BME Operational Amplifiers

BME 301

•  An amplifier produces an output signal with the same

wave shape as the input signal but usually with a

larger amplitude.

v

_o

(t)=A

_v

v

_i

(t)

•  A

_v

is called the voltage gain and if <0 then the

amplifier is inverting; otherwise non-inverting.

Basic Amplifier Types

Load R_L Amplifier A_v vo(t) v_i(t) + + - - Common Ground Node

Voltage-Amplifier Model

•  Circuit Parameters:

–  Starting from the left

•  v_s is the source input voltage and represents a microphone of an audio amplifier or the action potential of a muscle.

•  R_S is the resistance of the source voltage device •  v_i is the voltage to the input of the amplifier

•  i_i is the current flowing in the input of the amplifier •  R_i=v_i / i_i is the resistance at the input to the amplifier

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + - - + -

Voltage-Amplifier Model

•  Circuit Parameters:

–  Moving to the right side of the model

•  We see on the right part of the model a new icon called dependent voltage source.

–  It’s voltage which depends on a voltage at left side of the model, the input voltage at the amplifier

–  In particular, it’s the voltage across the input resistor R_i.

–  And the gain of the amplifier when nothing is connected to it’s output

•  Open Circuit Voltage Gain A_vo

•  R_o is the resistance at output of the amplifier

•  V_o is the voltage at the output of the ampilier

•  i₀ is the current flowing in the output of the amplifier

•  R_L=v_o / i_o represents the device connect to the output of the amplifier.

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + +

Voltage-Amplifier Model

•  Performance parameters:

–  Voltage Gain A

_v

=v

_o

/ v

_i

from the input of the amplifier to the

output of the circuit

–  Voltage Gain A

_vs

=v

_o

/ v

_s

from the source of the amplifier

model to the output of the circuit

–  Current Gain A

_i

=i

_o

/ i

_i

–  Power Gain G=P

_o

/ P

_i

from the input of the amplifier to the

output of the circuit

–  Power Gain G=P

_o

/ P

_s

from the source of the amplifier

model to the output of the circuit

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + - - + -

Ideal Amplifiers

•  Some calculations:

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + - - + - Starting from the output (right side of the model) and using voltage division:

v_o = RL R_L + R_O Avovi Or A_v = vo v_i = R_L R_L + R_O Avo Now

looking at the left part of the model and using voltage division:

v_i = Ri

R_i + R_s vs

Combining the two:

v_o = RL R_L + R_O Avo R_i R_i + R_s vs Or A_vs = vo v_s = R_i R_i + R_s Avo R_L R_L + R_O

Ideal Amplifiers

•  Some calculations:

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + - - + - Starting from the output (right side of the model) and using voltage division:

i_O = vo R_L = 1 RL R_L R_L + R_O Avovi = 1 R_L + R_O Avovi substituting for vo i_O = 1 R_L + R_O Avovi = 1 R_L + R_O AvoiiRi = R_i R_L + R_O Avoii substituting for vi Or A_i = io i_i = R_i R_L + R_O Avo

Now the power Gain

P_i = voiO v_ii_i = v_o v_i i_O iii = A_vA_i P_i = voiO v_si_i = v_o v_s i_O iii = A_vsA_i

Ideal Amplifiers

•  Performance Parameters:

v_{s v} o v_i R_o R_i R_s R_L A_vov_i i_o i_i + + - - + -

Going back to the gain from the source to the output

v_o v_s = R_i R_i + R_s Avo R_L R_L + R_O

This states the gain of the amplifier depends on the external components. This is BAD!!!!!

However, if R_i → ∞ and R_O → 0; then Ri

R_i + R_s → 1 and

R_L

R_L + R_O → 1.

Therefore, vo

Another Amplifier

The Differential Amplifier

•  Output of the amplifier is a function of the difference of the inputs: v_o=A_d(v_i1 - v_i2)

•  However, most real Differential Amplifier are affected by the average of the input and we define the common mode input signal

v_icm =1/2(v_i1 + v_i2)

•  Therefore, v_o=A_d(v_i1-v_i2)+A_m v_icm and the ratio of A_d to A_m is called the common mode rejection ratio

v_o v_id=v₁-v₂ v₁ A_vov_i v₂ v_o v₁ v₂ + - Inverting input Non-Inverting input Model Icon

Operational Amplifiers

•  An operational Amplifier is an ideal

differential amplifier with the following

characteristics:

– Infinite input impedance, R

_i

is infinite

– Zero output impedance, R

_o

is zero

– Infinite gain for the differential signal, A

_d

is

infinite

– Zero gain for the common-mode signal

– Infinite Bandwidth

v_o

-

+

v₁

v₂

Inverting input Non-Inverting input

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.

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, etc.to solve for the

Example: Inverting Amplifier

1.  Verify Negative Feedback: Note that a portion of v_o is fed back via R₂ to the

inverting input. So if v_i increases and,

therefore, increases v_o, the portion of v_o fed

back will then have the affect of reducing v_i (i.e., negative feedback).

2.  Use the summing point constraint.

3.  Use KVL at the inverting input node for both the branch connected to the source and the branch connected to the output

inverted) : input the to opposite is output that the (note of t independen is which -zero is since 0 constraint point -summing the to due constraint point -summing the to due zero is since 0 1 2 2 2 0 2 1 1 1 L in i i in R v R R v R i v i i v R i v = + − = = + = R_L v_in R2 R₁ v_i

v_o -

+

0 i₁=v_in/R₁ i₂ -

+

+

+

-

-

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.

Non-inverting Amp

1.  First check: negative feedback?

2.  Next apply, summing point constraint 3.  Use circuit analysis

+ -- vo + -- R_L R₂ R₁ + -- v_i + -- v_f i_o I_in= 0 + - v_in v_in = v_i + v_f = 0 + v_f = v_f v_f = R1 R₁ + R₂ vo = vin; A_v = vo v_in = R₂ + R₁ R₁ = 1+ R₂ R₁ Note: 1.  The gain is always greater than one 2.  The output has the same sign as the input

Medical Instrumentation Amplifier

Differential Amplifier

+ -- R₂ R₁ + - v₂ + -- R₂ R₁ + - v₁ R₃ R₄ + - v_o R₅ R₆

Non-inverting Amplifier

Non-inverting Amplifier

Medical Instrumentation Amplifier

v

_o

=

R

5

R

₆

(1

+

R

₂

R

₁

)(v

1

− v

2

)

+ -- R2 R₁ + - v₂ + -- R₂ R₁+ - v₁ R₃ R₄ v_o R₅ R₆ Differential Amplifier Non-inverting Amplifier Non-inverting Amplifier + - - +

Integrators and Differentiators

∫ ∫ =− − = = = t in t o in dx x v RC dx x i C v t i R t v t i 0 0 2 2 1 ) ( 1 ) ( 1 ) ( ) ( ) ( v_in

C R v_i v_o

-

+

0 i₂ i₁ v_in

R C v_i v_o

-

+

0 i₂ i₁ dt t dv RC R t i v t i dt t Cdv t i in o in ) ( ) ( ) ( ) ( ) ( 2 2 1 − = − = = =

Frequency Analysis

v_in vi v_o -

+ 0 i1=vin/Z1 i2 1 in o L in 1 o 1 1 1 in Z Z V V Z V Z Z Z I V I I Z I V 2 2 2 2 2 ) ( ) ( of t independen is which ) ( -zero ) (virtually is since 0 ) ( ) ( constraint point -summing the to due ) ( ) ( zero ) (virtually is since 0 ) ( ) ( ) ( = = + − = = + = ω ω ω ω ω ω ω ω ω ω j j j v j j j j v j j j i i Z_L v_in C R vi vo -

+ 0 i1=vin/Z1 _i 2 Z_L v_in R C vi vo -

+ 0 i1=vin/Z1 _i 2 integrator an 1 ) ( ) ( ₂ RC j j j ω ωω = 1 = − in o Z Z V V tor differenia a ) ( ) ( ₂ RC j j j _ω ω ω _{= = −} 1 in o Z Z V V Z2 Z1

21

Frequency Response

v_in

C₂ R₁ v_i v_o

-

+

0 i₂ i₁ v_in

R₂ C₁ vi v_o

-

+

0 i₂ i₁ R₂ V_out V_in = − Z₂ Z₁ = − R₂ R₁ 1 (1+ jωC₂R₂)= R₂ R₁ 1 1+ (ωC₂R₂)2 ∠π− tan −1_(ωC 2R2) 0 500 1,000 1,500 2,000 2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08

HZ R1 51 C1 10mf R2 100k R₁ 1 2 2 1 1 2 1 1 1 1 2 1 1 1 1 1 _{1 1} tan ( ) (1 ) _{1 ( )} 2 out in V Z R j C R R C R C R V Z R j C R R _{C R} ω ω π _ω ω _ω − = − = − = ∠ − − + ₊ 0 500 1,000 1,500 2,000 2,500

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08

HZ

Homework

1.  What is the summing point constraint?

2.  Calculate the gain for this amplifier (in terms

of R1, R2, and R3.

R_L v_in R₂ R1 v_i

v_o -

+

i₁ i₂ -

+

+

+

-

-

RP R₃

Homework

3.  Calculate and plot the voltage gain of the

following circuit as function of frequency, ω.

+ -- vo + -- R_L 1 + -- v_i + - v_s 1 1