• Caused by the unbalanced Q1

(1)

Output Offset Voltage

• Non-zero V_out even though two inputs are grounded

• Caused by the unbalanced Q₁ and Q₂

Input Offset Voltage

• Application of a finite input offset voltage (V_io) can nullify the output

Offset Null Adjust

• The offset null input pins are used to eliminate the output offset voltage

(2)

Common Mode Rejection Ratio (CMRR)

• Common-mode signals ~ identical signals applied simultaneously to the two inputs of an op-amp

• The ideal op-amp is expected to produce zero output for the common-mode signal inputs

• How to measure the ability of an op-amp to “ignore” common-mode signals ~ Common Mode Rejection Ratio (CMRR)

• CMRR ~ ability to reduce unwanted external noisy signals

10

(differential)

20 log [dB]

(common mod

d

)

an

e

v v

CMRR A CMRR

 A 

(3)

Slew Rate

• The ability to measure how fast the output signal responds to the input signal

• Typically 0.5 V/ms for #KA741

• For example,

• The maximum operation frequency is determined by [Ex.15-5]

An Op-amp operates with the maximum output voltage of 10 V_pk. When its slew rate is given by 0.4 V/ms, determine the maximum operation frequency of the Op-amp.

The output goes from -10 V to +10 V in 25 ms.

20 V 0.8 V/

Slew ra

25 s te = V

^out

t   s



 m

m

max

Slew rate 2

_pk

f  V



max

Slew rate 0.4V/

6.37 kHz 2 _pk 2 (10 _pk)

f s

V V

  m 

 

(4)

Ideal OP Amp

An ideal amplifier should have

• an infinitely high input impedance ~ I_in = 0

• an output impedance of zero ohms

• an infinitely high gain ~ V_diff = 0 with negative feedback

• an infinitely wide bandwidth

Op Amp Golden Rules Rule 1. The Voltage Rule :

The op amp output will change as necessary to keep the two input voltages identical.

If any input signal or voltage tries to make the input potentials different, the op amp output will change in the opposite polarity to the input and, via the feedback loop, keep the difference between the two inputs at 0V.

Rule 2. The Current Rule:

Because the input impedance is infinitely high, no current can flow into either input.

(5)

Table 1: The Ideal Amplifier vs The Practical Op Amp

Parameter The Ideal Amplifier 741 TLC271 LMC660

Input Impedance

(Z_IN) infinity. 2 MΩ 1 TΩ >1 TΩ

Input Bias Current

(I_IN) Zero. 80 nA 60 pA 0.002 pA

Large Signal Volta

ge Gain (A_V) Infinity 316 to 200,000 (50 dB to 106dB)

5,000 to 46,000 (7 4dB to 93dB) mini

mum

40,000 to 990,000 (92dB to 126dB) Output Impedance

(Z_OUT) 0Ω Depends on gain and feedback but typically less than 100Ω to more than 1KΩ

Ideal vs. Practical OP Amp

(6)

Inverting Amplifier (1)

• Assume ideal case of V₁ = V₂ or V_diff = 0

• Remember A_OL = infinite & Z_in = infinite ~ no current flow between two inputs

• If non-inverting input is grounded (V₊ = 0), V_- = 0 ← “virtual ground”

0 ( )

,

Since , ~ minus sign for the180 phase difference

Amplifier

f

CL

in out

in f

f f out

in f v

in in in

I in

CL CM

V V

R A R

i i

R R

i

A

V R i i A

C RR A

V

M

i R

  

 



  

 

+

̶

(7)

Inverting Amplifier (2)

( )

( ) ( )

( )

( ) ( )

Using Miller's theorem, ,

1 1

|| ||

1 Since 1, , / ,

||

f OL

in M out M f

OL OL

f

in I in in M in in in

OL

f

OL in M in in f OL in

OL

OL f out

in I

I out M out

OL

in

R A

R R R

A A

Z R R Z R R Z

A

A R Z R R A R R

A Z R Z A R

A

Z R

 

       

 

        

     

 

( )

1 ||

Since 1, , ||

out

OL f out out I f out out

Z

A R Z

Z

R Z

Z



   

R_in(M)

R_out(M)

(8)

Ex. 15-7

Q. Conduct a circuit analysis for the inverting amplifier.

( ) ( )

max

100 10

10 10

80 10 10, 000 0.001

Since 1 and 10, (10)(1 ) 10

Slew rate 0.5 500 2 2 (5 ) 31.4 15.9

f in

in out

CL CM

in pp CL CL

CL

in I out I

out in pp pp

pk pk

A Z Z CM

R k

Z A A

V V A A V V V

V s kH

R

z kHz

V R

V

V f

   



  

  

    

  m  

 

(9)

Noninverting Amplifier (1)

• The input is applied to the non-inverting input

• The input and output signals are now in phase

• Negative feedback resistor (R_f) together with R_in set the closed loop gain of the amplifier

• The circuit does not have a virtual ground point, but the bottom end of R_in is connected to ground

( )

Voltage division rule, Voltage difference,

Since ( ), ( ) (1 )

1

in

out out

f in

diff in

out OL diff OL in out OL in out out OL OL in

f CL

out OL OL

in OL OL

NI

in

V R V KV

R R

V V V V V

V A V A V V V A V KV V A K A V

V A A

V A K A K K

A R

R



  



 



   

       

    



( ) ( )

Generally, Z

_{in NI}

 Z

_in

~ 1 M  and Z

_{out NI}

 Z

_out

~ 100 

(10)

Noninverting Amplifier (2)

• The input is applied to the non-inverting input

• The input and output signals are now in phase

• Negative feedback resistor (R_f) together with R_in set the closed loop gain of the amplifier

• The circuit does not have a virtual ground point, but the bottom end of R_in is connected to ground

( )

Since ,

Note that (noninvertingamp gain) (inverting amp gain )

1

in in

in in in

out in

out in in f in f f

f in f in in

out

in in

f CL NI

in in

V V V V V V

V i R

V V

R

i R and i i i

R i R i R

V

A

i R

V

R





 

 





    

   

 



(11)

Ex. 15-8

Q. Conduct a circuit analysis for the noninverting amplifier.

( )

max

1 11 1

80 11 11, 000 0.001

(11)(1 ) 11

Slew rate 0.5 500

2 34.6 34.6 14.5

out f

in in

in out

CL CM CL in

pk CL NI

out

V R

Z M

Z

A A A

A

CMRR

V

V Vpp Vpp

V s kHz

V V kHz

f

   

 

 

  

  m  



Inverting Noninverting

A_CL 10 11

Z_in 1 k > 1 M

Z_out < 80  < 80 

CMRR 10,000 11,000

f_max 15.9 kHz 14.5 kHz

(12)

(13)

Voltage Follower

• Remove R_in and R_f from the noninverting amplifier

• Output is applied to the inverting input

• Very high Z_in and low Z_out

• Input and output voltages are in-phase

• A_v = 1 ~ unit voltage gain

( )

max

1 0 1

1 Slew rate

1

2

f in

C CL V

L

CM CM

CL in in

p F

k out

R R A

A A

A

CMR

A V V

V R

V f

     



 

 

(14)

Ex. 15-9

Q. Conduct a circuit analysis for the voltage follower.

max

1 0 1 1

1 80

1 1

1, 000 0.001

Slew rate 0.5 500 2 2 (3 ) 18.8 26.6

f CL

in

in in OP

out out OP

CL

CM CM

out CL in in

pk pk

A R

R

Z Z M

Z Z

CMRR A

A A

V A V V

V s kHz

f kHz

V V



     



  

   

 

  m  

 

(15)

OP-amp frequency response

• OP amp does not have lower cutoff frequency (f_c,low = 0 Hz) ~ dc amplifier

• Higher cutoff frequency (f_c) is located at A_V = -3 dB

• -20 dB/decade rate abouve f_c

• Decreasing voltage gain increases maximum operation frequency, or vice versa ~ “trade-off”

• Unit gain frequency (f_unity) ~ A_CL = 0 dB or A_V = 1

• Gain bandwidth product is constant to be

CL C unity

A f  f

10 100 000 10 1

100 10 000 100 1

1 1 000 1 1

CL C CL C CL C

Hz, A f ( , )( Hz ) MHz Hz, A f ( , )( Hz ) MHz kHz, A f ( , )( kHz ) MHz

 

(16)

Ex. 15.10

Gain bandwidth product of LM318 Op amp is given by 15 MHz. When A_CL = 500, determine the bandwidth. What is the maximum A_CL with f_c = 200 kHz.

15 30 and 30

500

When 200 15 75

200

unity

C C

CL

unity

C CL

C

f MHz

f kHz BW f kHz

A

f MHz

f kHz, A

f kHz

    

   

Ex. 15.11

We want to design an Op amp with A_CL = 500 and BW = 80 kHz. Can it be done by using LM318 ?

500 80 40

The answer is No!

CL C

A f ( )( kHz ) MHz

(17)

HW #7 Due = 12. 6. (Friday) 15:00