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Op amp problems and solutions, Exercises of Analog Electronics

Madurai Kamaraj University Analog Electronics

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Typology: Exercises

2023/2024

Uploaded on 02/27/2024

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EXERCISE IDEAL OP AMP ANALYSIS

Ideal Op Amp Exercise Rev. 1/6/2003 C. Sauriol Page 1

EXERCISE IDEAL OP AMP ANALYSIS

No.1 Assuming ideal op amps, determine Vo for each and every circuit shown below.

2k

10k

1k

30k

1k

2k

10k

2k 24k

3k

2K

10K

14k

16k

15k

10k

3.9K

1,5k

3k

5,1k

2k

-2 V

Vo

1,8k

Vo

+0,1V

Vo

-0,5V

Vo

+1V

-1,5V

Vo

+15V

-15V

Vo

-2 V

+5V

+8V

A) B)

C) D)

E) F)

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EXERCISE IDEAL OP AMP ANALYSIS

No.1 Assuming ideal op amps, determine Vo for each and every circuit shown below.

2k

10k

1k

30k

1k

2k

10k 2k 24k

3k

2K

10K

14k

16k

15k

10k

3.9K

1,5k

3k

5,1k

2k

-2V

Vo

1,8k

Vo

+0,1V

Vo

-0,5V

Vo

+1V

-1,5V

Vo

+15V

-15V

Vo

-2V

+5V

+8V

A ) B)

C) D)

E) F)

No.2 Assume typical op amp data for circuits A through E and worst case values for circuit F.

Op amp parameters for VSUP =±15V minimum typical maximum

O/P voltage swing ±12V ±13,5V -

I/P voltage range ±11V ±12,5V -

Short circuit current ±12 mA ±20 mA -

30K

4,7k

3,3k

1,8k

10k

V x

6 V

1 V
- 8 V

V o

1 5 V

1 5 V

2 V
3 V

10k 20k

6,8k V^ o

1 6 V

1 6 V

Vi n

1K 2k

680 V o

1 6 V

1 6 V

2 0 0

2 V

10K

6,8k (^) Vo

1 0 K

1 2 V

1 2 V

5 V

10k 100k

10k (^) V o

1 6 V

1 6 V

V 1

V 2

R 3 1 0 0 K

R 3

A) Determine V x

B) Determine V (^) o for V (^) in =+6V and V (^) in =-6V.

C) Determine V (^) o.

D) Determine V (^) o.

E) Determine V (^) o.

F) Determine the maximum value of R 3 if we do not want to saturate the inputs of the op amp given that V and V2 range from 80V to100V.

SOLUTIONS

+ 10V -

2k

10k

1k

30k

1k

2k

10k 2k 24k

3k

2K

10K

14k

16k

15k

10k

3.9K

1,5k

3k

5,1k

2k

- 2 V

Vo

1 , 8 k

Vo

+0 , 1 V

Vo

- 0 , 5 V

Vo

+1 V

- 1 , 5 V

Vo

+1 5 V

- 1 5 V

Vo

- 2 V

+5 V

+8 V

B)

C) D)

E) F)

0 A

1 mA

1 mA 2 V

- 2 V

- 1 2 V

0 A

0 V

0 , 1 mA

0 , 1 mA 0 A 0 A

+ 3 V -

- 3 V

0 A

- 0 , 5 V

0 , 2 5 mA

mA

+ 2 , 5 V -

- 3 V

0 , 1 2 5 mA

0 , 3 7 5 mA

1 , 1 2 5 V

- 4 , 1 2 5 V

- 1 , 2 5 V

0 , 1 2 5 mA

+ 2 , 2 5 V -

mA

+ 1 1 , 2 5 V -

- 1 2 , 5 V

0 A

1 mA

1 6 V

1 4 V

+1 V

0 , 1 3 3 mA

mA

0 , 1 3 3 mA

- 1 , 3 3 3 V +

+2 , 3 3 3 V

1 , 6 mA

+3 , 2 V

+ 1 , 8 V -

- 5 , 2 V + - 0 , 6 8 V +

+3 , 8 8 V

1 , 2 mA 1 , 2 mA

mA (^) 0 , 1 3 3 3 mA

No.1 A )

4,7k

3,3k

1,8k

10k

Vx

- 3 .6 V

+6 V

- 1 V - 8 V

- 1 V

0 A

mA

0 , 7 8 9 mA

mA

- 2 .6 V +

+ 7 V -

No.2 A)

10k 20k

6,8k

Vo

+1 6 V

- 1 6 V

Vi n

0 A

+6 V 0 A

+6 V

- 1 4 , 5 V

mA

- 4 , 8 3 3 V

B)

10k 20k

6,8k

Vo

+1 6 V

- 1 6 V

Vi n

0 A

mA

- 6 V

+1 4 , 5 V

+4 , 8 3 3 V

Vo

+1 5 V

- 1 5 V

+2 V

+3 V

0 A

+3 V

- 1 3 .5 V

C)

No feedback, the output saturates with

a polarity determined by the sign of the

differential I/P voltage:

Vo = A d (V + -V - ) = (2-3) = -

Vo = -V sat = -13,5V

Positive feedback will make the output

saturate with a polarity determined by

the sign of the differential I/P voltage.

With +6V applied to the -ve I/P of the

op amp, the O/P should be of the opposite

polarity, therefore assume

Vo = -V sat = -14,5V and determine the

V+ and verify the sign of the differential

I/P voltage in order to validate your

assumption of V o = -V sat, that is:

Vo = A d (V + -V - ) = (-4,83-6) = -

therefore the assumption was valid.

Same procedure here, except now V -^ is

negative, therefore the O/P polarity is

expected to be positive:

Vo = +V sat = +14,5V

Verify assumption with sign of V d = (V + -V - )

Vo = A d (V + -V - ) = (+4,83-(-6)) = +

No.

If F 〉

10 2 π RF C (^) F

=

10 2 π 100 k × 0,1 μ

= 159 Hz then ∆ V (^) o ( PP ) = − (^) R^1

( ) E C F

Vin ( AC ) dt t 1

t 2

A)

V ( ave ) =^ V

(^) PW

( ) T +^ V^

− (^) SW

( ) T

Vin(DC) = +0,2V

V T

V area

V V dt

o PP

t

o PP in AC

1000 0. 8 0. 6

1000

( )

( ) ( )

2

1

∆ =− × ×

∆ =− ×

Integral does not apply for 50

Hz and 100 Hz.

0 , 6 T 0 , 4 T

-2V DC

Vin

+1V

-1V

Vout

∆V out = 0,48Vpp at 1 kHz

and 48 mVpp at 10 kHz

+0,2V

DC

AREA

AREA

B)

P PP

o PP

t

o PP in AC

V F

V

T

V area

V V dt

1 2

1000 1 2

1000

( )

( ) ( )

2

1

= × × = =

∆ =− ×

F = 500 Hz

F > 159 Hz integral OK.

2 V (^) pp DC level

T 2

AREA

I N P U T

1 V (^) pp DC level

O U T P U T

C)

F Hz

V F

T

V area

V V dt

PP

o PP

t

o PP in AC

375

1

375 4

3 1000 0 , 5

1000

( )

( ) ( )

2

1

=

= = 

  

 × ×

∆ =− × =

F > 159 Hz integral OK.

2 V DC level pp 3 T 4

I N P U T

1 V (^) pp DC level

AREA A R E A O / P

+1V

-1V

V (^) ( ave ) = V

(^) PW

( ) T +^ V^

− (^) SW

( ) T Vin(DC)^ = +0,5V

No.3 D)

∆ Vo ( PP ) = − 1000 Vin ( AC ) dt t 1

t (^) 2

∫

∆ Vo ( PP ) = − 1000 × area

= 1000 ×

100 μ × 5 2

 

 

∆ Vo ( PP ) = 0,25 VPP

O/P is a parabolic wave, not a

sine wave.

DC level

10 VPP

0. VPP

100 μμμμ s I N P U T O U T P U T

No.4 A)

V (^) o = − RF CE

dVin dt

V (^) o = − 10 K × 0,1μ ±

10 V 2 ms

 

 

V (^) o = m 5 V (^) P^2 ms

10 V (^) pp

-5V p

+5V p

I

N

P

U

T

10 V (^) pp

O

P

+5V p

-5V p

B) On edges we have:

V o = − 10 K ×^ 0,1μ^ (±∞ )

V (^) o =^ m∞^ ⇒^ m^ V (^) sat

On flat portions we have:

V o = − 10 K × 0,1μ (± 0 )

V (^) o = 0

The O/P spikes settle down

in 5τ = 5RE C E = 50 μs

2 ms

1 V p

+1Vp

0V

50 μs

- V sat

+ V sat

I N P U T O U T P U T

C) To stabilise negative feedback in order to avoid self oscillations from the circuit.

No. 5 Phase shifter C o

X P ATAN 180 0 1 0 2 = 

  

 = Φ= ×

( )

( ) ( )

nF nF std FC TAN k TAN

P

P FCP

X TAN P

X ATAN P

X ATAN

o o

o C o C o C

026 9. 1 2 1000 100 10

1 2 10

1

2

1 20 2 10 10

1

1 1 1 1

= ⇒ × ×

= ×

=

 ⇒ = = 

  

  ⇒ = 

  

 Φ = = ×

π π

π

Op amp problems and solutions, Exercises of Analog Electronics

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Partial preview of the text

Download Op amp problems and solutions and more Exercises Analog Electronics in PDF only on Docsity!

EXERCISE IDEAL OP AMP ANALYSIS

No.1 Assuming ideal op amps, determine Vo for each and every circuit shown below.

2K

2K

10K

10K

3.9K

-2V

+0,1V

-0,5V

+1V

-1,5V

+15V

-15V

-2V

+5V

+8V

A ) B)

C) D)

E) F)

No.2 Assume typical op amp data for circuits A through E and worst case values for circuit F.

Op amp parameters for VSUP =±15V minimum typical maximum

O/P voltage swing ±12V ±13,5V -

I/P voltage range ±11V ±12,5V -

Short circuit current ±12 mA ±20 mA -

30K

SOLUTIONS

+ 10V -

2K

2K

10K

10K

3.9K

- 2 V

+0 , 1 V

- 0 , 5 V

+1 V

- 1 , 5 V

+1 5 V

- 1 5 V

- 2 V

+5 V

+8 V

B)

C) D)

E) F)

0 A

- 2 V

- 2 V

- 1 2 V

0 A

0 V

0 V

+ 3 V -

- 3 V

0 A

0 A

- 0 , 5 V

- 0 , 5 V

+ 2 , 5 V -

- 3 V

1 , 1 2 5 V

- 4 , 1 2 5 V

- 1 , 2 5 V

- 1 , 2 5 V

+ 2 , 2 5 V -

+ 1 1 , 2 5 V -

- 1 2 , 5 V

0 A

0 A

0 A

0 A

0 A

0 A

1 4 V

+1 V