EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
QUESTION
1.
Describe briefly the following terms as applied to an ideal operational amplifier:
i)
iii)
v)
vii)
Virtual ground;
Input dc offset voltage;
Input bias current;
Slew rate.
ii)
iv)
vi)
Virtual short;
Output dc offset voltage;
Input offset current;
SOLUTION
1.
i)
In the inverting amplifier shown in the following figure, since the open-loop gain is very
large, the two inputs v1 and v2 must be nearly equal. And since v2 is connected to ground, v1
must be at zero potential. The fact that v1 is at zero potential does mean that the terminal is
connected to ground. Rather, the terminal is said to be at virtual ground.
RF
R1
vI
v1
v2 = 0
ii)
vO
+
In the non-inverting amplifier shown in the following figure, the negative feedback forces v1
to track v2. Thus v1 and v2 are essentially equal. Such condition is referred to as virtual
short.
RF
R1
v1
v2
~
1
vI
-
+
vO
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
iii)
The input dc offset voltage (VOS) is defined as the input differential voltage that must be
applied to the open-loop op-amp to produce zero output voltage.
iv)
The output dc offset voltage is defined as the measured output voltage when the input
voltage is zero
v)
If the input stage is symmetrical with all corresponding elements matched the input bias
currents IB1 and IB2 are equal i.e. I B1  I B 2 . However, in a practical op-amp, due to
component (input transistors) mismatched, I B1  I B 2 . In such a case, the input bias current
IB is the average of the two i.e.
IB 
vi)
I B1  I B 2
2
The input offset current IOS is the net difference between the input currents IB1 and IB2 i.e.
I OS  I B1 - I B 2
2
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
vii)
The slew rate is defined as the maximum rate of change in the output voltage per unit of time
Slew rate 
Vout
t
V
t
0
Vout  Vmax - - Vmax 
V in
t  t 2 - t1
+V max
V
0
t1
t2
t
-V max
V out
2.
List down five parameters of an op-amp and give comparisons between an ideal and a practical opamp with regards to these parameters and hence draw the equivalent circuit of an ideal op-amp and
a simplified equivalent circuit of a practical op-amp.
SOLUTION
PARAMETERS
IDEAL OP-AMP
PRACTICAL OP-AMP
Input resistance
Infinite
500 k – 2 M
Output resistanced
Zero
20  – 100 
Open-loop voltage gain
Infinite
20k – 200k
Bandwidth
Infinite
Few kHz
Output dc voltage
zero
1 mV – 2 mV
Input bias current
zero
pA – 10 A
3
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
The equivalent circuit representation of an ideal op-amp
A simplified equivalent circuit representation
of a practical op-amp
4
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
3.
With the aid of suitable diagrams, give brief description of the terms DIFFERENTIAL MODE
INPUT and COMMON-MODE INPUT as applied to an op-amp and hence give the mathematical
definition for the term COMMON-MODE REJECTION RATIO (CMRR).
SOLUTION
In the differential mode input, the input signal is applied across the inverting and non-inverting
terminals of an op-amp as shown in Figure 3(a). Since an op-amp responses to the difference
between the two input signals at its input terminals, differential mode input may also be represented
as shown Figure 3(b) but we must ensure that vi1  vi2.
vd
~
+
-
vO
Figure 3(a) : Differential mode input
Figure 3(b) : Differential mode input when vi1  vi2
5
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
From Figure 3(a), the differential gain Ad is;
vO
vd
From Figure 3(b); vd  vi1 - vi 2 . Hence;
Ad 
Ad 
vO
vi1 - vi 2
Since, in both Figures 3(a) and (b), the op-amp is under open loop condition, the differential gain
Ad is also known as the open-loop gain AOL. Thus.
AOL 
vO
vi1 - vi 2
which is normally extremely large.
In Figure 3(b), if we make vi1 = vi2, it is known as common mode input i.e. both input terminals
receive identical input signal. Another way to illustrate this condition is shown in Figure 3(c).
+
v CM
-
~
vO
Figure 3(c) : Common mode input
The common-mode gain ACM is defined as;
ACM 
vO
vCM
which is normally very small.
The common-mode rejection ratio CMRR is defined as the ratio of differential gain Ad to commonmode gain ACM namely;
6
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
CMRR 
Ad
ACM
CMRR is normally expressed in dB, thus;
 A 
CMRR dB  20 log  d 
 ACM 
4.
Consider an operational amplifier with ideal parameters and the op-amp is operated as an inverting
constant gain multiplier with the input signal voltage V1 applied through R1 and the output signal
Vo is fed back to the input through RF.
a)
Draw the circuit diagram.
b)
Draw the ac equivalent circuit for practical op-amp, ideal op-amp and redrawn the equivalent
circuit.
c)
Prove that the gain is controlled by 𝑅𝐹 & 𝑅1 using superposition theorem and appropriate
equivalent circuit.
SOLUTION
a) & b)
7
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
c) Using superposition theorem to find the value of Vi.
Consider the 1st voltage source (V1), so set (-AvVi=0)
R1
1
V1
2
2
+
Rf Vi1
-
1
𝑉𝑖1 = 𝑅
𝑅𝑓
1 +𝑅𝑓
× 𝑉1
Consider the 2nd voltage source (-AvVi), so set V1=0
8
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
Rf
2
+
Vi2 R1
-
2
- Av Vi
1
𝑉𝑖2 =
1
𝑅1
× (−𝐴𝑣𝑉𝑖 )
𝑅1 + 𝑅𝑓
𝑉𝑖 = 𝑉𝑖1 + 𝑉𝑖2
𝑅𝑓
𝑅1
𝑉𝑖 = [
× 𝑉1 ] + [
× (−𝐴𝑣 𝑉𝑖 )]
𝑅1 + 𝑅𝑓
𝑅1 + 𝑅𝑓
𝑉𝑖 =
𝑅𝑓 𝑉1
𝑅1 + 𝑅𝑓 + 𝑅1 𝐴𝑣
Usually, Av>>1 and AvR1>> Rf & R1, therefore, Rf & R1 can be neglected
𝑉𝑖 ≅
𝑅𝑓 𝑉1
𝑅1 𝐴𝑣
From ac equivalent circuit, we know that Vo= -AvVi
Therefore, to find gain,
𝐴=
𝑉𝑜 𝑉𝑜 𝑉𝑖
= ×
𝑉1 𝑉𝑖 𝑉1
𝐴 = −𝐴𝑣 ×
𝐴=−
𝑅𝑓
𝑅1 𝐴𝑣
𝑅𝑓
𝑅1
9
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
5.
Consider an operational amplifier in Figure 1 with ideal parameters and the op-amp is operated as
non-inverting constant gain multiplier.
a)
Draw the ac equivalent circuit.
b)
Prove that the gain is A  1 
RF
using voltage divider rule.
R1
RF
R1
vo
+
vi
~
Figure 1
SOLUTION
a) Ac equivalent circuit
b) Since Vi=0, using voltage divider rule
𝑉1 =
𝑅1
× 𝑉𝑜
𝑅1 + 𝑅𝑓
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
𝐴=
𝑉𝑜 𝑅1 + 𝑅𝑓
=
𝑉1
𝑅1
𝐴=1+
6.
𝑅𝑓
𝑅1
Perform an ac analysis on the circuit in Figure 2 to obtain an expression for the voltage gain Av
where;
.
v
Av  O
vI
V CC
RC
RC
.
vO
Q2
Q1
vI
~
RE
.
SOLUTION
The equivalent circuit may be labeled as follows;
11
Figure 2
V EE
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
Assuming perfectly matched transistors;
iB1  iB 2  iB ;
iE1  iE 2  iE ;
1   2   ;
ri1  ri 2  ri
iE  1   iB
and
iC1  iC 2  iC
Considering that;
iC  i B
Referring to the equivalent circuit above;
Ib 
Rearrangin g,
Vi - 2 (   1) I b R E
ri
Vo  I c Rc  I b Rc 
Ib 
Vi
ri  2(   1) RE
Vi Rc
ri  2(   1) RE
1
The common-mode voltage gain;
Av 
7.
Vo
Rc

Vi
ri  2(   1) RE
The inverting amplifier in Figure 3 receives an input signal from the source vs. The source is
capable of delivering a maximum current of 4 A at a peak voltage of 0.2 V. Design the amplifier
for a gain |Av| of 20. Assume that the op-amp is ideal.
R2
R1
~
-
Figure 3
vo
vs
+
SOLUTION
12
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
To limit the current drawn from vs at 4 A at a peak voltage of of 0.2 V, the amplifier must have an
input resistance Ri;
Ri 
0.2
 50 k
4  10 -6
R1  Ri  50 k
Av 
R2
 20
R1
R2  20R1  1 M
13
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
8.
Calculate the output voltage V2 & V3 based from Figure 4.
Figure 4
SOLUTION
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
9.
Determine the output voltage of the DAC in Figure 5(a). The sequence of 4 digit binary
codes represented by the waveforms in Figure 5(b) are applied to the inputs. A high level
is a binary 1, and a low level is a binary 0. The least significant binary digit is D0. Draw a
graph of binary input versus Vout.
Figure 5
SOLUTION
First, determine the current for each of the weighted inputs. Since the inverting input of the opamp is at 0V(virtual ground) and a binary 1 corresponds to a high level (+5V), the current
through any of the input resistor equals 5V divided by the resistance value.
There is almost no current at the inverting op-amp input because of its extremely high
impedance. Therefore, assume that all of the input current is through Rf. Since one end of Rf is
at 0V (virtual ground), the drop across Rf equals the output voltage, which is negative with
respect to virtual ground.
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
From Figure 5(b), the first binary input code is 0000, which produces an output voltage of 0V.
The next input code is 0001,( it stands for decimal 1).For this, ,this the output voltage is -0.25V.
The next code is 0010, which produces an output voltage of -0.5. The next code is 0011, which
produces an output voltage of -0.25 V + [-0.5V] = -0.75V. Each successive binary code increases
the output voltage by -0.25V. So, for this particular straight binary sequence on the inputs, the
output is a stair step waveform going from 0V to -3.75V in -0.25V steps, as shown in figure
below. If the steps are very small, the output approximates a straight line (linear).
16
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
10. Based from Figure 6, find the output voltage, Vout and illustrate the respective equivalent
circuit when the binary input is 0001.
Figure 6
SOLUTION
Vout= -0.625V
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
11.
For the circuit in Figure 7;
a)
b)
derive the expression for the output voltage vO in terms of the input voltages vI1 and vI2, and
the circuit parameters R1, R2, R3, and R4;
Show that under certain condition, the output voltage vO is proportional to the difference of
the input voltages vI1 and vI2 and determine the constant of proportionality under this
condition.
R2
R1
vI1
vO
R3
vI2
+
R4
Figure 7
SOLUTION
Using superposition theorem and setting vI2 to zero;
i2
vI1
R1 i1
0
v 1a
R2
vO1
v 1b
R3
+
0
R4
v1b  0
vO1  -
R2
vI1
R1
(Inverting amplifier)
18
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
Setting vI2 to zero;
i2
i1
R1
0
v 2a
R3
vI2
R2
vO2
v 2b
+
0
R4
vO 2
R
 1  2 (Non-inverting amplifier)
v 2b
R1
 R 
vO 2  1  2 v2b
R1 

 R4 
v I 2
v2b  
 R3  R4 
(Voltage divider rule)
Hence;
 R  R4 
v I 2
vO 2  1  2 
R1  R3  R4 

The total response;
vO  vO1  vO 2  -
 R  R4 
R2
v I 2
v I 1  1  2 
R1
R
R

R
1  3
4 

The above expression can be written as;
vO  -
 R  R / R 
R2
v I 1  1  2  4 3 v I 2
R1
R1  1  R4 / R3 

19
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
b)
For a difference amplifier, the output voltage vO is zero if the two input voltages are equal
i.e.;
vO  0
for
vI1  vI 2
This is achievable in the above expression if the following condition is met;
R4 R2

R3 R1
Under this condition, the expression for vO becomes;
vO 
R2
v I 2 - v I 1 
R1
In other words, the output voltage is proportional to the difference of the input voltages and
the constant of proportionality is;
R2
R1
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
12.
Design the circuit in Figure 8(a) to produce an output voltage vO in Figure 8(c) when the input
voltage vI is as shown in Figure 8(b). Assume the op-amp is ideal and use C = 220 pF.
R
C
-
vI
Figure 8(a)
vO
+
v I (V)
5
0
t (s)
-5
Figure 8(b)
v O (V)
8.8
0
0
5
10
15
20
t (s)
-8.8
Figure 8(c)
21
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
SOLUTION
For a differentiator;
vO t   -CR
dv I t 
dt
R-
0t 5
dv I t 
 2 V/μs
dt
 2  10 6 / s
and
vO t   -8.8 V
vO t 
dvI t  / dt C
Hence;
R
8.8
8.8

6
6
2  10 C 2  10  220  10 -12
R  20 k
13.
For the circuit in Figure 9(a), C = 0.1 F and R = 10 k. Both the output and input voltages are
zero at t = 0. An input voltage vI shown in Figure 9(b) is applied to the circuit. Sketch the
resulting output waveform vO(t).
C
R
-
vI
Figure 9(a)
Ao
vO
+
v I (V)
1
Figure 9(b)
0
1
2
3
4
22
-1
t (ms)
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
SOLUTION
4
-6
Time constant,   RC  10  0.1  10  1 ms
v I t   1 for 0  t  1
Thus;
vO t   VO -
t
1 t
vI t 'dt '  -103  1dt '  -103 t

0
CR 0
(Since vO = 0 at t = 0)
Hence at t  1 ms  vO  -1 V
Also;
v I t   -1 for 1  t  2 and vO  -1 V at t  1 ms
Thus;
vO t   VO -
1 t
- vI t 'dt '
CR 10-3
 -1  103 
t
10 -3

1dt '  -1  103 t 10-3
t
 -1  103 t - 10- 3



2
3
Hence at t  2ms  vO  -1  103 2  10-3 - 10-3  0
v O (V)
1
0
1
-1
23
4
t (ms)
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
14.
The input signal in Figure 10(a) is applied to the comparator in Figure 10(b). Draw the output
showing its proper relationship to the input signal. Assume the maximum output levels of the
comparator are ±14𝑉.
Figure 10
SOLUTION
The reference voltage is set by 𝑅1 and 𝑅2 as follows:
As shown in figure below, each time the input exceeds +1.63V, the output voltage switches to its +14V
level, and each time the input goes below +1.63V, the output switches back to its -14V.
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EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
15.
Determine the output voltage waveform for Figure 11. (Hint: Find the values of upper
trigger point (UTP) and the lower trigger point (LTP)).
Figure 11
SOLUTION
This comparator has both hysteresis and zener bounding. The voltage across 𝐷1 and 𝐷2 in either direction
is 4.7V+0.7V =5.4V. This is because one zener is always forward-biased with a drop of 0.7V when the
other one is in breakdown.
The voltage at the inverting (-) op-amp is Vout ± 5.4V. Since the differential voltage is negligible, the
voltage at the non-inverting(+) op-amp input is also approximately Vout ± 5.4V. Thus,
Since the non-inverting input current is negligible,
25
EKT 214 ANALOG ELECTRONIC CIRCUIT II
SEM II 2012/2013
SOLUTION TUTORIAL 1 – OPERATIONAL AMPLIFIER & OP-AMP APPLICATIONS &
FREQUENCY RESPONSE
The upper trigger point (UTP) and the lower trigger point (LTP) are as follows:
The output waveform for the given input voltage is shown in figure below:
26