Operational Amplifiers

Lesson #5 Chapter 2

BME 372 Electronics I – 138 J.Schesser Operational Amplifiers • An operational Amplifier is an ideal differential with the following characteristics: – Infinite input impedance – Infinite gain for the differential signal – Zero gain for the common-mode signal Inverting input – Zero output impedance v 1 v – Infinite Bandwidth - o v2 + Non-Inverting input

BME 372 Electronics I – 139 J.Schesser Operational Amplifier Feedback • Operational Amplifiers are used with negative feedback • Feedback is a way to return a portion of the output of an amplifier to the input – Negative Feedback: returned output opposes the source signal – Positive Feedback: returned output aids the source signal • For Negative Feedback – In an Op-amp, the negative feedback returns a fraction of the output to the inverting input terminal forcing the differential input to zero. – Since the Op-amp is ideal and has infinite gain, the differential input will exactly be zero. This is called a virtual short circuit – Since the input impedance is infinite the current flowing into the input is also zero. – These latter two points are called the summing-point constraint.

BME 372 Electronics I – 140 J.Schesser Operational Amplifier Analysis Using the Summing Point Constraint • In order to analyze Op-amps, the following steps should be followed: 1. Verify that negative feedback is present 2. Assume that the voltage and current at the input of the Op-amp are both zero (Summing- point Constraint 3. Apply standard circuit analyses techniques such as Kirchhoff’s Laws, Nodal or Mesh Analysis to solve for the quantities of interest.

BME 372 Electronics I – 141 J.Schesser Example: Inverting Amplifier 1. Verify Negative Feedback: Note that a

portion of vo is fed back via R2 to the R2 inverting input. So if vi increases and, i 2 therefore, increases vo, the portion of vo fed i1=vin/R1 0 R1 back will then have the affect of reducing vi - - v (i.e., negative feedback). + i + + + RL vin 2. Use the summing point constraint. - vo - 3. Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output

vin  i1R1  0 since vi is zero due to the summing - point constraint vin i1R1 Zin    R1 i1  i2 due to the summing - point constraint i1 i1 v0  i2 R2  0 since vi is zero

R2  - vin which is independent of RL (note that the output is opposite to the input : inverted) R1 BME 372 Electronics I – 142 J.Schesser Op-amp

• Because we assumed that the Op-amp was ideal, we found that with negative feedback we can achieve a gain which is: 1. Independent of the load 2. Dependent only on values of the circuit parameter 3. We can choose the gain of our amplifier by proper selection of resistors.

BME 372 Electronics I – 143 J.Schesser Another Example: Inverting Amplifier

1. Verify Negative Feedback: i 4 2. Use the summing point constraint. R2 R4

i2 3. Use KVL at the inverting input node for the i3 R3 i1=vin/R1 0 R1 branch connected to the source and KCL & - KVL at the node where the 3 resistors are vi + R vo L connected vin

vin  i1R1  0 since vi is zero due to the summing - point constraint

i1  i2 due to the summing - point constraint vi  i2 R2  i3R3  0  i2 R2  i3R3 since vi is zero i4  i3  i2 vo  R4i4  R3i3 BME 372 Electronics I – 144 J.Schesser Another Example: Continued R i R  i R  i  2 i i 2 2 3 3 3 2 4 R3 R2 R4 vin vin  i1R1  0; i1  i2  i2  i2 i3 R3 R1 i =v /R 0 1 in 1 R R R 1 1 i  i  i  ( 2 1)i  ( 2  )v - 4 3 2 2 in R3 R3R1 R1 vi + R R2 1 R2 vin v L v  R i  R i  R (  )v  R v o o 4 4 3 3 4 in 3 in R3R1 R1 R3 R1

R2 R4 R4 R2  vin (   ) R3R1 R1 R1

vo R2 R4 R4 R2 Av   (   ) vin R3R1 R1 R1 v v  i R  0 in in 1 1 Zin  Rin   R1 i1 i1  i2 Why would we use this design over the simpler one? i2 R2  i3R3 1. Same gain but with smaller values of resistance i4  i3  i2 2. Higher gain vo  R4i4  R3i3 BME 372 Electronics I – 145 J.Schesser Non-inverting Amp

1. First check: negative feedback? I = 0 2. Next apply, summing point constraint in + io 3. Use circuit analysis + + vi - -- RL vvv 0 vv + R2 in i f f f vin vo -- + R1 v R vvvfoin; f 1 -- RR12 -- v R  RR A o 21 1 2 v vR R in 11Note: v Since iZ 0; in 1. The gain is always greater than one in in i in 2. The output has the same sign as the input

BME 372 Electronics I – 146 J.Schesser Non-inverting Amp Special Case

What happens if R2 = 0? Iin= 0

+ io vin  vi  v f  0  v f  v f + v + R i - 1 -- RL v f  vo  vo  vin ; + vin vo R1  0 -- v 0  R Av  o  1 1 vin R1 --

vin Since iin  0;Zin    iin

This is a unity gain amplifier and is also called a voltage follower.

BME 372 Electronics I – 147 J.Schesser Some Practical Issues when Designing Op-amps

• Since ratios of resistor values determines the gain, choosing the proper resistor values is crucial – Too small means large currents drawn – Too large yields another set of problems • Open Loop Gain is not constant but a function of frequency • Non-linearities of the amplifier – Voltage clipping – Slew rate • DC imperfections – Offsets – Bias Currents

BME 372 Electronics I – 148 J.Schesser Selecting Resistor Values

• Let say we want a gain of 10. This

means that R2 = 9R1. Iin= 0

+ io • If we chose R1=1Ω, then for a 10 volt + + output, there will be 1 A flowing threw vi - R and R . -- 1 2 + R2 vin vo -- • THIS IS DANGEROUS!!!! + vf R1 • On the other hand if R1=10 MΩ, then -- there may be unwanted effects due to -- R pickup of induced signals Av 1 2 R • Therefore choosing values between 1 100Ω and 1MΩ is optimum

BME 372 Electronics I – 149 J.Schesser Frequency Issues • Let’s assume that the open loop gain of our Op-amp is a function of frequency. 20 log |AOL( f )| dB 20 log |A0OL| AoOL AfOL () 1(/ jf fBOL )

where AfoOL is the open-loop gain for  0,

fBOL is called the open-loop break

frequency since when ff BOL

then AfOL ( ) A oOL / 2 or at the half power point.

fBOL Note that when fAf oOL BOL , AOL ()f  1 This is will define an important relationship for the amplifier when feedback is used

BME 372 Electronics I – 150 J.Schesser Frequency Issues Iin= 0

+ io + A + oOL Vi - AfOL () 1(/ jf fBOL ) -- + R2 Using phasors: Vin Vo -- RR + VVV11 ; where  fOO Vf R1 RR12 RR 12 --  -- V O VVin i V O   V O; since V O  V iA OL AoOL AOL AfOL() Af OL () 1 jff ( / BOL ) VOOLA AfCL ()  ACL  1()1()Af Af  A OL OL 1 oOL Vin1 A OL  1(/ jf fBOL ) A  (Note for the ideal Op-amp A  oOL OL 1(/ jf f ) A 1 RR R BOL oOL and A 12  1 2 1(/ jf f ) A 1(/)Ajff CL BOL  oOL oOL BOL RR11 1(/)1(/)jf fBOL jf f BOL which what we expected.) AoOL 1 AA oOL oCL ff 1(jj )1( ) fABOL(1 oOL ) f BCL BME 372 Electronics I – 151 J.Schesser Iin= 0

+ io Frequency Issues + + Vi - -- 20 log |A0OL| 20 log |AOL( f )| dB + R2 Vin Vo -- + Vf R1 -- 20 log |AOL( f )| dB 20 log |A0CL| --

fBOL fBOL(1+AoOL)

 A interesting factor now comes to light: Where:  A oOL AoOL AoCLffAAf BCL BOL(1 oOL ) oOL BOL AffA and (1 ) 1 A oCL BCL BOL oOL oOL 1 AoOL The gain bandwidth product is constant!!!

BME 372 Electronics I – 152 J.Schesser Example • For an amplifier with Open-loop dc gain of 100 dB with Open-loop breakpoint of 40 Hz the Bandwidth product for β=1, 0.1, 0.01

β AOCL AOCL(dB) fBCL 1 0.9999 0 4 MHz 0.1 9.9990 20 400 kHz 0.01 99.90 40 40 kHz

BME 372 Electronics I – 153 J.Schesser 20

10

0 Iin= 0 0 5 10 15 20 Voltage clipping

-10 + io + -20 + Vi - -- • The Op-amp has a limit as to how + 3k Vin Vo large an output voltage and current -- can be produced. (See figure 2.28) + V 1k • For example for the circuit shown it f -- has the following limitations: ±12 V -- and ±25 mA a) If the load resistor is 10k Ω, what is Vomax Vomax 12 12 the maximum input voltage which can b)Iomax     123mA be handled without clipping RL R1  R2 100 3000 1000 b) Repeat for 100 Ω For this case, maximum output current will occur

AvCL 1 3/1  4 before maximum output voltage is reached. V V 12 12 V V V V omax omax omax omax 25mA omax omax a)Iomax    4  Iomax      RL R1  R2 10 3000 1000 RL R1  R2 100 3000 1000

 4.2mA Vomax  2.44V

For this case, maximum output voltage will occur Vinmax  2.44 4  0.61V before maximum output current is reached. V  12 4  3V BME 372 Electronics I – 154 inmax J.Schesser Slew Rate • Slew Rate is a phenomenon which occurs when the Op-Amp can not keep up the change in the input. • Therefore, we identify the maximum rate of change of the Op-amp as the Slew Rate - SR

BME 372 Electronics I – 155 J.Schesser DC imperfections

• We saw that we have to provide DC voltages to an amplifier in order to provide it with the power to support amplification. – For a differential amplifier which must handle both positive and negative voltages • The process of designing this DC circuitry is called biasing • As a result, biasing currents flow through the amplifier which affects its performance. • In particular, a voltage during to the biasing will appear at the output without any input signal. • These extra voltage can be due to: – Bias currents flowing in the feedback circuitry – Bias current differentials – Voltage offsets due to the fact that the Op-amp circuitry is not ideal BME 372 Electronics I – 156 J.Schesser Special Amplifiers

• Summer (Homework Problem) • Instrumentation Amplifier – Uses 3 Op-amps – One as a differential amplifier – Two Non-inverting Amps using for providing gain

BME 372 Electronics I – 157 J.Schesser Medical Instrumentation Amplifier Non-inverting Amplifier

+ R5 -

R6

+ R2 - v2 --

R + 1 vo R3

R1 R4

R2

- + Differential Amplifier

+ v 1 -- Non-inverting Amplifier

BME 372 Electronics I – 158 J.Schesser Medical Instrumentation Amplifier Non-inverting Amplifier

+ R5 -

R6

+ v2D R2 - v2 --

R + 1 vo v1D R3

R1 R4

R2

- + Differential Amplifier

+ v 1 -- Non-inverting Amplifier

BME 372 Electronics I – 159 J.Schesser Medical Instrumentation Amplifier vvvv Differential Amplifier 2Dx xo RR R5 65 vx v 11 v 2D v ()o R RRRRx

v 6 6655 2D - v RR v

2D v ()56 o =0 + x vi RRRR v 6655 o R v R3 vvvvv4 1D vy yDxix1 RR34 vR RR v R4 24D 56 o v1D () RRR634 RRR 655 RRR R vv556()4 v R  R R oDRRRR12 D Chose 5 6 4 1 6534 R5 R3  R4 RR R Chose 56 4  1 R4 R5  R4 R6  R5 R3  R5 R4 RRR534

R4 R6  R5 R3 R5 vvvoDD(12 ) R R R 6  3 6 R R BME 372 Electronics I – 5 4 160 J.Schesser Medical Instrumentation Amplifier Non-inverting Amplifier

+ R5 -

R6

+ R2 - v2 --

R + 1 vo R3

R1 R4

R2

- + Differential Amplifier

+ v 1 -- Non-inverting Amplifier

BME 372 Electronics I – 161 J.Schesser Medical Instrumentation Amplifier Non-inverting Amplifier

+ v  v v  v 2 D 2  2 A - R R v2D 2 1 + R2 v2 -- 1 1 R2 v2 D  R2 (  )v2  vA R1 R2 R1 R1 v vA 1D R1  R2 R2 R1 v2 D  v2  vA R1 R1

R2

- Likewise + R1  R2 R2 v1D  v1  vA + v R1 R1 1 -- Non-inverting Amplifier

BME 372 Electronics I – 162 J.Schesser Medical Instrumentation Amplifier

Non-inverting Amplifier

+ R R 5 v  5 (v  v ) - o 1D 2 D R6

R6 + - v2 R2

-- R1  R2 R2 + v2 D  v2  vA v R R R1 o 1 1

R - 3 R1  R2 R2

v1D  v1  vA R1+ R1 R1 R4

R2 R5 R1  R2 R2 R1  R2 R2 vo  [ v1  vA  ( v2  vA )] - Differential Amplifier R6 R1 R1 R1 R1 + R5 R1  R2 R5 R2 + vo  ( )(v1  v2 )  (1 )(v1  v2 ) v1 R R R R -- Non-inverting Amplifier 6 1 6 1

BME 372 Electronics I – 163 J.Schesser Wheatstone Bridge as a sensor

R2

V R

v2D 1 -

rC rD +

vo R rB rA v1D 3

R4

BME 372 Electronics I – 164 J.Schesser Wheatstone Bridge

V V

rC rC rD rD

V2 V2 V1 rB r rA V rA B r r V1 C D Using Thevinin's Theorem on the Wheatstone Bridge Left side Right side V V2 1 rB rBrC rA rArD rB rA V2  V;r2  V1  V;r1  rB  rC rB  rC rA  rD rA  rD

+ r2 + r1 V2 V1 -- --

BME 372 Electronics I – 165 J.Schesser Wheatstone Bridge

V

rC rD

V2 V1 rB rA

rB rA VBridge  V2 V1  (  )V rB  rC rA  rD

rB rA rA rB When bridge is balanced   rBrD  rArC   rB  rC rA  rD rD rC

(nominally, rA  rB  rC  rD ) and VBridge  0

BME 372 Electronics I – 166 J.Schesser Wheatstone Bridge

V  v v  v 2 x  x o R2 R1  r2 R2 V 1 1 v 2  v (  )  o R  r x R  r R R

vx 1 2 1 2 2 2 - V R  R  r v 2  v ( 1 2 2 )  o R  r x (R  r )R R

R 1 2 1 2 2 2 + r 1 + 2 R v  v  4 V V2 vo x y R  r  R 1 + r R 3 1 4 -- 1 3 V R R  R  r v 2  4 V ( 1 2 2 )  o V1 1 R1  r2 R3  r1  R4 (R1  r2 )R2 R2 -- R V R R  R  r V v 4 2  4 ( 1 2 2 ) 1  o R1  r2 R3  r1  R4 R2 (R1  r2 ) R2

R2 R4 R1  R2  r2 vo  [( )( )V1 V2 ] R1  r2 R3  r1  R4 R2 R  R  r R Chose 1 2 2 4  1 R2 R3  r1  R4

R2 R2 rA rB vo  (V1 V2 )  (  )V R1  r2 R1  r2 rA  rD rB  rC

BME 372 Electronics I – 167 J.Schesser Wheatstone Bridge

R2

vx - R  R  r R Chose 1 2 2 4  1 R R  r  R

R 2 3 1 4 + r2 1 + R1  R2  r2 R3  r1  R4 V2 v  o R R + r R 2 4 -- 1 3 r r r r V1 B C A D R1  R2  R3   R4 rB  rC rA  rD -- R4  R2 R4

Hard to deal with; then let's make sure that r2  R1  R2 and r1  R3  R4

R1  R2 R3  R4   R2R3  R2R4  R4R1  R4 R2  R2R3  R4R1 R2 R4 R R 4  2 R3 R1

R2 rA rB vo  (  )V R1 rA  rD rB  rC

Note that the output depends on the unbalance of the bridge and when the bridge is balanced vo  0 R Gain is still 2 R1 BME 372 Electronics I – 168 J.Schesser Integrators and Differentiators

v (t) C i (t)  in  i (t) 1 R 2

i2 i 0 1 R 1 t 1 t vo   i2 (x)dx   vin (x)dx - C 0 RC 0 vi + vo vin

R

i2 0 i1 C Cdvin (t) i1 (t)   i2 (t) - dt vi + vo vin dv (t) v  i (t)R  RC in o 2 dt

BME 372 Electronics I – 169 J.Schesser Frequency Analysis

i1=vin/Z1 Z i 2 Vin ( j)  I1 ( j)Z1 ( j)  0 since vi is (virtually) zero 2 0  Z I1 ( j )  I2 ( j ) due to the summing - point constraint 1 -  vi  + v V ( j )  I ( j )Z  0 since v is (virtually) zero v o o 2 2 i in  Z2  - Vin ( j ) which is independent of ZL Z1 V (j ) Z  o  - 2 Vin ( j) Z1 C R i =v /Z 1 in 1 i =v /Z i2 0 1 in 1 R i2 C 0 - vi - + vi vo ZL + vin vo ZL vin   V ( j) Z V ( j) Z o  - 2   1 an integrator o  - 2   jRC a differeniator Vin ( j ) Z1 jRC Vin ( j ) Z1

BME 372 Electronics I – 170 J.Schesser Frequency Response

 V ( j) Z 1 o  - 2  

Vin ( j ) Z1 jRC i2 0 C i1 R

- vi + vo vin ω  V ( j) Z o  - 2   jRC i2 R 0 V ( j ) Z i1 C in 1

- vi + vo vin ω

BME 372 Electronics I – 171 J.Schesser Frequency Response

R V Z R 1 R 1 2 out  2  2  2   tan1(C R ) V Z R (1  jC R ) R 2 2 2 in 1 1 2 2 1 1 (C2R2 ) i 2 R1 51 C1 10mf R2 100k i 0 C2 1 R1 2,500 2,000 - vi 1,500 + 1,000 vo vin 500  0 1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 HZ V ZRjCRR  CR  out 22  112  11  tan1 (CR ) 2 11 VZRjCRRin 11(1 111 )1( CR ) 2 i2 R 11 0 2 i1 R1 51 C1 10mf R2 100k

- 2,500 v R C i 2,000 1 1 + vo 1,500 vin 1,000 500 0 1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 BME 372 Electronics I – HZ 172 J.Schesser Integrators and Differentiators • Integrators and Differentiators are used in analog computers • An analog computer solves a differential equations • By using Integrators and Differentiators one can “program” a particular differential equation to be solved • Usually only integrators are used since the gain of a differentiator occurs at high frequencies while the opposite is true for the integrator. • Since the frequency response of an real Op-amp attenuates high frequencies, using a differentiator conflicts with the characteristics with a real Op-amp • Also noise (high frequencies) are amplified by differentiators

BME 372 Electronics I – 173 J.Schesser R-Wave Detector

BME 372 Electronics I – 174 J.Schesser Positive Feedback • What happens to our Op-amp circuit if positive feedback is used? Using KCL at the input : R

i1  i2  iin  0 i2 iin iin  0; since Amplifier has infinite input impedance i1 R v  v v  v + i in  i o  0 vi - R R v RL v o in vo  AOLvi where AOL is the open loop gain 1 1 v  (v  v )  (v  A v ) i 2 in o 2 in OL i

• Even if vin = 0 and there is a slight voltage at vi then – vo = AOLvi will increase vi – vo will grow even larger – Eventually this will reach an extreme since there is not infinite energy in the circuit

BME 372 Electronics I – 175 J.Schesser Positive Feedback • Then our positive feedback design will operate between its positive and negative extremes, say ±5V R

i 1 2 vvv() iin iino i1 2 R + For vVvoi to be at  5 , 0 vi - R vo L 11 vin vvvv()(5)0 iinoin22

This is true as long as vVin  -5

Vo For vVvoi to be at  5 , 0 11 +5 vvvv()(5)0 iinoin22

-5 +5 Vin This is true as long as vVin  5

-5

BME 372 Electronics I – 176 J.Schesser Homework • Probs 2.2, 2.5, 2.6, 2.10, 2.22, 2.24, 2.25, 2.28 • Calculate and plot the output vs frequency

for these circuits. R1=1k, R2=3k, C=1μf Use

Matlab to perform the plot R2

v - C R2 R C i 1 + v - v o vi in R1 + v vo in R2

R1 R2

C - C vi + - vo vi vin R1 + vo vin BME 372 Electronics I – 177 J.Schesser Homework

V rC= R-ΔR rD=R+ΔR

V2 V1 rB = rA= R+ΔR R-ΔR

Wheatstone Bridge - Strain Gauge • A strain gauge shown here is used with a difference amplifier. Calculate the amplifier output signal as a function of ΔR and difference amplifier resistors. Assume that ΔR<

BME 372 Electronics I – 178 J.Schesser Homework Calculate and plot the output vs frequency for

this circuits. R1=1k, R2=3k, C=1μf. Use Matlab to perform the plot.

C

R1 R2 +

vi - C vo vin

BME 372 Electronics I – 179 J.Schesser