EXERCISE
IDEAL OP AMP ANALYSIS
EXERCISE
No.1
IDEAL OP AMP ANALYSIS
Assuming ideal op amps, determine Vo for each and every circuit shown below.
A)
B)
1,8k
30k
-2V
Vo
1k
+0,1V
Vo
10k
2k
1k
C)
D)
2k
2K
-0,5V
10K
+1V
Vo
3k
2K
Vo
-1,5V
10k
2k
24k
10K
F)
E)
+15V
3.9K
10k
14k
5,1k
-2V
15k
1,5k
+5V
Vo
3k
Vo
+8V
16k
2k
15k
-15V
Ideal Op Amp Exercise
Rev. 1/6/2003
C. Sauriol
Page 1
EXERCISE
No.2
IDEAL OP AMP ANALYSIS
Assume typical op amp data for circuits A through E and worst case values for circuit F.
Op amp parameters for VSUP=±15V
O/P voltage swing
I/P voltage range
Short circuit current
A)
Determine V
minimum
±12V
±11V
±12 mA
.
typical
±13,5V
±12,5V
±20 mA
D)
x
4,7k
maximum
-
Determine V o.
10k
1K
2k
+6V
Vx
+1 6 V
3,3k
- 8V
1,8k
- 1V
+2 V
680
Vo
200
- 16V
B)
E)
Determine V o for V in =+6V and V in =-6V.
10k
Determine V o.
10K
30K
+5V
20k
+1 2V
+1 6V
6,8k
Vo
Vo
6,8k
Vi n
10K
- 12V
- 16V
C)
F)
Determine V o.
Determine the maximum value of R3
if we do not
want to saturate the inputs of the op amp given that V1
and V2 range from 80V to100V .
+1 5V
10k
100k
V1
+2 V
+16V
Vo
R3
+3 V
10k
Vo
V2
- 15V
R3
100K
- 16V
Ideal Op Amp Exercise
Rev. 1/6/2003
C. Sauriol
Page 2
EXERCISE
IDEAL OP AMP ANALYSIS
No.3
100k
t2
( )∫ V
∆Vo (PP) = Vo (t2 ) − Vo (t1 ) = −
Vin (ave) = Vin+
1
RE C F
( )+ V ( )
PW
T
− SW
in
T
A)
Draw
the
output
waveform with respect to Vin
shown for frequencies of 50
Hz, 100 Hz, 1 kHz and 10 kHz
- label waveforms with AC and
DC values as well as PW and
SW.
in (AC )
dt
t1
if
ω
0, 1 µF
10k
10
〉
RF CF
Vi n
for a squarewave
Vo
+1V
Vin
-1V
0,6T
0,4T
B)
1 Vpp
If Vin is a 2 Vpp squarewave with a 50% duty cycle, calculate the frequency of Vin that will produce Vo =
C)
Repeat step B for 75% duty cycle.
D)
If Vin is a 10 Vpp triangular wave with a frequency of 5 kHz, draw the expected O/P waveform with
respect to Vin.
No.4 A) Determine the output waveform
relative to an input triangular wave with a
10 VPP amplitude and a frequency of 250
Hz.
0,1 µF
10k
100
Vin
B)
Determine the output waveform
relative to an input square wave with a 2
VPP amplitude and a frequency of 250 Hz.
Vo = −RF CE
C)
What is the function of the 100Ω
resistor?
if
PHASE SHIFTER
AVF =
P1 + jX C
P1 − jX C
/ AVF = 2arctan
Vo
ω 〈 R0,1C
E
No.5
dVin
dt
E
R1
R1
AVF = 1
XC
P1
C
Vo
Design the circuit in order to
o
obtain a phaseshift of 20 to
o
180 with a 100K pot (P1) at a
frequency of 1 kHz.
Ideal Op Amp Exercise
Vin
Rev. 1/6/2003
P1
C. Sauriol
Page 3
EXERCISE
IDEAL OP AMP ANALYSIS
SOLUTIONS
- 2V
No.1 A )
B)
0A 1, 8k
0 , 1 mA
- 2V
0V
Vo
- 2V
2V
+
+ 3V -
0 , 1 mA
- 12V
0A
1k
0A
+0 , 1 V
+ 10V -
Vo
1 mA 10k
2k
- 3V
0A
1k
0V
1 mA
C)
D)
- 0, 5V
0A
30k
2k
1, 125
2K
mA
- 4, 125V
- 0, 5V
0A
3k
+ 2, 5V -
Vo
1, 125V
+
0 , 3 7 5 mA
+1 V
- 3V
- 1, 5V
- 0, 5V
0, 25
mA
10k
2k
- 1, 25V
1, 125
mA 10K
+ 2, 25V -
+ 11, 25V 0A
2K
Vo
- 12, 5V
0 , 1 2 5 mA
0A
0 , 1 2 5 mA
24k
10K
- 1, 25V
0 , 2 5 mA
0 , 1 2 5 mA
+1 5 V
E)
14k
1 mA
+
14V
-
F)
+1 V
15k
10k
0 , 1 3 3 mA
1 , 2 mA
16k
1, 066
mA
1,5k
+ 1, 8V -
Vo
0A
+
16V
-
+2 , 3 3 3 V
0 , 1 3 3 mA
1 , 2 mA
+3 , 2 V
0A
Vo
3k
+3 , 8 8 V
0A
2k
+3 , 2 V
- 15V
Rev. 1/6/2003
0 , 1 3 3 3 mA
+8 V
1 , 6 mA
15k
Ideal Op Amp Exercise
5,1k
+5 V
+1 V
1 mA
- 0, 68V +
- 2V
- 1, 333V +
0A
0, 933
mA
1, 333
mA
- 5, 2V +
3.9K
C. Sauriol
Page 4
EXERCISE
IDEAL OP AMP ANALYSIS
No.2 A)
4,7k
1, 489
mA
- 1V 0, 7
mA
10k
+6 V
+ 7V -
+2 V
0 , 7 8 9 mA
Vo
3,3k
Vx
+3 V
0A
- 2 .6 V +
- 3 .6 V
+1 5 V
C)
- 1 3 .5 V
0A
+3 V
1,8k
- 1V
- 8V
0A
- 1V
B)
- 15V
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) = V o = -V sat = -13,5V
- 4, 833V
10k
+
20k
-
+1 6 V
0, 483
mA
0A
Vo
6,8k
Vi n
- 14, 5V
+6 V
0A
V+ and verify the sign of the differential
I/P voltage in order to validate your
assumption of V o = -V sat , that is:
+6 V
- 16V
Vo = A d (V + -V - ) = (-4,83-6) = therefore the assumption was valid.
+4 , 8 3 3 V
10k
-
20k
+
+1 6 V
0A
0, 483
mA
Vo
6,8k
Vi n
- 6V
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
V o = -V sat = -14,5V
and determine the
+1 4 , 5 V
0A
Same procedure here, except now V - is
negative, therefore the O/P polarity is
expected to be positive:
V o = +V sat = +14,5V
Verify assumption with sign of V
Vo = A d (V + -V - ) =
- 6V
d = (V
+ -V - )
(+4,83-(-6)) = +
- 16V
Ideal Op Amp Exercise
Rev. 1/6/2003
C. Sauriol
Page 5
EXERCISE
2 D)
2
mA
IDEAL OP AMP ANALYSIS
+2 V
1K
2
mA
1, 25
mA 1K
2k
Vo
+6 V
+1 6 V - 4 V +
0A
+1 , 2 5 V 1 , 2 5
mA
32
mA
Vo
2k
+3 , 7 5 V
+1 6 V
0A
2 mA
680
20
mA
1 , 2 5 mA
680
+2 V
+
6V
-
0A
+2 V
- 16V
3 0 mA
+2 V
1 8 , 7 5 mA
+
3, 75V
-
0A
200
+2 V
- 16V
200
Output of op amp has reached current limit,
notice that V - V + and -ve feedback is
- = V+
rendered ineffective not forcing V
E)
0, 5
mA
1, 125V
0, 5
0 V mA
10K
+ 15V Vo
+5 V
- 15V
+1 2 V 30k
0A
0A
0, 3875
mA 10K
0, 3875
mA +
Vo
+5 V
+1 2 V
+ 3, 875V -
- 10, 5
30k
0A
0, 5
mA
6,8k
0A
0, 5
mA
6,8k
1 , 5 mA
15V
+
0V
- 12V
1 , 0 5 mA
10K
0V
- 12V
10, 5V
+
Output of op amp has reached saturation,
notice that V - V + and -ve feedback is
Impossible voltage, therefore
Vo = -V sat = -10,5V typical.
rendered ineffective not forcing V
F)
10k
R3
100V
max
10k
8, 8
mA
R3
V- or V + max = 16 - 4 = 12V
R3 < 12V/8,68 mA = 1382
+1 6 V
0A
R3 should be less than 1382
1 2 V max 0 A
8 , 6 8 mA
Ideal Op Amp Exercise
- = V+
100k
V1
V2
10K
Vo
in order to keep V - and V+ inside a safe range of ±12V.
0, 12
mA
100K
- 16V
Rev. 1/6/2003
C. Sauriol
Page 6
EXERCISE
IDEAL OP AMP ANALYSIS
No.3
If
F〉
10
10
=
= 159 Hz then ∆V o (PP) = −
2π RF CF 2 π 100k × 0,1µ
A)
t2
( )∫ V
1
RE C F
dt
in ( AC)
t1
+1V
V (ave) = V
( )+ V ( )
+ PW
T
AREA
− SW
T
Vin
+0,2V
DC
AREA
Vin(DC) = +0,2V
-1V
t2
0, 6T
∆Vo ( PP ) = −1000∫ Vin ( AC ) dt
0, 4T
t1
∆Vo ( PP ) = −1000 × area
Vout
∆Vo ( PP ) = −1000 × 0.8 × 0.6T
Integral does not apply for 50
Hz and 100 Hz.
-2V DC
∆Vout = 0,48Vpp at 1 kHz
and
B)
I
N
P 2 V pp
U
T
t2
∆Vo ( PP ) = −1000∫ Vin ( AC ) dt
t1
∆Vo ( PP ) = −1000 × area
= 1000 ×
T
1000
× 1VP =
= 1VPP
2
2F
O
U
T
P
U
T
F = 500 Hz
F > 159 Hz integral OK.
48 mVpp at 10 kHz
AREA
DC level
AREA
T
2
1 V pp
DC level
C)
+1V
t2
I
N
P
U
T
∆Vo ( PP ) = −1000∫ Vin ( AC ) dt
t1
∆Vo ( PP ) = −1000 × area =
3T  375

1000 ×  0,5 ×  =
= 1VPP
4  F

F = 375Hz
F > 159 Hz integral OK.
Ideal Op Amp Exercise
AREA
2 V pp
3T
4
A
R
E
A
DC level
-1V
O
/
P
DC level
1 V pp
V (ave) = V +
Rev. 1/6/2003
( )+ V ( )
PW
T
− SW
T
Vin(DC) = +0,5V
C. Sauriol
Page 7
EXERCISE
IDEAL OP AMP ANALYSIS
No.3 D)
100 µs
t2
∆Vo ( PP) = −1000 ∫ Vin (AC ) dt
t1
∆Vo ( PP) = −1000 × area
= 1000 × 

100 µ × 5 

2
∆Vo ( PP) = 0,25VPP
O/P is a parabolic wave, not a
sine wave.
I
N
P 10
U VPP
T
DC level
O
U
T 0.25
P VPP
U
T
DC level
No.4 A)
+5V p
V o = − RF CE
dVin
dt
 10V 
V o = −10K × 0,1µ ±
 2ms 
10 V pp
-5V p
+5V p
O 10 V
pp
/
P
V o = m5V P
B)
I
N
P
U
T
On edges we have:
V o = − 10K × 0,1µ( ±∞ )
V o = m∞ ⇒ m V sat
On flat portions we have:
V o = −10K × 0,1µ( ±0 )
Vo = 0
The O/P spikes settle down
2
ms
-5V p
+1V p
I
N
P
U
T
O
U
T
P
U
T
2
ms
-1Vp
+ V sat
50 µs
0V
50 µs
- V sat
in 5τ = 5RE CE = 50 µs
C)
To stabilise negative feedback in order to avoid self oscillations from the circuit.
No. 5
Phase shifter
X 
P1 = 0 Φ = 2 × ATAN  C  = 180 o
 0 
X 
X
1
⇒ 10 o = ATAN  C  ⇒ TAN 10 o = C =
P1
2π FCP1
 P1 
X 
Φ = 20 o = 2 × ATAN  C 
 P1 
1
1
P1 =
=
= 9.026 nF
o
2π FC × TAN 10
2π 1000 ×100k × TAN 10 o
( )
( )
( )
⇒ 9.1 nF
std
The +ve input sees 0 to 100k DC wise, average of 50k, and the –ve input sees R1IIR1 = R1/2 = 50k
R1 = 100K. This means that O/P DC offset will not be minimized for all P1 settings.
Ideal Op Amp Exercise
Rev. 1/6/2003
C. Sauriol
Page 8