EE 230 Lecture 17. Nonideal Op Amp Characteristics

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1 EE 3 Lecture 17 Nonideal Op Amp Characteristics

2 Quiz 11 The dc gain of this circuit was measured to be 5 and the 3dB bandwidth was measured to be 6KHz. Determine as many of the following as possible from this information if it is known that the op amp can be modeled as a single-pole lowpass amplifier. Ao (dc gain of the Op Amp) P (pole of the Op Amp) (gain-bandwidth product of Op Amp) 1 OUT IN 3

3 And the number is?

4 And the number is?

5 Quiz 11 Solution: The dc gain of this circuit was measured to be 5 and the 3dB bandwidth was measured to be 6KHz. Determine as many of the following as possible from this information if it is known that the op amp can be modeled as a single-pole lowpass amplifier. A o (dc gain of the Op Amp) p (pole of the Op Amp) (gain-bandwidth product of Op Amp) Insufficient information to determine A o or 1 R =K BW= 1+ O R 1 BW OUT 3 IN

6 Quiz 11 Solution: 1 The dc gain of this circuit was measured to be 5 and the 3dB bandwidth was measured to be 6KHz. Determine as many of the following as possible from this information if it is known that the op amp can be modeled as a single-pole lowpass amplifier. A o (dc gain of the Op Amp) p (pole of the Op Amp) (gain-bandwidth product of Op Amp) OUT Insufficient information to determine A o or R =K BW= 1+ BW O R 1 =5 6KHz = 3MHz =(18.8M Rad / Sec) IN 3

7 Review from Last Time Instrumentation Amplifier Differential Amplifiers Can reduce effects of dc offset if gain must be very large Must pick C to that frequencies of interest are in passband

8 Review from Last Time Impedance Converters V1( G 1+G ) = VXG I= 1 ( V-V 1 X) G3 ZZ 1 3 Z IN= - Z Observe this input impedance is negative!

9 Review from Last Time Impedance Converters I 1 Z 5 IN V 1 Z 3 Z 4 One Port Z 1 Z Z = IN ZZZ ZZ Z = R C s If Z 1 =Z 3 =Z 4 = Z 5 =R and Z =1/sC IN ( ) If Z =R, Z 3 =R 3, Z 4 =R 4, Z 5 =R 5 and Z 1 =1/sC RR 3 5 Z IN= scr R 4 This is an inductor of value L=R C This is a capacitor of value RR 4 C EQ= C (can scale capacitance up or down) RR 3 5 If Z =Z 4 = Z 5 =R and Z 1 =Z 3 =1/sC IN ( ) 3 Z = R C s 3 This is a super capacitor of value This circuit is often called a Gyrator RC

10 Review from Last Time Nonideal Properties of Operational Amplifiers In even the most basic applications, the laboratory performance of the circuit often differs dramatically from what is predicted for some op amps. With proper knowledge of the characteristics of the op amp, designers can usually design circuits that behave almost like what is expected with ideal op amps Essential to know nonideal properties of the op amp and how to manage them to be an effective design engineer

11 Review from Last Time Some of the more common nonideal effects in Op Amp circuits V IN (t) V OUT (t) VM t t V OUT (t) VDD V OUT (t) t t VSS VDD VDD t t VSS VSS VDD VDD t t VSS VSS VDD t VDD t Will try to identify the source cause of all of these problems and how they can be resolved VSS VSS

Review from Last Time Inventor of two-stage Op Amp Robert Widlar (considered by many as the

semiconductor in Sept 63 at age of approx 6 where he designed several pioneering op amps.

He was turned down, and jumped ship to the fledgling National Semiconductor.

12 Review from Last Time Inventor of two-stage Op Amp Robert Widlar (considered by many as the most brilliant integrated circuit designer ever) Widlar began his IC career at Fairchild semiconductor in Sept 63 at age of approx 6 where he designed several pioneering op amps. By 1966, the commercial success of his designs became apparent, and Widlar asked for a raise. He was turned down, and jumped ship to the fledgling National Semiconductor. At National he continued to turn out amazing designs, and was able to retire just before his 3th birthday in 197.

13 Review from Last Time Inventor of the internally-compensated Op Amp Dave Fullagar (from posted www site) Joined Fairchild in Jan 1966 and asked to design an op amp His design was the first internally-compensate op amp, the 741 Fullagar was 6 years old when this was designed (introduced) in 1968 Largest selling integrated circuit ever Still in high-volume production even though over 4 years old Fullagar later started the linear design activities at Intersil Cofounder (catalyst) of Maxim

14 Review from Last Time Nonideal Op Amp Characteristics Absolute Maximum Ratings Electrical Characteristics AC DC These are in the data sheets of the op amps along with connection information, occasionally application information, connection information, and sometimes even information about the design Application notes, available from almost all manufacturers, often give more general information, definitions, more extensive application information, and other useful details.

15 Review from Last Time Widely Varying Performance Characteristics (selected comparison) Model Type Supply Min (V) Supply Max (V) Output Current (ma) Supply Current/ Channel (ma) (MHz) Power (mw) Min Price $ Max Price $ LMP34 (quad) LM741 Micropower General Purpose LM3886 Power , LMP 31 LMH 664 Low Voltage High Speed Minimum Maximum

16 Review from Last Time Nonideal Op Amp Characteristics Critical Parameters Usually Less Critical Parameters Gain-Bandwidth Product () Offset Voltage DC voltage gain, A 3dB Bandwidth, BW =A BW Input Voltage Range Output Voltage Range Output Saturation Current Slew Rate Common Mode Rejection Ratio (CMRR) Power Supply Rejection Ratio (PSRR) R IN and R OUT Bias Currents Full Power Bandwidth Compensation

17 Review from Last Time Gain, Bandwidth and 1 OUT V OUT A( s ) = V-V 1 BW = -p ( ) A -p A( s ) = s-p Since Alternatively ( ) =A -p A( s ) = s-p Almost all op amps are designed to have a first-order response down to unity gain is one of the most important parameters in many op amp applications!

18 Review from Last Time Gain, Bandwidth and Summary of Effects of on Basic Inverting and Noninverting Amplifiers V V 1 A 1 s = s+bw ( ) =A BWA V OUT A R K=1+ R 1 A( s ) = s Adequate model for most applications R K= R 1 BW = K db/ decade 1 BW = 1+K A FB K ( s)= K 1+s db A FB (s) BW A BW A FB ( s) K = ( 1+K ) 1+s

19 Example: Compare the closed-loop BW for an inverting and noninverting amplifier where the magnitude of the closed loop gain is 1. Assume the op amp is MHz. BW = K BW = 1+K BW MHz = = MHz 1 BW MHz = = 1MHz 1+1 Note the difference in BW is significant when the gain is small!

20 Measurement of Most direct: measure A o measure ω b =A o ω b A o is difficult to measure (and exact value usually not of concern) ω b is difficult to measure (and exact value seldom of concern) Direct method of determining is not practical If a circuit is adversely affected be a parameter, then this circuit is often useful for measuring that parameter provided relationship between performance and parameter is determined/known.

21 Strategy for Measuring 1. Build FB noninverting amplifier with gain K o. Measure BW 3. =(K )(BW) Keep gain (K ) quite large (maybe 1) and amplitude small enough so there is no SR distortion. With large K, frequency where gain drops 3dB will be small enough that it can be accurately measured.

22 Example: If an op amp has a of 1MHz and a dc gain of a closed loop amplifier of 1, what is the BW of the closed loop amplifier? Solution: BW 1MHz = = = 1kHz K 1 o Example: Determine the maximun dc gain of a noninverting FB amplifier if designed with an OA with =1MHz, if the closed loop BW must be greater than khz. 1MHz Solution: KBW = Ko = = = 5 BW khz

23 Example: If the input to the amplifier is.1sin(π1t), determine the actual and desired output if the op amp is the LMP31 biased with +/-.5V supplies. R R The desired output is 1+ V =1V = sin ( π 1t ) BW = = K 1 1 IN Need to determine if BW is limiting performance of circuit IN

26 Example: If the input to the amplifier is.1sin(π1t), determine the actual and desired output if the op amp is the LMP31 biased with +/-.5V supplies. VOUTDesired = sin(π 1 t) ωsig = π 1 13KHz BW = = = 1.3KHz 1 1 Thus input frequency is somewhat higher than band edge FB K K ( ) A s = 1+ s A db/ decade FB ( ) A s = s π 1 db p BW ω sig ω

27 Example: If the input to the amplifier is.1sin(π1t), determine the actual and desired output if the op amp is the LMP31 biased with +/-.5V supplies. VOUTDesired = sin(π 1 t) ωsig = π 1 FB ( ) A s = FB ( ω) A j = ( ω) s π j 1 π 13 1 AFB j = 1+j7.7 FB 1 ( ω ) = 1.9 A j = ( π ) V OUT V OUT db A p db/ decade BW ω sig j7.7 AFB j = -tan = 1-1 ( ω) 8.6 o o =.1*1.9sin(π 1t 8.6 ) o =.1sin(π 1t 8.6 ) ω

28 Addressing Bandwidth Limitations If both amplifiers have the same total gain of A V =1, compare the bandwidths of the two amplifiers if the Op Amps all have the same

29 Addressing Bandwidth Limitations R B K= 1+ R A V V V A CASCADE= = V V V OUT OUT 1 IN 1 IN the frequency-dependent noninverting amplifier was derived earlier: FB ( )= A CASCADE A K K = K K 1+s 1+s ( s) CASCADE ( jω) K = K 1+jω the dc gain is given by K ACASCADE ( j) = K 1 K = = 1+j A s K K 1+s to find the 3dB BW, must solve following ( ) ACASCADE jω K =

30 Addressing Bandwidth Limitations A A CASCADE CASCADE ( jω) ( jω) ( ) ACASCADE jω K = but the gain magnitude of the cascade is given by K K = K = K 1+jω 1+jω K K == = K K 1+ω 1+ω where RB K= 1+ Av 1 R = = K A so, BW CASCADE is the solution of K K = 1+ω 1 ω = K 1 BWCASCADE = K

31 Addressing Bandwidth Limitations 1 BW = A CASCADE V BW = A V (derived before) Comparing the bandwidths with the same A V =1, obtain 1 BW CASCADE= =.64 BW = = A major improvement in BW is possible if a large gain is required! Even more improvement if more stages cascaded if gain is large

the good circuit Usually the bad circuit Lets see what

32 Determination of proper Op Amp orientation??? If ideal op amps both have gain A FB R =1+ R 1 Usually the good circuit Usually the bad circuit Lets see what happens to the bad to determine if that will give insight into what the problem is

33 Determination of proper Op Amp orientation If ideal op amps both have gain A FB R =1+ R 1 Usually the good circuit Usually the bad circuit If configured as above, the circuit on the left will amplify the signal as desired and that on the right will latch up to either V DD or V SS Sounds like it might be a stability problem associated with a pole on the positive real axis in the RHP

34 Determination of proper Op Amp orientation If ideal op amps both have gain R 1 A FB=1+ = R β 1 Usually the good circuit Usually the bad circuit Check stability with model of Op Amp A( s ) = V1( G 1+G ) = VOUTG s-p V OUT= ( VIN -V1 ) V OUT= - ( VIN -V1 ) s-p s-p V1( G 1+G ) = VOUTG AFB ( s ) = R1 s-p+ R+R 1 - AFB ( s ) = R1 s-p- R+R 1 A FB s = s-p+β ( ) p FB= p-β A FB - s = s-p- β ( ) p FB= p+β

35 Determination of proper Op Amp orientation If ideal op amps both have gain R 1 A FB=1+ = R β 1 Usually the good circuit Usually the bad circuit Check stability with model of Op Amp p FB= p-β p FB= p+β Recall: A system is stable iff all poles of the system lie in the open LHP

36 Determination of proper Op Amp orientation If ideal op amps both have gain R 1 A FB=1+ = R β A( s ) = Usually the good circuit s-p Usually the bad circuit Check stability with model of Op Amp p FB= p-β p FB= p+β 1 Since p is negative, p FB << Pole far in LHP on negative real axis Although p negative, p<<β Pole far in RHP on positive real axis!! Thus, circuit on left is stable but one on right is unstable and thus not useful as an amplifier Would have observed same results with simpler model A( s ) = s

37 Determination of proper Op Amp orientation The good circuit A( s ) = s The bad circuit If we put in the model for A(s) we will find the circuit on the left has all poles in the LHP and the one on the right has a pole far in the RHP on the positive real axis

38 Determination of proper Op Amp orientation Put in frequency dependent model for op amp A( s ) = s in the OVERALL CIRCUIT and determine which orientation of the op amp has all poles in LHP In almost all op amp circuits of interest, there will be a unique op amp orientation that will provide a stable circuit This can be somewhat tedious if there are several op amps because they must all be oriented correctly Experience is useful at providing guidance on how to orient the op amps An unstable circuit can be embedded in a larger circuit that is stable and a stable circuit can be embedded in a larger circuit and make it unstable so can not consider only the stability of a subcircuit but rather must consider the overall circuit One of the major reasons the concept of stability was discussed in this course was to have a method of correctly orienting the op amps in op amp circuits

39 Stability Problems V IN(t) V OUT (t) VM t t V OUT (t) VDD V OUT(t) t t VSS VDD VDD t t VSS VSS VDD VDD t t VSS VSS VDD VDD t t Stability Problem VSS VSS

40 End of Lecture 17

EE 435 Lecture 15. Two-Stage Op Amp Design

EE 435 Lecture 15. Two-Stage Op Amp Design EE 435 Lecture 15 Two-Stage Op Amp Design Review from Last Time Cascaded Amplifier Issues A A 0 p s p Single-stage amplifiers -- widely used in industry, little or no concern about compensation Two amplifier