FEEDBACK, STABILITY and OSCILLATORS

Transcription

1 FEEDBACK, STABILITY and OSCILLATORS à FEEDBACK, STABILITY and OSCILLATORS - STABILITY OF FEEDBACK SYSTEMS - Example : ANALYSIS and DESIGN OF PHASE-SHIFT-OSCILLATORS - Example 2: ANALYSIS and DESIGN OF WIEN BRIDGE OSCILLATORS à FEEDBACK, STABILITY OF FEEDBACK SYSTEM, OSCILLATORS ü STABILITY OF FEEDBACK SYSTEM V out = A vo.hv in -b.v out L A vf = V out ÄÄÄÄÄÄÄÄÄÄÄÄ V in = A vo.hv in -b.v outl ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ fi A vf = V in A vo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ +b.a vo The negative feedback effectively reduce the gain. Note that if» b.a vo» >> then, A vf > A vo ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ A vo.b > ÄÄÄÄÄ b For non - inverting opamp amplifier A vf = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ R ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ R +R F Univ. of Southern Maine Prof. M. G. Guvench

2 Phase at w Ø (number of zeros -number of poles). 90 Even if A vo (0) may be very large, at high frequencies» A vo HjwL» can become very small; making the approximation above invalid. A vf HjwL = A vo HjwL ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ + A vo HjwL.b HjwL For DC, the negative feedback results in a (+) sign in the denominator. Therefore,» A vf HjwL» will always be smaller than» A vo HjwL. But, if the inputs are reversed or somehow A vo HjwL.bHjwL changes its sign, the feedback becomes positive, the denominator can become <, therefore, A vf with feedback becomes even greater than A vo of the operational amplifier. Interesting Point: Denominator ª 0 fl "Instability" fl Oscillations Conditions for Instability HOscillationsL + A v HjwL.b HjwL 0 Equivalently J + A vo HjwL.b HjwLD = 0 + A vo HjwL.b HjwLD = 0 a. + vo HjwL.b HjwLD = 0 or Re@A vo HjwL.b HjwLD =- b. vo HjwL.b HjwLD = 0 fi Phase of HA vo.bl = 0, 80, 360 Note that, it is impossible to make + A vo HjwL.b HjwLD = 0 for phase = 0or 360 Univ. of Southern Maine 2 Prof. M. G. Guvench

3 For instability : J Angle@A vo.bd = 80 Re@A vo.bd =- N or, J Angle@A vo.bd = 80» A vo.b» = N Univ. of Southern Maine 3 Prof. M. G. Guvench

4 Stable system since» A vo.b < at the critical frequency(2) which satisfies the phase condition. Definitions/Conditions for Stability: Loop Gain = A vo.b Worst case: b= for passive circuits Univ. of Southern Maine 4 Prof. M. G. Guvench

5 f M (Phase Margin) > 30 ~ for a good internally compensated opamp Gain Margin > 0 db xxxxxxxxxxxxxxxxxxxxx OSCILLATOR ª Unstable but at a unique frequency so that it generates a sine wave with minimal distortion and minimal drift of oscillation frequency. Univ. of Southern Maine 5 Prof. M. G. Guvench

6 à Example : ANALYSIS and DESIGN OF PHASE-SHIFT-OSCILLATORS Note that for an oscillator to oscillate it does not need a signal, therefore, the signal input shown in the block diagrams can be eliminated. Amplifier Gain = A v Beta HjωL =βhjω ) b (jw) circuit Beta HjwL =b HjwL = V 2 HjwL ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ V HjwL Univ. of Southern Maine 6 Prof. M. G. Guvench

7 Educated conclusion: use higher number of RC's than 3 to achieve higher slope for better frequency V is known, I is unknown stimuli i j k V 0 0 y z = { i j k R + ÄÄÄÄÄÄÄÄÄÄÄ j wc -R 0 -R 2 R + ÄÄÄÄÄÄÄÄÄÄÄ j wc -R 0 -R 2 R + ÄÄÄÄÄÄÄÄÄÄÄ j wc and output voltage, V 2 = R.I 3 y i j z k { I I 2 I 3 y z { One can apply Cramer's Rule to solve for I 3 and the output voltage V 2 = R. I 3. In Mathematica the same calculation can be done by using linear solver which yields { I, I 2, I 3 } and pick out I 3, the third term from the list, and multiply it by R, as done below. V 2 = ExpandAllAR TakeALinearSolve ω > A i j k jrcy E R + j ω C R 2 R + j ω C R 0 R 0 R 2 R + j ω C y i, j k z { V 0 0 y E, 83<E ê. z { V y+ 5y 2 + y 3 == V + 6y+ 5y 2 + y ê. y > 3 j ω RC V 5 6 C 3 j 3 R 3 ω + 3 C 2 j 2 R 2 ω + 2 CjRω + Rearrange V 2 = V 5 H L + H6 L ω 2 R 2 C 2 jω RC ω 2 R 2 C 2 For the phase to reach 0 or 80 imaginary term should disappear, Univ. of Southern Maine 7 Prof. M. G. Guvench

8 Therefore, 6 ω 2 R 2 C = 0 ω 2 0 = $%%%%%%%%%%%%%%%% 6 R 2 C 2 Then, If Summary: V 2 V A vo i k j y z β { = i y j k 5. z { ω 0 2 = ω 2 R 2 C 2 ω 2 = = 6 R 2 C = 30 = 29 6 R 2 C 2 then oscillations will start and grow at ω 0. If» A vo» 29, If A vo < 0 (i.e. inverting) Oscillations will start and grow up at ω 0 = è!!! 6 RC A vf = A vo = A vo +β HjωL A vo = Univ. of Southern Maine 8 Prof. M. G. Guvench

9 à Example 2: ANALYSIS and DESIGN OF WIEN BRIDGE OSCILLATORS Univ. of Southern Maine 9 Prof. M. G. Guvench

10 Univ. of Southern Maine 0 Prof. M. G. Guvench

11 Beta =β HjωL = = = Z b Z a + Z b = R b +jω C b IR a + M + J jω C a IR a + M I + jω C jω C a R b M + b N R b +jω C b A R a + j Iω R a C b M + C b + E ω R b C a R b C a = + C b + M + j Iω R a R b I R a R b C a ω R b C a M When ω 2 = ω 0 2 = R a R b C a C b the imaginary term disappears β Hj ω 0 L = R a + C b + C a R b If H+L feedback is provided A vf = A vo β HjωL A vo at ω=ω 0 A vf Hjω 0 L A vo β Hjω 0 L A vo or A vf Hjω 0 L if A vo For C a = C b = C, R a = R b = R A vo 3 β Hjω 0 L = R a R b + C b C a + Univ. of Southern Maine Prof. M. G. Guvench

12 Wien - Bridge Oscillator i j R f + y z R a k R { R b + C b C a + Amplifier Design : Condition : Z in >> Z ab where Z ab = HZ a êê Z b L ω=ω0 Condition 2 : A vo > i k j y z β { ω=ω 0 In order to control the amplitude () Small Signal Gain > minimum needed for oscillation, (2) Average gain reduces and becomes less than ( M at the amplitude of oscillations it is required to stabilize. β ω= ω 0 Amplifer Chs. with Saturating Gain Univ. of Southern Maine 2 Prof. M. G. Guvench

13 Inverting Amplifier with Gain Saturation Low amplitudes Mid amplitudes High amplitudes R F = R F R F R F êê R F2 R F = R F êê R F2 êê R F3 i y β A vf = j k + z > β { β A v Univ. of Southern Maine 3 Prof. M. G. Guvench

14 H b.a v L b.a v > b.a v ª HA v L Large Signal HA v L Small signal Univ. of Southern Maine 4 Prof. M. G. Guvench

15 ACTIVE FILTERS à ACTIVE FILTERS - PASSIVE FILTER + AMPLIFIER - FILTER IS INTEGRATED IN THE FEEDBACK LOOP - TRUE ACTIVE FILTERS IMPLEMENTED WITH DEPENDENT SOURCES a. ACTIVE TWO-POLE SALLEN-KEY LOW-PASS FILTER b. ACTIVE TWO-POLE SALLEN-KEY HIGH-PASS FILTER c. ACTIVE TWO-POLE SALLEN-KEY BAND-PASS FILTER à ACTIVE FILTERS "Active" Filter Passive Filter Components + OpAmp as amplifier + OpAmp as dependent source à Example : PASSIVE FILTER + AMPLIFIER Advantage:. R L does not load the filter; therefore, it does not affect the frequency response. 2. The filter can actually amplify the signal in its path. (Insertion gain vs insertion loss of a passive filter) Univ. of Southern Maine 5 Prof. M. G. Guvench

16 R Thevenin.C observe that ω 3dB = and depends on loading resistance R RC L. At higher frequencies, frequency response of "real opamp" will alter the frequency response of the active filter. * Note that the danger of instability at the frequency where phase angle ª 80 * If the opamp is internally frequency compensated it can introduce no more than 90 phase while its gain is still greater than "0 db". This implies that: Overall filter will never have a gain greater than 0 db when phase = -80 ï stable. Univ. of Southern Maine 6 Prof. M. G. Guvench

17 à Example 2: FILTER IS INTEGRATED IN THE FEEDBACK LOOP If the opamp has wide enough GBW and low frequency gain; A vf HjωL = V out V in = A vo > +βa vo β HjωL This has the effect of Low Pass ö Low Reject (High Pass) High Pass ö High Reject Band Pass ö Band Reject A Good "Band Reject" Filter Remark: A good Low / High / Band Reject actually reduces the signal in the reject band significantly, i.e. gain <<. Univ. of Southern Maine 7 Prof. M. G. Guvench

18 à Example 3: TRUE ACTIVE FILTERS IMPLEMENTED WITH DEPENDENT SOURCES a. ACTIVE TWO-POLE SALLEN-KEY LOW-PASS FILTER b. ACTIVE TWO-POLE SALLEN-KEY HIGH-PASS FILTER c. ACTIVE TWO-POLE SALLEN-KEY BAND-PASS FILTER a. ACTIVE TWO-POLE SALLEN-KEY LOW-PASS FILTER Equivalent circuit: Univ. of Southern Maine 8 Prof. M. G. Guvench

19 Node Equations: Univ. of Southern Maine 9 Prof. M. G. Guvench

20 i j k V in R + V O sc 0 y z = JHG + G 2 + sc L G 2 { G 2 HG 2 + sc 2 L N JV N V 0 V i in R j y z k 0 = JHG + G 2 + sc L HG 2 + sc L { H G 2 L HG 2 + sc 2 L N JV N V 0 Using Cramer' s Rule to get V O V in R V 0 V in R = DetA i V HG + G 2 + sc L in R j y z I V in M. k H G 2 L 0 E { R DetAJ HG + G 2 + sc L HG 2 + sc L H G 2 L HG 2 + sc 2 L N E = I V in M H G I V in M. R 2 L HG + G 2 + sc L HG 2 + sc 2 L G 2 HG 2 + sc L R V O V in = H HsL = G.G 2 s 2 C C 2 + sc G 2 + sc 2 HG + G 2 L + G 2 HG + G 2 L G 2 2 G 2 sc = G.G 2 s 2 C C 2 + sc 2 HG + G 2 L + G.G 2 = HR C L HR 2 C 2 L s 2 + s I + M + I M I M R C R 2 C R C R 2 C 2 Rename : 2 i =ω 0 j + R C R 2 C 2 k R C y z = 2 ζ ω 0 R 2 C { H HsL = ω 0 2 s ζω 0 s +ω 2 0 No inductor! The response of this circuit is similar to the response of an RLC - Low Pass passive filter shown below. Except that there is no need to employ bulky, lossy and expensive inductors to implement it. Univ. of Southern Maine 20 Prof. M. G. Guvench

21 V out ê V in = ê sc êhr + sl + ê scl = êhsrc + s 2 LC + L = H ê LCLêHs 2 + R ê Ls + ê LCL where ω 0 2 ê LC is the "resonant frequency" and 2 ζ ω 0 R ê L is the "damping factor" H = ω 0 2 s ζ ω 0 s +ω 2 0 a =, b = 2 ζ ω 0, c =ω 0 2 = constant Hs p L Hs p 2 L p, p 2 = è!!!!!!!!!!!!!!!! b ± b 2 4 ac 2 a = b 2 ± $%%%%%%%%%%%%%%%%%%% i k j b y z c 2 { 2 = ζ ω 0 ± "######################### Hζ ω 0 L 2 ω 02 = ζω 0 ± jω 0 "############ ζ 2 complex conjugate H =» constant»» s p»» s p 2» Univ. of Southern Maine 2 Prof. M. G. Guvench

22 b. ACTIVE TWO-POLE SALLEN-KEY HIGH-PASS FILTER Insertion Loss Univ. of Southern Maine 22 Prof. M. G. Guvench

23 Univ. of Southern Maine 23 Prof. M. G. Guvench

24 c. ACTIVE TWO-POLE SALLEN-KEY BAND-PASS FILTER Univ. of Southern Maine 24 Prof. M. G. Guvench

25 Node Equations : V IN i R j k + KV 2 R 0 y z = i j { k + R R + sc + sc sc y sc + sc z JV N V 2 R { or V i IN y R j z k 0 = i j { k 2 H R + scl HsC + K R L sc H + scl R y z JV N V 2 { Using Cramer' s rule, KV 2 = V OUT = K DetA i j 2 H R + scl V IN y R z k sc 0 E { DetA i2 H K + scl HsC + y R R j k sc H + scl z E R { = K + scv IN R 2 H + scl H K + scl sc HsC + L R R R Univ. of Southern Maine 25 Prof. M. G. Guvench

26 V OUT V IN = K sc R after dividing by C2, 2 HsCL H sc L + 2 H R R L2 HsCL 2 sc K R H HsL = K sc R HsCL 2 + H4 KL sc + 2 H R R L2 H HsL = K s RC è!!!! 2 s 2 + H4 KL RCÆ s + J RC N2 Æ 2 γ = 2 ζ ω n ω n 2 ω n = è!!! 2 RC ω n =ω 0 2 +γ 2 The damping factor : ζ= H4 KL RC 2 2 è!!!! RC = H4 KL 2 è!!! 2 Poles = γ±jω 0 Resonance is at ω=ω 0 BandWidth = BW =ω 2 ω = ω 0 +γ Hω 0 γl = 2 γ Quality Factor of a Resonant Circuit : Q = ω peak BW > ω 0 2 ζ = è!!! 2 4 K Univ. of Southern Maine 26 Prof. M. G. Guvench

27 K RC» H max» = H+jωL 2 + H 4 K L jω+hω RC 0L 2 jω = K 4 K = K è!!! 2 è!!! 2 4 K = K Q è!!! 2 K 4» H» max unstable! Oscillator Univ. of Southern Maine 27 Prof. M. G. Guvench

28 Univ. of Southern Maine 28 Prof. M. G. Guvench