OPERATIONAL AMPLIFIER ª Differential-input, Single-Ended (or Differential) output, DC-coupled, High-Gain amplifier

Transcription

1 à OPERATIONAL AMPLIFIERS à OPERATIONAL AMPLIFIERS (Introduction and Properties) Phase relationships: Non-inverting input to output is 0 Inverting input to output is 180 OPERATIONAL AMPLIFIER ª Differential-input, Single-Ended (or Differential) output, DC-coupled, High-Gain amplifier IDEAL OP-AMP REAL OP-AMP A VO, Gain (Differential) R id Bipolar input >> 10 6 MOS or JFET input R out C.M.R.R db Bandwidth 100 Hz. Univ. of Southern Maine 1 Prof. M.G. Guvench

2 GBW HGain - Bandwidth ProductL 1 MHz MHz Offsets 0 (none) Finite Univ. of Southern Maine 2 Prof. M.G. Guvench

3 I BQ1 and I BQ2 I input bias >» I BQ1» >» I BQ2» Input Offset Current I OS = HI BQ1 - I BQ2 L Real Operational - Amplifiers V OS = V OS HTL, I OS = I OS HTL, A VO = A VO HTL Temperature Coefficient of Input Offset Voltage i.e. H KL dv OS ÄÄ dt > V OS ÄÄÄÄ T where T is absolute temperature, V OS d 5 mv (Bipolar) 1-25 mv (FET), expensive HI OS, I BIAS ) d 1 ma (Bipolar) ~ pa -na (JFET) ~ pa (MOSFET) à OPERATIONAL AMPLIFIER CIRCUITS (APPLICATIONS) In the following analyses, unless mentioned otherwise, the operational amplifiers are treated as if they are "ideal" or near ideal. Univ. of Southern Maine 3 Prof. M.G. Guvench

4 ü I / V (CURRENT / VOLTAGE) CONVERTER J I Bias + I Bias - N 0 V OS 0 Ideal Operational Amplifier If A VO ô v OUT = A VO Hv + - v - L Hv + - v - L = v OUT ÄÄÄÄÄÄÄ A VO ô0 v + 0 î v - = 0 HVirtual GroundL For v - = 0 H v - L - Hv OUT L = R.i R = R.i IN ô v OUT =-R.i IN Therefore, this current-voltage converter circuit behaves just like a "Current-Controlled-Voltage-Source" with zero input resistance. (see Figure below) Univ. of Southern Maine 4 Prof. M.G. Guvench

5 ü INVERTING AMPLIFIER Since v - = 0 Hvirtual groundl i IN = i 1 = v 1 - Hv - L = v 1 Therefore i IN = v 1 flows into the I - V converter which creates v OUT =-R F.i IN, resulting in v OUT = -R F ÄÄÄÄÄÄ v 1 or, A vo = - R F R OUT = 0 R IN = v 1 ÄÄÄÄÄÄÄ = v 1 = for the "inverting" amplifier. i 1 i IN Univ. of Southern Maine 5 Prof. M.G. Guvench

6 ü THE SUMMING AMPLIFIER Univ. of Southern Maine 6 Prof. M.G. Guvench

7 Since v - = 0 Hvirtual groundl i 1 = v 1-0 ÄÄ = v 1 i 2 = v 2-0 ÄÄ R 2 = v 2 ÄÄÄ R i N = v N - 0 ÄÄÄ = v N ÄÄ R N R N N i IN = i i i=1 and v OUT =-R F.i IN v OUT = - R F.v 1 - R F.v R F ÄÄ R 2 R N.v N v OUT = -Ha 1 v 1 + a 2 v a N v N L where a i = - R F R i Comment: (1) This is equivalent to, mathematically, a weighted sum of inputs or simply an addition operation except for the (-) signs. (2) To change the sign of the coefficient of an element of the sum simply employ a unity gain inverting amplifier at its input. v OUT = - R ÅÅÅÅÅ R v IN =-v IN Univ. of Southern Maine 7 Prof. M.G. Guvench

8 ü THE NON-INVERTING AMPLIFIER A vo >> 1 v + = v 1 v - = ÄÄÄÄÄ.v OUT + R F v + - v - = v OUT ÄÄÄÄÄÄÄ A vo ª 0 î v + = v - Therefore, v 1 = ÄÄÄÄÄ.v OUT î v OUT = i j R F + 1 y z.v 1 or A v = i j R F + 1 y z + R F k { k { From the resulting equation observe that: 1. The voltage gain is +I R F ÅÅÅÅÅÅ + 1M positive. R 2. If ÅÅÅÅÅÅ F ö 0 ( R F ö0 (short), ö (open) ) then the gain becomes (+1), i.e. unity gain (absolute unity). v OUT = v 1 UnityGainBuffer Amplifier 3. Since operational amplifier does not draw any current i + º 0 ï R IN > Univ. of Southern Maine 8 Prof. M.G. Guvench

9 ü Effect of Finite A VO on Feedback Amplifier Gain BETA ª b is a transfer relationship. In general b =b (jw) For finite A VO, v OUT = A vo Hv + - v - L, v + = v 1, v - =b.v OUT v OUT = A vo H v 1 -b.v OUT L î H1 +ba vo L v OUT = A vo.v 1 î v OUT ÄÄÄÄÄÄÄ v 1 = A vo ÄÄÄÄÄÄ H1 +ba vo L î v OUT = 1 ÄÄÄÄÄ b ÄÄ 1 + ÄÄÄÄÄ 1 b.a vo v 1 Therefore, as long as b.a vo >> 1 v OUT > 1 ÄÄÄÄÄ b v 1 Example: Univ. of Southern Maine 9 Prof. M.G. Guvench

10 Special Case : v OUT = 1 v 1 = i j1 + R F y z v 1 R 1 k { R F + β= Vout Vin = R F + Non inverting amplifier ü FREQUENCY RESPONSE OF REAL OPERATIONAL AMPLIFIERS Internally Frequency Compensated Op.Amp. (One pole / capacitor dominates) A vo HjwL > A vo H0L ÄÄÄÄÄÄÄ w 1 + j ÄÄÄÄÄÄÄ w 0 A vo HjwL w<<w0 > A vo H0L A vo HjwL w>>w0 > A vo H0L ÄÄ w j ÄÄÄÄÄÄÄ w 0 = A vo H0L.w 0 ÄÄÄ jw = GBW Ä jw Note that when at unity gain frequency A vo Hjw 1 L > 1 and w=w 1 = GBW Notes/Definitions: Unity Gain Frequency = Gain Bandwidth Product (if single pole, -20dB/dec) Unity Gain Frequency GBW (if not -20dB/dec) Definitions: A vo The Open Loop Gain (unloaded, no feedback) BETA ª b The Feedback Ratio b A vo The Closed Loop Gain Univ. of Southern Maine 10 Prof. M.G. Guvench

11 ü Non-Inverting Amplifier's Gain BandWidth Product In the feedback amplifier utilizing this "Real Operational Amplifier" 1 V out HjwL = ÄÄ b HjwL. 1 Ä.V 1 HjwL ÄÄÄÄÄÄ b HjwL.A vo HjwL For the non - inverting amplifier b HjwL = ÄÄÄÄÄ = constant + R F V out ÄÄÄÄ = ÄÄÄÄÄ 1 V 1 b. 1 ÄÄ ÄÄ A vo H0L w 1+j Ä w 0 b. ÄÄÄÄÄÄÄ = 1 ÄÄÄÄÄ b. 1 ÄÄ j ÄÄÄÄÄÄÄ w w ÄÄ 0 b.a vo H0L = 1 ÄÄÄÄÄ b. 1 ÄÄÄÄÄ Ä 1 I1 + ÄÄ b.a vo H0L M + jw b.w 0 A vo H0L For GBW =w 0.A vo H0L and, as long as b.a vo H0L >>> 1 V out ÄÄÄÄ = ÄÄÄÄÄ 1 V 1 b. 1 ÄÄÄÄ 1 + ÄÄÄÄÄÄÄ jw i j GBW ÄÄÄÄÄÄ y 1 z ÄÄÄÄ k b { 1 ÄÄÄÄÄ b = ÄÄÄ w 1 + j I ÄÄÄÄÄ where Hw -3 db L of the feedback amp = GBW ÄÄÄÄÄÄ ÄÄÄÄÄ I 1 b M w -3 db M and i j 1 y ÄÄÄÄÄ z is low frequency gain of the feedback amplifier k b { HGain. Bandwidth L feedback amplifier = HGBWL operational amplifier Univ. of Southern Maine 11 Prof. M.G. Guvench

12 à FREQUENCY RESPONSE OF AN OPAMP AMPLIFIER WITH FREQUENCY DEPENDENT FEEDBACK Definitions: A vo The Open Loop Gain (of the opamp, unloaded, no feedback) BETA ª b The Feedback Ratio b A vo The Closed Loop Gain A v The Feedback Gain (of the overall amplifier) What if A vo =A vo (jw) and b(jw) the frequency response of the operational amplifier is more complicated? Answer: As long as b A vo >>> 1 the response will be 1/b(jw) as shown with the example below. Univ. of Southern Maine 12 Prof. M.G. Guvench

13 à THE DIFFERENCE AMPLIFIER Univ. of Southern Maine 13 Prof. M.G. Guvench

14 Univ. of Southern Maine 14 Prof. M.G. Guvench

15 Use the idea of superposition for 1 & 2 : 1. v 1 0 v 2 π 0 v OUT =- R F.v 2 2. v 1 π 0 v 2 0 v 1 ' = R B.v 1 R A + R B R B v OUT = i j R F + 1 y z v ' 1 = i j R F + 1 y z.v 1 k { k { R A + R B v OUT =- R F v 2 + i j R F + 1 y R B ê R A z ÄÄÄÄÄ k { HR B ê R A + 1L.v 1 If R B ÄÄ R A = R F v OUT = R F Hv 1 - v 2 L Pure Differential Response A Different Approach : HA v L Op.Amp ô î v + > v - v + = R B v 1 v - = R A + R B R F ÄÄÄÄÄ v 2 + ÄÄÄÄÄ v OUT + R F + R F R B v 1 > R A + R B R F ÄÄÄÄÄ v 2 + ÄÄÄÄÄ v OUT + R F + R F v OUT = i j + R F y ÄÄÄÄÄ z i y j v 1 - ÄÄÄÄÄ v 2 z k { k R A + R B + R F { R B R F If R B = R F and R A = î R B ÄÄ R A = R F Then v OUT = + R F i y ÄÄÄÄÄ j ÄÄÄÄÄ v 1 - ÄÄÄÄÄ v 2 z k + R F + R F { R F R F v OUT = R F Hv 1 - v 2 L Choose R B = R F and R A = to make the resistances ( R A êê R B ) and ( êê R F ) equal so that equal I B+ and I B- bias currents (i.e. zero I OS ) create balanced shifts in v + and v -. Otherwise, the difference will get amplified and result in a large offset at the output. Univ. of Southern Maine 15 Prof. M.G. Guvench

16 à LOGARITHMIC AMPLIFERS L og Amplifier for v 1 > 0 If v - is virtual ground then i 1 > v 1 - v - ÄÄÄÄ î i 1 > v 1 i 1 = i C = I S e vbe ÄÄÄÄÄÄ kt ê q for v BE =-v OUT i 1 = v 1 = i C = I S e -vout Ä kt ê q Then for v 1 > 0 output becomes v OUT =- HkTL ÄÄÄÄÄ q v 1 ln i y j ÄÄÄÄÄÄÄ z < 0 k.i S { Lo g Amplifier for v 1 < 0 Univ. of Southern Maine 16 Prof. M.G. Guvench

17 For v 1 < 0 v BE = v OUT i C = I S e vbe ktääääää ê q ÄÄÄÄ = IS e v OUT kt ê q i 1 =-i C Then v OUT =+ HkTL ÄÄÄÄÄ q ln i j -v 1 y ÄÄÄÄÄÄÄ z > 0 k.i S { à ANTI-LOGARITHMIC AMPLIFIERS v 1 > 0 ÄÄÄÄ i C = I S e +veb kt ê q ÄÄÄÄ v OUT =-R F.i C =-R F.I S e +veb kt ê q ÄÄÄÄ For v EB = v 1 v OUT =-R F.I S e + v 1 kt ê q works for v 1 > 0 Univ. of Southern Maine 17 Prof. M.G. Guvench

18 Anti Log Amplifier for v 1 < 0 ÄÄÄÄ v OUT =+R F.I S e - -v 1 kt ê q Hworks if v 1 < 0L ü Applications of Log / Anti-Log Amplifiers: Log HA * BL = Log A + Log B Log HA ê BL = Log A - Log B e a Log A + b Log B = e a Log A.e b Log B = A a.b b Therefore, all of the following mathematical functions, multiplication, division, logarithms, exponentiation, power including roots can be implemented using the circuit shown above. Univ. of Southern Maine 18 Prof. M.G. Guvench

19 à DIFFERENTIATING AMPLIFER (The OpAmp Differentiator) Since v - ô0 Hvirtual groundl then, v OUT =-R.i R + v - ô v OUT =-R.i R i C = C. dv dt = C. d Hv 1 - v - L ÄÄÄÄ dt ô i C = C. dv 1 ÄÄÄ dt v OUT =-RC. dv 1 ÄÄÄ dt To change the sign use unity gain inverter at the input (or the output). Frequency Response: Univ. of Southern Maine 19 Prof. M.G. Guvench

20 v OUT H+L =-RC ÄÄÄÄÄÄÄ d dt v 1 HtL in H s or jw domainl ô V OUT HsL =-RC ÄÄÄÄÄÄÄ d dt V 1 e st V OUT HsL =-src.v 1 e st V 1 e st = V 1 HsL ô V OUT HsL =-src.v 1 HsL V OUT HjwL =-jw RC.V 1 HjwL î H HjwL =-j i w y j ÄÄ z k 1 ê RC { à INTEGRATING AMPLIFIER (The OpAmp Integrator) For v - = virtual ground and i R = i C i R = v 1 - v - ÄÄÄÄ R = v 1 ÄÄÄÄÄÄÄ R and i C = C ÄÄÄÄÄÄÄ d dt Hv - - v OUT L =-C dv OUT ÄÄÄ dt v 1 ÄÄÄÄÄÄÄ R =-C dv OUT ÄÄÄ dt î v OUT =-ÄÄÄÄ 1 t R C v - Ht'L t' + v OUT H0L 0 By introducing a switch the initial value, v OUT H0L set at zero. Univ. of Southern Maine 20 Prof. M.G. Guvench

21 The switch, S keeps shorting the terminals of the capacitor for t < 0. When it opens at t = 0 the capacitor's initial voltage is set to be zero. Frequency Response: V OUT HsL =- 1 ÄÄÄÄÄÄÄ src.v 1 HsL V HjwL =- 1 ÄÄ jw R C V 1 HjwL Univ. of Southern Maine 21 Prof. M.G. Guvench

22 à POWER OPERATIONAL AMPLIFIERS Complementary pairª Q 1 and Q 2 have similar V BE - I C chs. and identical b's but one is NPN, the other is PNP. I OUT = Hb +1L.I OA HI OUT L MAX Hb +1L.HI OA L MAX HI OA L MAX ~ 10 ma then HI OUT L MAX ~ 1A For higher currents you need to use a Darlington transistor. Complementary Emitter Follower Chs. Univ. of Southern Maine 22 Prof. M.G. Guvench

23 Note that Darlington outputs will have more crossover distortion because of doubling of VBE drops. Effect of Negative Feedback on Cross-Over Distortion If A VO of Op Amp. is very high v + = v - v 1 = ÄÄÄÄÄ.v OUT ô v OUT = i j1 + R F y z.v 1 + R F k { Note that the voltage v 1 will need to be ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 0.7 V A VO ~ 7 mv to bring the output out of the cross-over distortion. With an output swing of 10 V, Gain ~ 100 input swing is ÅÅÅÅÅÅÅÅÅÅ ~ 0.1 V. 7 mv Therefore the ratio of (input referred cross - over voltage ê peak voltage) ~ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ~ , which is very small mv fraction. Univ. of Southern Maine 23 Prof. M.G. Guvench

24 POWER AND OUTPUT RANGE LIMITATIONS 1. I OA I OAMAX ~ 10 ma, I L I LMAX = I OUTMAX > Hb +1L.I OAMAX 2. Typically, V OA V OAMAX, V OAMAX > V CC - 1V V LMAX V OAMAX - V BE1 3. P DQ1 = I L HV CC - V L L for the worst case : V L > 0 P DMAXQ1 I LMAX.V CC For a resistive load: V L = R L.I L P Q = V L.V CC - V 2 L ÄÄÄÄ R L R L Transistor Power Dissipation L V L HV CC V L Lvs. DC Load Voltage For non-resistive loads and for AC waveforms the transistor's power dissipation versus load voltage will be different depending on the waveforms and the phase between voltage and current. Univ. of Southern Maine 24 Prof. M.G. Guvench

25 à OPAMP DC-VOLTAGE REGULATOR ü Simple Zener Diode Regulator (Review) Disadvadvantages to overcome: 1. Fixed output voltage, 2. It cannot handle large I L because it needs high P DMAX zeners, 3. Output voltage varies because of finite r Z ~ 1-10 W, 4. Output resistance º r Z, cannot be smaller, 5. Ripple reduction H r Z R o +r Z L requires higher values of R O, which increases the DC voltage drop across R O, and the power wasted on it. Univ. of Southern Maine 25 Prof. M.G. Guvench

26 ü OPAMP DC-VOLTAGE REGULATOR WITH ADJUSTABLE OUTPUT v + > v - k 1 V Z = k 2 V L ô V L = k 1 k 2.V Z where 0 k 1 1 k 2 = ÄÄÄÄÄ + R F Power is supplied by V CC. To make V UR the source that supplies power to R L simply eliminate V CC use V UR for it. Limitations : I LMAX < Hb +1L I OAMAX and V LMAX < V URMIN - V BE1 - V O.A.DROP For V CC = V URMAX Q 1 has to be able to handle P DMAX > I LMAX HV URMAX - V LMIN L Univ. of Southern Maine 26 Prof. M.G. Guvench

27 ü OPAMP DC-VOLTAGE REGULATOR WITH ADJUSTABLE OUTPUT AND CURRENT LIMIT Current limit is turned on when I RCL.R CL > V BE ON Q2 ª 0.7 V I RCL = I L + I ÑÖÖÖÖÖÖÖÖÖÖÖÖÖÖÖÖ FEEDBACK ÖÖÖÖÖÖÖÜ î I LMAX.R CL > 0.7 V small Comment : Design of zener diode regulator is much simpler since its load is a constant (does not vary) and is also a high resistance load (does not need high current to operate). à PRECISION RECTIFIERS Input : v IN HtL Output : v OUT HtL = 9 Kv IN HtL for v IN > 0 0 for v IN < 0 Variations : HAL Responds to : Hthe case abovel v IN > 0 and either inverts HA1L or does not invert HA2L HBL Responds to : v IN < 0 and either inverts HB1L or does not invert HB2L Univ. of Southern Maine 27 Prof. M.G. Guvench

28 Circuit (A1) Analysis of Precision Rectifier Circuit (A1) : Precision Rectifier When v IN > 0 v OUT = Hv - L - i IN.R 2, i IN = v IN - Hv - L ÄÄÄ and Hv - L ô0 v OUT =- R 2.v IN When v IN < 0 v OUT = Hv - L - i R2.R 2 = 0 for Hv - L ô0 and i R2 = 0 v OUT = 9 - R 2 ÄÄÄÄÄÄ v R IN 1 if v IN > 0 0 if v IN < 0 Therefore, "Precision Half-Wave Rectifier Circuit (A1)" given above responds to positive v in. It creates an output waveform which is the rectified positive half of v in, amplified and inverted. (B1) Analysis of Precision Rectifier Circuit (B1) Precision Rectifier Circuit Univ. of Southern Maine 28 Prof. M.G. Guvench

29 Reversing the diodes makes the circuit HA1L respond to the signal with opposite polarities. From above, v OUT = 9 - R 2 ÄÄÄÄÄÄ v R IN 1 if v IN < 0 0 if v IN > 0 Therefore, "Precision Half-Wave Rectifier Circuit (B1)" given above responds to negative v in. It creates an output waveform which is the rectified negative half of v in, amplified and inverted. Comment: The other versions of the precision rectifier can be implemented by employing inverting, and summing amplifers in combination with the circuits A1 and B1 discussed above. Homework: Design a "Precision Full-Wave Rectifier" which will output the signal full-wave rectified, all positive, and amplified 10 times. Univ. of Southern Maine 29 Prof. M.G. Guvench