Input and Output Impedances with Feedback

Transcription

1 EE 3 Lecture Basic Feedback Configurations Generalized Feedback Schemes Integrators Differentiators First-order active filters Second-order active filters

2 Review from Last Time Input and Output Impedances with Feedback R INF =? R β = R+R R OF =? Exact analysis : Consider amplifier as a two-port and use open/short analysis method V R IN R O A V V Will find R INF, R OF, A V almost identical to previous calculations Will see a small A VR present but it plays almost no role since R INF is so large (effectively unilateral) Two-Port Nonideal Op Amp R R A VF β R F R R +βa ( ) INF=RIN + AVβ A VRF V β

3 Review from Last Time Buffer Amplifier A V VOUT = = V IN R IN = One of the most widely used Op Amp circuits R OUT = Provides a signal to a load that is not affected by a source impedance This provides for decoupling between stages in many circuits Special case of basic noninverting amplifier with R = and R = V R = + V R OUT IN

4 Review from Last Time Basic Inverting Amplifier V IN + OUT = R V R IN OUT V OUT V R = - R IN R OUT = R IN =R Input impedance of R is unacceptable in many (but not all) applications This is not a voltage feedback amplifier (it is a feedback amplifier) of the type (note R IN is not high!) Feedback concepts could be used to analyze this circuit but lots of detail required

5 Review from Last Time Summing Amplifier F OUT k k VOUT V V Vk = R R R R F k R R R V = - V - V V F F F OUT k R R Rk Output is a weighted sum of the input voltages Any number of inputs can be used Gains from all inputs can be adjusted together with R F Gain for input V i can be adjusted independently with R i for I k All weights are negative Input impedance on each input is R i

6 Generalized Inverting Amplifier s-domain representation T s = Z Z F ( s) ( s) Z and Z F can be any s-domain circuits If Z =R, Z F =/sc, obtain T( s) = s What is this circuit?

7 Generalized Inverting Amplifier What is this circuit? T s = s Consider the differential equation ( τ ) t y = K x dτ Taking the Laplace Transform, obtain Xs Y( s ) = K s Y s T s = = Xs K is the frequency where T(jω) = and is termed the Integrator Unity Gain Frequency K s Thus, this circuit is an inverting integrator with a unity gain frequency of K = () -

8 Inverting Integrator T s = s T jω = jω T jω = ω 9 T jω = o Unity gain frequency is ω =

9 Inverting Integrator Integrators are widely used! R IN =R R OUT = t V = V + V ( τ) dτ OUT IN IN T s = s The integrator function itself is ill-conditioned and integrators are seldom used open-loop The ideal integrator has a pole at s= which is not in the LHP If the input has any dc component present, since superposition applies, the output would diverge to ± as time increases The offset voltage (discussed later) will also cause an integrator output to diverge

10 Inverting Integrator t V = V + V ( τ) dτ OUT IN IN What is the output of an ideal integrator if the input is an ideal square wave? Amplitude of output dependent upon product

11 Inverting Integrator t V = V + V ( τ) dτ OUT IN IN What is the output of an ideal integrator if the input is an ideal sine wave? Amplitude of output dependent upon product

12 Inverting Integrator Ill-conditioned nature of open-loop integrator DC OUT t V = V + V ( τ) dτ OUT IN IN If V IN () =, V OUT = V t t DC DC V VOUT = dτ V V t ( τ ) dτ DC OUT = = VDC t For any values of V DC, R, and C, the output will diverge to ±

13 Inverting Integrator Ill-conditioned nature of open-loop integrator t V = V + V ( τ) dτ OUT IN IN Any periodic input that in which the average value is not EXACTLY will have a dc component () V t A sin kωt+θ = + IN k k = This dc input will cause the output to diverge!

14 Noninverting Integrator t V = V + V ( τ) dτ OUT IN IN Obtained from inverting integrator by preceding or following with inverter Requires more components Also widely used Same issues affect noninverting integrator

15 Lossy Integrator F OUT IN Add a large resistor to slowly drain charge off of C and prevent divergence Allows integrator to be used open-loop Changes the dc gain from - to -R F /R But the lossy integrator is no longer a perfect integrator

16 What if R F is not so large? F OUT IN T s Z = Z F ( s) ( s)

17 What if R F is not so large? T s = Z Z F ( s) ( s) T s R sc R+ sc R = = R R +scr

18 First-order lowpass filter with a dc gain of R /R R R T( jω) R = R +scr T s R controls the pole (and also the dc gain) R controls the dc gain (and not the pole)

19 Summing Integrator C R T s = s V V R V OUT V OUT = V s IN V k R k V = V V V By superposition... OUT k sr C sr C sr kc All inverting functions Can have any number of inputs Weights independently controlled by resistor values Weights all changed by C V OUT k = Vi sr C i= i

20 I ve got a better noninverting integrator! R C V OUT V V R = - R OUT IN V IN VOUT = - V s IN R R OUT V IN V V R = + R OUT IN V OUT IN VOUT = + V s IN?

21 I ve got a better noninverting integrator! IN VOUT = + V s IN? OUT But is this a useful circuit? IN V sc+g =V sc OUT V V OUT IN +s = s It has a noninverting transfer function But it is not an noninverting integrator!

22 First-Order Highpass Filter But is this a useful circuit? OUT IN V V OUT IN +s = s This is a first-order high-pass amplifier (or filter) but the gain at dc goes to so applications probably limited.

23 Two-capacitor noninverting integrator Requires matched resistors and matched capacitors V sc+g = V Actually uses a concept called pole-zero cancellation OUT V( sc+g ) = VING VOUT = + V s Generally less practical than the cascade with an inverter IN sc

24 Generalized Inverting Amplifier s-domain representation If Z =/sc, Z F =R, obtain T( s) What is this circuit? = s T s = Z Z F ( s) ( s)

25 Generalized Inverting Amplifier What is this circuit? Consider the differential equation dx( t) y = K dt Taking the Laplace Transform, obtain Y( s ) = Ks X( s) K - is the frequency where T(jω) = T( s) = s Y s T s = = Ks Xs Thus, this circuit is an inverting differentiator with a unity gain frequency of K - = () -

26 Inverting Differentiator T( s) = s Differentiator gain ideally goes to at high frequencies Differentiator not widely used Differentiator relentlessly amplifies noise Stability problems with implementation (not discussed here) Placing a resistor in series with C will result in a lossy differentiator that has some applications

27 First-order High-pass Filter R sr T( s ) = - C = - R+ +R Cs sc R R T( jω) T jω = ωr C + ω

28 Applications of integrators to solving differential equations IN OUT Standard Integral form of a differential equation X = b X + b X + b X a X + X + X +... OUT OUT OUT 3 OUT IN IN IN Standard differential form of a differential equation X = α X + α X + α X β X + β X + β X +... ' '' ''' ' '' OUT OUT OUT 3 OUT IN IN 3 IN Initial conditions not shown Can express any system in either differential or integral form

29 Applications of integrators to solving differential equations Linear X IN X OUT Consider the standard integral form System X = b X + b X + b X a X + X + X +... OUT OUT OUT 3 OUT IN IN IN X IN a a a a 3 am b b b 3 b n X OUT This circuit is comprised of summers and integrators Can solve an arbitrary linear differential equation This concept was used in Analog Computers in the past

30 Applications of integrators to solving differential equations Linear X IN X OUT Consider the standard integral form System X = b X + b X + b X a X + X + X +... OUT OUT OUT 3 OUT IN IN IN Take the Laplace transform of this equation X = b X + b X + b X b + a X + a X + a X + a X a s s s s s s s s OUT OUT OUT 3 3 OUT n n IN IN IN 3 3 IN m m Multiply by s n and assume m=n (some of the coefficients can be ) s X = bs X + b s X + bs X b + a s X + as X + a s X + a s X a n n- n- n-3 n n- n- n-3 OUT OUT OUT 3 OUT n IN IN IN 3 IN n X n n- n- n-3 n n- n- n-3 ( s bs bs b3s... b ) = X ( as + as + as + a3s a ) OUT n IN n n n- n- n-3 a s + as + a s + a s a s bs b s bs... b OUT 3 = = T s X X IN n n- n- n-3 3 n n

31 Applications of integrators to solving differential equations Linear X IN X OUT Consider the standard integral form System X = b X + b X + b X a X + X + X +... OUT OUT OUT 3 OUT IN IN IN n n- n- n-3 a s + as + a s + a s a s bs b s bs... b OUT 3 = = T s X X IN n n- n- n-3 3 n n This can be written in more standard form n n- α s + α s +... α s + α s + β s β s + β n n = T s n n- n

32 Applications of integrators to filter design Linear System T s n n- α s + α s +... α s + α = s + β s β s + β X IN X n m OUT n n- n α n X IN α n- α n- α n-3 α X OUT -β n- -β n- -β n-3 -β Can design (synthesize) any T(s) with just integrators and summers! Integrators are not used open loop so loss is not added Although this approach to filter design works, often more practical methods are used

33 End of Lecture

34 Applications of integrators to filter design + I s I s This is a two-integrator-loop filter I X = - X +X +αx s OUT IN OUT OUT I s X OUT= XOUT These are -nd order filters X X I OUT=T s = X IN s + αis+i I I OUT =T s = X IN s + αis+i I s I If I =I =I, these transfer functions reduce to T Is ( s ) = s + αis+i T I ( s) = s + αis+i

35 Applications of integrators to filter design + I s I s T Is ( s ) = s + αis+i Consider T (jω) ( jω) T = T ( jω) = jωi ( I -ω ) α +jω I ωi ( I ) ( ) -ω + ωαi T T( jω) I ( s) = s + αis+i This is the standard nd order bandpass transfer function Now lets determine the BW and ω P

36 T Applications of integrators to filter design Determine the BW and ω P T ( jω) Is ( s ) = s + αis+i T ( jω) = ωi ( I ) ( ) -ω + ωαi To determine ω P, must set d T d ( jω) ω ( jω) = T = T ( jω ) P ( jω ) P This will occur also when d T and the latter is easier to work with dω ω I T jω = I -ω + ωαi ( ) ( ) d T ( I -ω ) + ( ωαi) I -ω I ( I -ω ) + ( αi) jω dω = ( I -ω ) + ( ωαi ) =

37 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P Is T T ( jω) d T d ( s ) = s + αis+i = = ( I -ω ) + ( ωαi ) ( I -ω ) + ( ωαi) I -ω I ( I -ω ) + ( αi) jω ω It suffices to set the numerator to ( I -ω ) + ( ωαi) I =ω I ( I -ω ) + ( αi) Solving, we obtain ω = I P ( jω ) T P II = = α ( I ) ( ) -I + I α = ωi ( I ) ( ) -ω + ωαi ( jω ) P Substituting back into the magnitude expression, we obtain Although the analysis is somewhat tedious, the results are clean T T ( jω ) P T( jω)

38 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P T( jω) Is T s + αis+i ( s ) = T ( jω) = ωi ( I ) ( ) -ω + ωαi To obtain ω L and ω H, must solve T ( jω) This becomes ( I -ω ) + ( ωαi) I -ω I ( I -ω ) + ( αi) = α ( I -ω ) + ( ωαi ) BW = ωh- ω L= αi ωω H L = I = α The expressions for ω L and ω H can be easily obtained but are somewhat messy, but from these expressions, we obtain the simple expressions α α

39 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P Is T ( s) = s + αis+i T ( jω) = ωi ( I ) ( ) -ω + ωαi T T ( jω ) P ( jω ) P T( jω) ω P = I T( jωp) = α T( jω) α α

40 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P T Is ( s ) = s + αis+i BW = αi ωω H L = I α α T( jω) Often express the standard nd order bandpass transfer function as T Is ( s ) = s +BWs+I

41 Applications of integrators to filter design The nd order Bandpass Filter These results can be generalized T BW = a ( s) = BP Hs s +as+b K K TBP ( jω) ωp = b H K= a

42 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P T T Is ( s ) = s + αis+i Is ( s ) = s +BWs+I + I s I s T( jω) α α Can readily be implemented with a summing inverting integrator and a noninverting integrator

43 Applications of integrators to filter design The nd order Bandpass Filter Determine the BW and ω P T T Is ( s ) = s + αis+i Is ( s ) = s +BWs+I + I s I s T( jω) α α α I= BW= Widely used nd order Bandpass Filter BW can be adjusted with R Q Peak gain changes with R Q Note no loss is added to the integrators ω P = I BW = αi

44 Applications of integrators to filter design The nd order Bandpass Filter Design Strategy Assume BW and ω P are specified T Is ( s ) = BP s + αis+i + I s T( jω) I s α α I= α BW= ω P = I. Pick C (use some practical or convenient value). Solve expression ω P= to obtain R 3. Solve expression α BW= to obtain α and thus R Q

45 Applications of integrators to filter design The nd order Lowpass Filter T I ( s) = s + αis+i + I s I s T ( jω ) P T ( jω) T jωp BW I= ω P ω Exact expressions for BW and ω P are very complicated but ω P I Widely used nd order Lowpass Filter BW can be adjusted with R Q but expression not so simple Peak gain changes with R Q Note no loss is added to the integrators Design procedure to realize a given nd order lowpass function is straightforward

46 Another nd -order Bandpass Filter V( sc +sc +G +G 3) = VOUTsC +VING 3 VsC +V G = OUT If the capacitors are matched and equal to C s 3 T( s ) = - s+s + ( R //R3) ω = R R //R C P ( 3) BW = R K= R T s = - Since this is of the general form of a nd order BP transfer function, obtain 3 s 3 s+s + + ( R //R3) C K K TBP ( jω)

47 Another nd -order Bandpass Filter Design Strategy Assume BW, ω P, and K are specified T s = - s 3 ( R //R3) s+s + BW = P ( 3) R K= R 3 ω = R R //R C K TBP ( jω). Pick C to some practical or convenient value. Solve expression BW = to obtain R R K= R 3. Solve expression to obtain α and thus R Solve expression P to obtain R ω = R R //R C 3 K

48 Another nd -order Bandpass Filter Termed the STAR biquad by inventors at Bell Labs V sc +sc +G +G = V sc +V G + V 3 OUT IN 3 OUT V H OUT ( sc+g ) =VsC+VOUTG sc H ω P= R ( R //R 3) C s H 3 H- T( s ) = - s+s + ( R //R3)( H-) ( R //R3) TBP ( jω) BW = R //R H- K = 3 H R3 H- R ( R //R3)( H-) For the appropriate selection of component values, this is one of the best nd order bandpass filters that has been published K K

49 STAR nd -order Bandpass Filter Implementation: s H 3 H- T( s ) = - s+s + ( R //R3)( H-) ( R //R3)??? But the filter doesn t work!

50 STAR nd -order Bandpass Filter s H 3 H- T( s ) = - s+s + ( R //R3)( H-) ( R //R3) Implementation: Works fine! If op amp ideal, V V OUT IN = H??? Will discuss why this happens later! Reduces to previous bandpass filter at H gets large Note that the H amplifier has feedback to positive terminal