location
Inversion of Hysteresis Loop? Inverting Comparator with Hysteresis Noninverting Comparator with Hysteresis?
Inversion of Hysteresis Loop θ = + 2 Noninverting Comparator with Hysteresis? Edges of hysteresis loop determined by condition where + = - For this circuit that is where =0 It follows from the 2-input voltage divider equation that =θ + ( -θ ) OUT IN Substituting the second equation into the first, the edges of the hysteresis loop can be obtained by solving for the two possible values of OUT, and SATL
Inversion of Hysteresis Loop θ = + 2 DD SATL SS Noninverting Comparator with Hysteresis? 0=θ + ( -θ ) OUT IN Solving, we obtain θ θ - θ = HYL θ - = HYH SATL 0=θ + ( -θ ) HYL 0=θ SATL θ= + -θ ( ) If we define θ as we obtain = -θ HYH = -θ HYL 2 SATL HYH
Inversion of Hysteresis Loop θ = 2 DD SATL SS Noninverting Comparator with Hysteresis = -θ HYH SATL = -θ HYL HYH If = DD, STAL = SS =- DD HYL = θ = -θ HYH DD HYL DD
Shifted Inverted Hysteresis Loop θ = DD 2 SATL SS Shifted Noninverting Comparator with Hysteresis ( ) = +θ θ HYH SATL CENT CENT = +θ - θ 2 ( ) SATL If = DD, STAL = SS =- DD HYL HYH ( ) = +θ -θ HYL ( ) + ( ) = +θ θ = +θ - θ HYH DD HYL DD
How about the other circuit? If ideal op amps both have gain A FB =+ 2 Usually the good circuit Usually the bad circuit This circuit is unstable! Will now analyze the usually good circuit using nonlinear analysis method
End of Lecture 2