Op Amp Circuit

Op Amp Circuit

2102-487

Industrial Electronics

Amplifier Fundamentals

V_i R_i V_L

-

+ +- _AV

i

R_o + +- _V -

S

R_S + -

R_L

Source Amplifier Load

V_i +- _AV

i

R_i

R_o

V_o

-

+ +

-

Amplifier is a two-port device that accepts an external applied signal, referred to as input, an in turn produces a signal, referred to as output, that is proportional to the input: output = A x input

Voltage amplifier model Voltage amplifier with source and load

S s i

i i V

R R V R

= + _i

L o

L L AV

R R V R

= +

s i

i L

o L S

L

R R A R R R

R V

V

+

= +

|

|

| /

|V_L V_S < A

This shows that due to the loading effect

Loading effect complicates life because each time we change the source or the load , we need to recompute the overall gain, and we also have signal loss.

Amplifier Fundamentals

V_i +- _AV

i V_o

-

+ +

-

V_i V_L

-

+ +- _AV

i

+ +- _V -

S

R_s + -

R_L

Source Amplifier Load

Ideal voltage amplifier model Ideal Voltage amplifier with source and load

To eliminate loading effect, the voltage across R_sandR_omust be zero regardless of R_s and R_L. The only way to achieve this goal is b imposing R_i = ∞ and R_o =0

V A V

S

L = regardless of R_s and R_L.

In practice, the loading effect can be eliminated by the conditions of R_i >> R_s and and R_o << R_L

Operational Amplifier (Op Amp)

The operational amplifier (op amp) is a voltage amplifier having extremely high gain. By combining with external components, op amp could be configured to perform a variety of operations such as addition, subtraction, multiplication, integration, and differentiation etc.

Op amp symbol Simple practical Op amp model

For the popular 741 op amp, R_d = 2 MΩ, a ~ 200,000, and R_o = 75 Ω

V_p = Non-inverting input voltage, V_n = Inverting input voltage and V_o = output voltage a = unloaded voltage gain.

d n

p V V

V − = V_o = aV_d = a(V_p −V_n) V

+

_n

- +

V

-

_p

+

V

-

_o

+

-

+V_CC

-V_CC

V_n ^Rd

+-

aV_d

+ -

V_d

+ + -

- +

V

-

_p

+

V

-

_o

+V_CC

-V_CC R_o

Ideal Op Amp

We define ideal op amp as being an ideal voltage amplifier with infinite gain.

For the ideal op amp, a → ∞, imply V_p = V_n, R_d = ∞, imply I_n = I_p = 0 R_o = 0

I_p = Non-inverting input current, I_n = Inverting input current

Ideal practical Op amp model

Ideal Op Amp Rules:

1. No current flows in to either input terminal

2. There is no voltage difference between the two input terminals

aV_d

+- +

-

V_d

+ -

V+_n -

+

V-_p

+

V-_o

+V_CC

-V_CC

CC o

CC V V

V ≤ ≤

−

The output voltages are constrained by the following relationship

Operational Amplifier (Op Amp)

Op Amp Characteristic Property Values

∞

∞ 0

0 (virtual short) 0 (virtual short)

|V_o| ≤ V_CC

>200,000

>2 MΩ

<75Ω

<0.1 mV

<50 pA

|V_o| < V_CC Gain, a

Input Resistance, R_i Output Resistance, R_o

Input voltage difference, V_p-V_n Input current, i₁ or i₂

Output voltage limits

Ideal Op Amp Typical Op Amp

Property

Analysis of Op Amp Circuit:

-

V_p

+

V_n

V_o

+ -

V_i

i

R

_in

R

_out

V_n ^Ri +- aV_d +

- V_d + + -

- + V-_p

+ V-_o R_o

Apply KVL: −V_i +i(R_i + R_o)+a(V_p −V_n) = 0

i n p

R V i V − But: =

And we have: _{Vo =Vn}

+ -

+ -

R_i 2 MΩ

75 Ω R_o

a(V_p-V_n) V_n

V_p V_i

+ -

V_o

i o

i i

o

R a R

R V

V

) 1 1 (

+

− +

=

Voltage gain for the voltage follower

p

i V

V = and

Find V_o/V_i

Analysis of Op Amp Circuit:

i

Apply KVL: −V_i +i(R_i + R_o)+a(V_p −V_n) =0 But:

And we have: _in ⁱ _{a Ri Ro} i

R = V = (1+ ) + R_in

+ -

+ -

R_i 2 MΩ

75 Ω R_o

a(V_p-V_n) V_n

V_p V_i

+ -

V_o

Find Input resistance: ^{Rin = V}_iⁱ

n p

i V V

iR = −

Find output resistance:

=0

=

Vi

t t

out i

R V

+ -

R_i 2 MΩ

75 Ω R_o

a(V_p-V_n) V_n

V_p

+ -

^Vt

Simple equivalent circuit for finding R_out Equivalent circuit for finding R_in

i

_t

A

Apply KCL at A:

i

_t

) 0

( − =

+ − + −

−

o n p t

i p n

t R

V V A V R

V i V

But: V_n = 0 and V_p =V_t

t i

o

i

t o V

R R

R a

i = R +( +1)

And we have:

i o

i o t

t

out R a R

R R i

R V

) 1 ( +

= +

=

Analysis of Op Amp Circuit: Ideal Op Amp

aV_d

+- +

-

V_d

+ -

V_n

+ -

+

V-_p

+

V-_o

Using Rule 2: (no voltage difference between inverting and non-inverting inputs) Using Rule 1: (no current flows into the op Amp inputs)

i

_p

=0

= _n

p i

i V_p =V_i

o n

p V V

V = = V_o =V_i

R

_in

Input Impedance and Output Impedance 0

and

=

∞

= _out

in R

R

-

V_p

+

V_n

V_o

+ -

V_i

R

_out

Analysis of Op Amp Circuit:

Using the parameters of 741 op amp, R_i = 2 MΩ, a ~ 200,000, and R_o = 75 Ω 1

10 5

1− × ⁶ ≈

= ⁻

i o

V Voltage gain: V

Ω

×

= 400 10⁹ Rin

Input resistance:

Ω

=375µ Rout

Output resistance:

Comparison between Ideal and Practical Voltage Follower

Voltage Gain Input Resistance Output Resistance

Ideal Op Amp Typical Op Amp

Property

) 1 1

1 ( ≈

+

− +

i o

i

R a R

R 1

GΩ 400 )

1

( +a R_i +R_o = ^∞

Ω + =

+ 375µ

) 1

( _i

o

i o

R a R

R R

0

Inverting Amplifier

KCL

A

Use KCL at point A and apply Rule 1:

1

0

A in A out

f

v v v v

R R

− + − =

(no current flows into the inverting input)

Rearrange

1 1

1 1

0

in out A

f f

v v

v R R R R

   

+ − + =

   

   

   

Apply Rule 2:(no voltage difference between inverting and non-inverting inputs) Since V+ at zero volts, therefore V- is also at zero volts too. v

_A

= 0

1

0

in out f

v v

R + R =

1 out f

in

v R

v = − R

Input Impedance and Output Impedance R

_i

= R

₁

and R

_o

= 0

-

+ V_out

+ -

⁺

-

R₁

R_f

V_in

Non-inverting Amplifier

KCL

A

Use KCL at point A and apply Rule 1:

1

0

A out

A

f

v v v

R R

+ − =

Apply Rule 2: v

in

= v

A

1

1

^f

out in

v R

v = + R

Input Impedance and Output Impedance R

_i

= R

_∞

and R

_o

= 0

-

+

V_out

+ -

+

-

R₁

R_f

V_in

Basic Application of the Op Amp

-

+ V_o

+ -

⁺

-

R₁

R₂

V_i

-

+ V_o

+ -

+

-

R₁

R₂

V_i

Inverting amplifier Non-Inverting amplifier

= 0 R

o

R

1

R

_i

=

1 2

R A

_v

= − R

= 0 R

o

∞

i

= R

1

1

2

R

A

_v

= + R

Summing Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

3

1 2 ^{A A out}

0

A A

f

v v v v v v v v

R R R R

− −

− + − + + =

_ +

R

_f

i

₁

R

v

_out

+ v

₁

-

v

₂

v

₃

i

₂

i

₃

R R

i

v

_A

v

_B

1 2 3

i i = + + i i

Since v

_A

= 0 (Rule 2)

(

1 2 3

)

f out

v R v v v

= − R + +

Sum of v

₁

, v

₂

and v

₃

_ +

Difference Amplifier: Mathematic Operation

Use KCL and apply Rule 1:

R

₁

v

_out

+ v

₁

-

v

₂

R

₃

v

_A

v

_B

Since v

_A

= v

_B

(Rule 2) and

Substitute eq. (2) into eq. (1), we get R

₄

R

₂

(1)

(2)

If R

₁

= R

₂

= R and R

₃

= R

₄

= R

_f

(

2 1

)

f out

v R v v

= R −

Difference of v

₁

and v

₂

0

2 1

1

+ − =

−

R v v R

v

v

_{A A out}

2 4 3

4

v

R R v R

v

_{A B}





 





= +

=

1 1 2 2

1 2 4 3

4

1 v

R v R R

R R R

v

_out

R   −

 



 +

= +

Differentiator and Integrator: Mathematic Operation

_ +

_ +

Differentiator

Integrator R

v

_in

C

C R

v

_in

i

v

_out

+ -

v

_out

+ - i

i

i

v

_c

+ -

v

out

= − iR But i C dv

^C

= dt and v

ⁱⁿ

= v

^C

in out

v RC dv

= − dt

out C

v = − v

But

0

( ) 1 (0)

t

C C

v t idt v

= C ∫ + and v

ⁱⁿ

= iR

0

1 (0)

t

out in C

v v dt v

= − RC ∫ +

Difference Amplifier: Superposition

-

+ V_o

+ -

R₃

+ -

+

-

R₁ R₂

R₄ V₁

V₂

-

+ V_o

+ -

R₃

+

-

R₁ R₂

R₄

V₁

-

+ V_o

R₃

+ -

+

-

R₁ R₂

R₄ V₂

1 1 2V R V_o = − R

2 1 2 4

3

4 1 V

R R R

R

V_o R 

 



 +

= +

2 1 2 4

3 4 1

1

2 1 V

R R R

R V R R

V_o R 



 



 + + +

−

=

0

and

1

1

=

_o

=

i

R R

R R

_i2

= R

₃

+ R

₄

and R

_o

= 0

Commode and Differential Mode

Differential mode input: V

^dm

= V

²

− V

¹

Common mode input: V

^cm

= V

²

2 + V

¹

-

+ V_o

+ -

R₃(=R₁)

+ -

+

-

R₁ R₂

R₄(=R₂) V₁

V₂

-

+ V_o

- +

R₃(=R₁)

- +

₊

-

R₁ R₂

R₄(=R₂) V_cm

+ -

^Vdm/2

V_dm/2

Difference amplifier, in terms of the common and differential-mode inputs

Rearrange: V

₁

= V

_cm

− V

_dm

/ 2 V

₂

= V

_cm

+ V

_dm

/ 2

Commode and Differential Mode

Output for difference amplifier:

-

+ V_o

+ -

R₃(=R₁)

+ -

+

-

R₁ R₂

R₄(=R₂) V₁

V₂

Using superposition, the output from difference amp can be expressed as

cm cm dm

dm

o

A V A V

V = +

A_dm = amplification of differential input A_cm = amplification of common mode input

2 2 1

1

V A V

A V

_o

= +

Difference amplifier

(

_{1 2}

) (

¹_{2 2 1}

)

2

1 1

V V V V

V = + − −

(

_{1 2}

) (

¹_{2 2 1}

)

12

2

V V V V

V = + + −

(

_{2 1}

)(

_{2 1}

) (

_{2 1}

) (

₂¹ _{2 1}

)

12

A A V V A A V V

V

_o

= − − + + +

We have

Therefore in this case,

(

_{2 1}

)

2

1

A A

A

_dm

= −

and

A

_cm

= ( A

₁

+ A

₂

)

Commode mode rejection ratio:

cm dm

A A log

10

20

CMRR =

Quality index

Commode and Differential Mode

-

+ V_o

+ -

R₃(=R₁)

+ -

+

-

R₁ R₂

R₄(=R₂) V₁

V₂

The design gain of this amplifier is

Ex The difference amplifier is constructed with an ideal Op Amp and 1% tolerance resistors of nominal values 2.2 kΩ and 5.1 kΩ. The resistors were measured and fond to have the following resistance values:

318 . kΩ 2 2 . 2

kΩ 1 . 5

1

2

= =

= R A R

Difference amplifier

4 3 2

1

R R R

R =

However, this value is based on the assumption of equal resistor ratios:

R₁ = 2.195 kΩ, R₃ = 2.215 kΩ, R₂ = 5.145 kΩ, R₄ = 5.085 kΩ

Determine the gain of the differential amplifier and its CMRR

The more exaction expression for the output voltage

1 1 2 2

1 2 4 3

4

1 V

R V R

R R R R

V

_o

R   −

 



 +

= +

Commode and Differential Mode

 

 



 +

= + 1

1 2 4 3

4

2

R

R R R A R

2 2 1

1

V A V

A

V

_o

= +

_{Here and}

1 2

1

R

A = − R

329 . 2 195 1

. 2

145 . 5 085 . 5 215 . 2

085 . 5

2

 =



 



 +

= + A

344 . 195 2

. 2

145 . 5

1

= − = −

A

Therefore

01464 .

0 337 . 2

−

=

=

cm dm

A A

The CMRR is

CMRR = 20 log

₁₀

= 20 log − 159 . 6 = 44 . 06 dB

cm dm

A A

This is only a moderately good differential amplifier. If physical resistors used for R₁ and R₃ were exchanged, the resistor ratios in each gain path would be more nearly exact:

321 . 2 215 1

. 2

145 . 5 085 . 5 195 . 2

085 . 5

2

 =



 



 +

= + A

323 . 215 2

. 2

145 . 5

1

= − = −

A

Therefore

00186 .

0 322 . 2

−

=

=

cm dm

A A

dB 92 . 61 0

. 1248 log

20 log

20

CMRR =

₁₀

= − =

cm dm

A

A

Non-Ideal Op Amp

aV_d

+- -

R_o

+

r_d V_OS

- +

1/2I_OS I_bias

I_bias

Non-ideal characteristics

•Finite input resistance

•Finite voltage gain

•Nonzero output resistance

•Output Saturation

•Maximum output current

•Input offset voltage, V_OS

•Input bias current, I_bias

•Input offset current, I_OS

V_os - the difference in voltage between the Op Amp input terminals when the output voltage is zero

I_bias - the average of the two input currents when the output voltage is zero

I_OS - the difference between the input currents

(

_{p n}

)

bias

I I

I =

₂¹

+

p n

OS

I I

I = −

Non-Ideal Op Amp

Apply KCL at the inverting input:

p p

p

I R

V = − -

+ E_o

+

-

R₁ R₂

R_p

I_n

I_p

V

_p

V

_n

0 0

2 1

=

− +

− +

o n n

n

I

R E V

R V

From Ohm’s Law:

By Op Amp action V_n = V_p , Eliminating V_n and V_p

( )

[

n p p

]

o

R R I R I

R

E R   −

 



 +

=

_{1 2}

1

2

//

1

R_p can be specified to cancel the two terms in the brackets Estimating the output error caused by the

input bias currents

2 1 // R R

R_p = In this case, this reduces the output error to

( )

[

_OS

]

o

R R I

R

E R

_{1 2}

1

2

//

1  

 



 +

=

Non-Ideal Op Amp

V_SAT + +

- V_d

+ -

+ V-_o V_o V_SAT

+ +

- V_d

+ -

+ -

aV_d +- +

- V_d

+ -

+ V-_o

V

_o

(V)

V

_d

(µV) V

_SATH

/a

V

_SATL

/a

V

_SATL

V

_SAT

Linear region:

Upper saturation region:

Lower saturation region:

a

Non-Ideal Op Amp

Ex The 741 inverting amplifier is driven by a ±10 V peak to peak triangular wave.

Sketch and label V_i, V_o and V_n. If 741 is supplied with ±15 V and this maximum output ±13 V.

-

+ V_o

+ -

+

-

R₁ R₂

V_i 10 kΩ 20 kΩ

-6.5V < V_i < 6.5V: the op amp is in the linear region V_o =- 2V_i

V_i > 6.5V: the op amp is Saturation V_o = -13 V

V_i < -6.5V: the op amp is Saturation V_o = 13 V

V

_n

SATL i

n V

R R V R R R V R

2 1

1 2

1 2

+ +

= +

SATH i

n V

R R V R R R V R

2 1

1 2

1 2

+ +

= +

10 V 6.5V

V_i

-10 V -6.5V

t

V_o

t 13 V

-13 V

V_n

2.33 V

-2.33 V