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56
Lecture No. 15
Ideal Operational Amplifiers
An ideal op-amp is a device which acts as an ideal voltage
controlled voltage source. The device will have the following
characteristics:
1. No current flows into the input terminals of the device. This is
equivalent to having an infinite input resistance Ri=∞. In practical
terms this implies that the amplifier device will make no power
demands on the input signal source.
2. Have a zero output resistance (Ro=0). This implies that the output
voltage is independent of the load connected to the output.
In addition the ideal op-amp model will have infinite open loop gain
(A →∞). The ideal op-amp model is shown schematically on
Figure.
The input marked “ + ” is called the noninverting input . . .
The input marked “ - ” is called the inverting input . . .
The output voltage of operation amplifier is: Vo = Ao (V+ - V-)
The gain of operation amplifier is:
Ao 
Vo
V - V-
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57
Ideal Operational Amplifier Operation
With Ao = ∞ , we can conceive of three rules of operation:
1. If V+ > V- then Vo increases . . .
2. If V+ < V- then Vo decreases . . .
3. If V+ = V- then Vo does not change . . .
In a real op amp, Vo cannot exceed the dc power supply
voltages. In normal use as an amplifier, an operational amplifier
circuit employs negative feedback - a fraction of the output
voltage is applied to the inverting input.
Op Amp Operation with Negative Feedback
Consider the effect of negative feedback:
1. If V+ > V- then Vo increases . . .
Because a fraction of Vo is applied to the inverting input, Vincreases . . .
The “gap” between V+ and V- is reduced and will eventually become
zero.
Thus, Vo takes on the value that causes V+ - V- = 0!!!
2. If V+ < V- then Vo decreases . . .
Because a fraction of Vo is applied to the inverting input, Vdecreases . . .
The “gap” between V+ and V- is reduced and will eventually
become zero.
Thus, Vo takes on the value that causes V+ - V- = 0!!!
In either case, the output voltage takes on whatever value that
causes V+ - V- = 0!!!
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58
Operational Amplifier Circuits
1- The Inverting Amplifier
Let’s put our ideal op amp concepts to work in this basic
circuit:
Voltage Gain
Because the ideal op amp has Ri = ∞ , the current into the inputs
will be zero.
This means i1 = i2 , i.e., resistors R1 and R2 form a voltage divider.
Therefore, we can use superposition to find the voltage V- .
Now, because there is negative feedback, vo takes on whatever
value that causes v+ - v- = 0 , and v+ = 0 !!!
Input Resistance
This means resistance “seen” by the signal source vi, not the input
resistance of the op amp, which is infinite.
Because V- = 0, the voltage across R1 is vi. Thus:
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59
Output Resistance
This is the Thevenin resistance which would be “seen” by a load
looking back into the circuit.
Our op amp is ideal; its Thevenin output resistance is zero:
2. The Current-to-Voltage Converter
The circuit diagram of a current-to-voltage converter is shown
in Figure below. The circuit is a special case of an inverting
amplifier where the input resistor is replaced with a short circuit.
- Because the V− terminal is a virtual ground, the input
resistance is zero. The output resistance is also zero.
- Because i1 +iF = 0 and vo = iF RF , it follows that the transresistance (trans-impedance) gain is given by:
vo
 RF
i1
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60
- Because RS connects from a virtual ground to ground, the
current through RS is zero. Then;
i1 = iS
and
vo = −RF iS
Thus the output voltage is independent of RS.
3-The Non-inverting Amplifier
If we switch the vi and ground connections on the inverting
amplifier, we obtain the noninverting amplifier:
Voltage Gain
This time our rules of operation and a voltage divider equation lead
to:
Input and Output Resistance
The source is connected directly to the ideal op amp, so:
A load “sees” the same ideal Thevenin resistance as in the
inverting case:
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61
Lecture No. 16
4-The Voltage Follower
Voltage Gain
This one is easy:
i.e., the output voltage follows the input voltage.
Input and Output Resistance
By inspection, we should see that these values are the same as for
the noninverting amplifier . . .
In fact, the follower is just a special case of the noninverting
amplifier, with R1 = ∞ and R2 = 0!!!
5-The Inverting Summer
This is a variation of the inverting amplifier:
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62
Voltage Gain
We could use the superposition approach as we did for the
standard inverter, but with three sources the equations become
unnecessarily complicated . . . so let’s try this instead . . .
Recall . . . vO takes on the value that causes v- = v+ = 0 . . .
So the voltage across RA is vA and the voltage across RB is vB :
Because the current into the op amp is zero:
Finally, the voltage rise to vO equals the drop across RF:
6- Differential Amplifier
The op amp is a differential amplifier to begin with, so of course
we can build one of these!!!
Linear electronic
Voltage Gain
Again, vO takes on the value required to make v+ = v- . Thus:
We can now find the current i1 , which must equal the current i2 :
Knowing i2 , we can calculate the voltage across R2 . . .
Then we sum voltage rises to the output terminal:
Working with just the v2 terms from equation, yield;
63
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64
And, finally, returning the resulting term to equation, yield;
7- Integrator
From our rules and previous experience we know that v- = 0 and
iR = iC , so . . .
Combining the two previous equations, and recognizing that
v O = - vC :
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65
Normally vC (0) = 0 (but not always). Thus the output is the integral
of vi , inverted, and scaled by 1/RC.
8- The Differentiator
From the i-v relationship of a capacitor:
Recognizing that vO = -vR :
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66
Lecture No. 17
9- The Comparator
• Open loop!
• Only two output states...
+vsat if vin > vref
-vsat if vin < vref
• Problems with noise!
vref
Vo
vin
vin
vref
t
Vo
+vsat
t
-vsat
10- The Schmitt trigger
• Positive feedback (not open loop)...
• Hysteresis.... There are now two distinct threshold voltages that
determine the state of the output...one threshold to go high if it's
low another threshold to go low if its high...
• Good for rejecting noise.
Voltage at non-inverting input By superposition:
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67
To find the trip-points let Vin = Vu Or VL and solve...
Subtract the trip-points to get the hysteresis...
To design:
1) Determine R1 and R2 based on trip-points:
2) Determine Vref (R3 and R4) from R1 and R2:
vin
Vu
VL
t
Vo
+vsat
t
-vsat
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Op Amp Circuits - Designing with Real Op Amps
1-Resistor Values
Our ideal op amp can supply unlimited current; real ones can’t . . .
Now, we can note the following:
1. To limit iF + iL to a reasonable value, we adopt the “rule of
thumb” that resistances should be greater than approx. 100 .
2. Larger resistances render circuits more susceptible to noise and
more susceptible to environmental factors.
To limit these problems we adopt the “rule of thumb” that
resistances should be less than approximately 1 M.
2- Source Resistance
- In some designs RS will affect desired gain.
- If we wish to ignore source resistance effects, resistances must
be much larger than RS (if possible).
Slew Rate of Op Amp Circuits
The slew rate (SR) is defined as the maximum rate of change of
the output of an op amp circuit. The SR in general describes the
degradation effect on the high frequency response of the active
amplifier (one with an op amp) near or at the rated maximum
output voltage swing. This effect is generally due to the
compensating capacitor and not to the transistor circuits internal to
the op amp. In short, the SR effect is due to the maximum supplied
current available for charging up the compensating capacitor.
We know that the current required to charge a capacitor is
The Slew Rate is found from
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69
Example (1): an Op Amp has a maximum output current of 1 mA.
If the compensating capacitor is 1000 pF, find the Slew rate.
Solution:
Example(2): Suppose that the input signal to a unity gain
amplifier configuration is a 20kHz sine wave and slew rate =
0.5V/μs. What is the largest possible amplitude of the input signal
to avoid distortion due to slewing?
Solution
If the input voltage is vi(t) =Msin2πft and the magnitude of the
gain at that frequency is unity,
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Lecture No. 18
Examples Sheet: Operational Amplifiers
Q1. An operational amplifier has an open loop gain of 160dB. If the
saturation voltage of the op-amp is ±12V, what is the
differential input voltage range for operation in the linear
region?
Answer:
Operation in the linear region is the output from the op-amp
satisfies the differential gain equation:
Where, Ao = 160dB = 100,000,000,
and
vo = vsat = ± 12V.
Hence, the input voltage range that gives linear operation is not
very large!
Q2. Plot the transfer characteristic of an op-amp with the following
characteristics: Ao = 1,000,000, vsat = ±10V.
vo  10

 10 V
Ao 10 6
A rough sketch is all that is required, with important characteristics
highlighted:
Answer: vd 
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71
Q3. Calculate the closed loop gain for the inverting amplifier shown
below. State any assumptions made in your analysis. What is
the input resistance of the circuit?
Answer:
Assumption: The op-amp is ideal,
Inverting amplifier closed loop gain equation:
Input resistance is given by:
Rin = R1 = 30k
Q4. Calculate the closed loop gain for the non-inverting amplifier
shown below. State any assumptions made in your analysis.
What is the input resistance of the circuit?
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72
Answer:
Assumption: The op-amp is ideal,
Non-inverting amplifier closed loop
gain equation:
Input resistance is infinite since the input is connected directly to the
non-inverting input of the Op Amp.
Q5. Design an amplifier circuit with a closed loop gain of 40dB and
an input impedance of 10k.
Answer:
Some thinking required here! You are asked to design an
amplifier with a closed loop gain of 100.
The non-inverting amplifier is not suitable to use here because
the non-inverting amplifier has infinite input impedance, we are
required to design an amplifier with finite input impedance. Hence,
we will have to use two inverting amplifiers with a total closed loop
gain of 100 – the signal inversion will cancel over the two stages.
Avt = Av1  Av2
R1 = 10kW since this forms the input impedance of the circuit.
Lets choose R2 = 100k,
Then
Av1
R2 100  10 3


 10
R1
10  10 3
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73
 Av2 =10
We choose similar values for stage 2, give a total gain of 100.
R2 = 100k and R1 = 10k
Q6. For the inverting amplifier shown below, plot its transfer
characteristics over the range -10V <vi< 10V, the op-amp
saturation voltage is equal to the supply voltage.
Answer:
The saturation voltage of the op-amp is given by Vsat = ±15V,
therefore the output from the op-amp cannot exceed this value.
We are required to plot the transfer characteristic (output voltage
against input voltage) over the input voltage range –10V<vi<10V.
1) We must determine the gain
When vo = Vsat
Hence, if the output is 15V, the input is –4.54V, and if the output is
–15V, the input 4.54V. For input signals greater than these the
output will saturate and vo=±15V. In between, the relationship will
be linear with slope given by the closed loop gain (negative).
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74
Q7. a) Drive the closed loop gain of the op-amp circuit illustrated below.
b) What is the gain when R1= R2=10kW, R3=5kW, and R4=15kW?
Answer:
a) In order to determine the gain of this circuit (in terms of vo and vs)
we need to simplify the input network by calculating the Thevenin
equivalent, i.e.
The Thevenin voltage is simply given by the voltage across R 2 (R1
and R2 form a voltage divider network), i.e.
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75
the resistance as seen at the output, Rth, is given by:
b) For the values given the gain is:
R4 R2
15  10 3  10  10 3
Av 

 0.75
R1 R2  R1 R3  R2 R3 ( 10  10  10  5  10  5 )  10 6
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76
Lecture No. 19
Q8. Derive an expression for the closed loop gain of the circuit
below.
Answer:
Assume that the currents will flow as illustrated and the op-amp is
ideal, and a virtual earth exists, v- = 0V and i- = 0A.
Kirchhoff’s Current Law at node A,
We know the inverting input is held at ground,
we know that vs is seen across R1
Kirchhoff’s current law at node B
And
We know that vB is seen across R3 hence
Using the equations for i2 and i3
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77
Using Kirchhoff’s voltage law at the output
Q9. a) Design a non-inverting amplifier to have a closed loop gain
of 10. The op-amp has an output saturation voltage of 10V,
what is its maximum signal amplitude that can be applied to
the input without saturation occurring?
b) If the output of the op-amp is connected to a load of 5k,
what will be the output current from the op-amp when the
maximum voltage is observed at the output of the op. Amp?
Note: State any assumptions made in your analysis.
Q10. For the noninverting amplifier illustrated below, what is?
a) The closed loop gain
b) The closed loop gain assuming R2>>R1
Note: State any assumptions made in your analysis.
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78
Q11. Consider the circuit below consisting of a bridge circuit and
amplifier. Determine the output voltage, vo, in terms of the
source voltage, vs. State any assumptions made in your
analysis.
Q12. Design a circuit to sum three signals, v1, v2 and v3, in the
following manner: vo = - (5 v1 +10 v2 + 20 v3 )
Answer:
we choose a summing amplifier with three inputs.
The output voltage
is given by:
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79
It must be choosing resistor values so that:
Let Rf = 100k
 R1= Rf / 5 = 20 k, R2= Rf / 10 = 10 k
and R3= Rf / 20 = 5 k
Q13. Design a circuit to sum three signals, v1, v2 and v3, in the
following manner: vo = - (5 v1 +10 v2 + 20 v3 )
Answer: the answer is similar to the previous question except you
must invert the output.
Q14. Design a circuit to sum three signals, v1, v2 and v3, in the
following manner: vo = 3 v1 +2 v2 - 4 v3
Answer: the answer is similar to the previous question except you
must invert inputs v1 and v2 prior to the inverting summing
amplifier.
Q15. Design a circuit to calculate the difference between two
signals, v1 and v2, where vo =10 (v1 -v2)
Answer:
We must be used the difference amplifier as shown below.
Choose resistor values
such that:
Let R1 = 10k  R2 =
R1 10=100 k
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80
Q16. In Figure below, Consider an op-amp connected to a power
supply of ±15V used without feedback. The op-amp saturates
at 80% of the power supply. Plot the transfer characteristics
and the output waveform if the input signal is vs=5sin (wt).
Answer:
vsat is given by 80% of ±15V, hence vsat= ±12V. The output will
switch at an input of 3V, since the 3V reference is connected to the
inverting input as shown in figure below.
vo
vo
12V
3V
vs
-12V
5V
-5V
t
vs
t
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81
Lecture No. 20
Oscillators
Oscillation: an effect that repeatedly and regularly fluctuates
about the mean value
Oscillator: circuit that produces oscillation
Characteristics: wave-shape, frequency, amplitude, distortion,
stability
Application of Oscillators
1 Oscillators are used to generate signals, e.g.
– Used as a local oscillator to transform the RF signals to IF
signals in a receiver;
– Used to generate RF carrier in a transmitter
– Used to generate clocks in digital systems;
– Used as sweep circuits in TV sets and CRO.
Oscillators divided to two main types:
1. Linear Oscillators: generate sinusoidal waves utilizing the
resonance phenomena.
1 Wien Bridge Oscillators
2 RC Phase-Shift Oscillators
3 LC Oscillators
2. Nonlinear Oscillators (function generators): generate the
square, triangle, and pulse waveform like Multivibrators.
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82
1- Linear Oscillators
Vs
+
V

Amplifier (A)
Vo
+
Vf
Positive
Feedback
Frequency-Selective
Feedback Network ()
For sinusoidal input is connected “Linear” because the output is
approximately sinusoidal
A linear oscillator contains:
- A frequency selection feedback network
- An amplifier to maintain the loop gain at unity
Vs
+

V
A(f)
+
Vf
SelectiveNetwork
(f)
Vo  AV  A( Vs  Vf )
Vf   Vo

Vo
A

Vs 1  A
If Vs = 0, the only way that Vo can be nonzero is that loop gain
A=1 which implies that
Vo
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83
| A | 1
Barkhausen Criterion
A  0
Wien Bridge Oscillator
Rf
R1

+
C
C
R
Vo
Z1
R
Z2
Z1
R1
Vi
Z2
C1
C2
R2
Vo
Linear electronic
XC 2 
84
1
C2
XC1 
and
1
C1
Z1  R1  jXC1
 1
1 
Z2  


 R 2  jXC 2 
1

 jR 2 XC 2
R 2  jXC 2
Therefore, the feedback factor,

Vo
Z2
(  jR 2 XC 2 / R 2  jXC 2 )


Vi Z1  Z 2 ( R1  jXC1 )  (  jR 2 XC 2 / R 2  jXC 2 )

 jR 2 XC 2
( R1  jXC1 )( R 2  jXC 2 )  jR 2 XC 2
 can be rewritten as:

R 2 XC 2
R 1 XC 2  R 2 XC1  R 2 XC 2  j( R 1R 2  XC1 XC 2 )
For Barkhausen Criterion, imaginary part = 0, i.e.,
R1R 2  XC1XC 2  0
or R1R 2 
1
1
C1 C2
   1 / R1R 2C1C2
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85
Supposing,
R1=R2=R
and
RXC
3RXC  j( R 2  XC2 )



XC1= XC2=XC,
1
RC
1
3
A v  1  A v  3  1 
Therefore,
Rf
2
R1
Rf
R1
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86
Lecture No. 21
RC Phase-Shift Oscillator
Rf
R1

C
+
C
R
C
R
R
Using an inverting amplifier
The additional 180o phase shift is provided by an RC phaseshift network
- Applying KVL to the phase-shift network, we have
C
C
C
V1
Vo
I1
R
I2
R
I3
R
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87
V1  I1 ( R  jXC )  I 2 R
0   I1R
 I 2 ( 2R  jXC )  I 3 R
0 
 I 2R
 I 3 ( 2R  jXC )
Solve for I3, we get
I3 
R  jXC
R
V1
R
2 R  jXC 0
0
R
0
R  jXC
R
0
R
0
2 R  jXC
R
R
2 R  jXC
V1R 2
I3 
( R  jXC )[( 2R  jXC )2  R 2 ]  R 2 ( 2R  jXC )
The output voltage,
V1R 3
Vo  I 3 R 
( R  jXC )[( 2R  jXC )2  R 2 ]  R 2 ( 2R  jXC )
Hence the transfer function of the phase-shift network is given
by:
Vo
R3


V1 ( R 3  5RXC2 )  j( XC3  6R 2 XC )
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88
For 180o phase shift, the imaginary part = 0, i.e.,
3
XC
 6R 2 XC  0 or XC  0 (Rejected)
2
 XC
 6R 2

1
6RC
and,

1
29
Note: The –ve sign mean the phase inversion from the voltage
Linear electronic
89
Lecture No. 22
LC Oscillators

Av Ro
~
+
2
Z1
Z2
1
Z3
Zp
1 The frequency selection network (Z1, Z2 and Z3) provides a
phase shift of 180o
2 The amplifier provides an addition shift of 180o
Two well-known Oscillators:
1 Colpitts Oscillator
2 Harley Oscillator
Av Ro
~
+
Vf
Z1
Z2
Vo
Z3
Zp
Linear electronic
90
Vf   Vo 
Z1
Vo
Z1  Z 3
Zp  Z 2 //( Z1  Z 3 )

Z 2 ( Z1  Z 3 )
Z1  Z 2  Z 3
For the equivalent circuit from the output
Ro
Io
+
+
 AvVi
Zp Vo

 A v Zp
 A v Vi
Vo
Vo

or

R o  Zp Zp
Vi
R o  Zp
Therefore, the amplifier gain is obtained,
A
Vo
 A v Z 2 ( Z1  Z 3 )

Vi R o ( Z1  Z 2  Z 3 )  Z 2 ( Z1  Z 3 )
The loop gain,
A 
 A v Z1 Z 2
R o ( Z1  Z 2  Z 3 )  Z 2 ( Z1  Z 3 )
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91
If the impedance are all pure reactances, i.e.,
Z1  jX1 , Z 2  jX 2 and Z 3  jX 3
The loop gain becomes,
A 
A v X1 X 2
jR o ( X1  X 2  X 3 )  X 2 ( X1  X 3 )
The imaginary part = 0 only when X1+ X2+ X3=0
- It indicates that at least one reactance must be –ve (capacitor).
- X1 and X2 must be of same type and X3 must be of opposite
type
With imaginary part = 0,
A 
 A v X1
A X
 v 1
X1  X 3
X2
For Unit Gain & 180o Phase-shift,
A  1  A v 
R
X2
X1
L1
R
C
L2
Harley Oscillator
C1
C2
Colpitts Oscillator
L
Linear electronic
92
Colpitts Oscillator
R
C1
L
C2
C2
+
V

I1
L
I2
gmV
node 1
I3
R
I4
In the equivalent circuit, it is assumed that:
1 Linear small signal model of transistor is used
2 The transistor capacitances are neglected
3 Input resistance of the transistor is large enough
At node 1,
V1  V  i1 ( jL )
V
C1
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93
where,
i1  jC2 V
 V1  V (1  2LC2 )
Apply KCL at node 1, we have
jC2 V  g m V 
V1
 jC1V1  0
R
1

jC2 V  g m V  V (1  2LC2 )  jC1   0
R

For Oscillator V must not be zero, therefore it enforces,



1 2LC2 
 gm  
  j (C1  C2 )  3LC1C2  0

R
R 

Imaginary part = 0, we have
o 
1
LCT
And
CT 
C1C2
C1  C2
Real part = 0, yields
gm 
C2
RC1
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94
Lecture No. 23
Nonlinear Oscillators generate square, triangle, and pulse
waveforms.
Generation of Square and triangle waveforms
using Multivibrator
The nonlinear oscillator devices make use of special class of
circuit known as Multivibrator.
The Multivibrator is divided into tree types:
- Bistable Multivibrator
- Astable Multivibrator
- Mono-stable Multivibrator
I- Operation of the Bistable Multivibrator
R1
R2
v+
v1
+

The voltage divider (R1, R2) will feed a fraction  
vo
R1
of
R1  R2
the output signal back to the positive input terminal of OP-Amp.
If A is grater than unity, the feedback signal will greater the
original increment in V+. This regenerative process continues
until Op-Amp saturation (Vsat+).When this happens, the V+ =
Vsat+ . Also, the Op-Amp would have ended up saturated in
Linear electronic
95
negative direction with (Vsat-) and V+ = Vsat- .
The transfer characteristic of bistable Multivibrator
When v1 increase from 0V, we can see from the circuit that
nothing happens until v1 reaches a value equal to V+ (i.e Vth =
Vsat+). Then vo goes to Vsat- because the difference between the
V+ and V- terminals becomes negative.
Now consider v1 is decrease and V+ =  Vsat-, we see that the
circuit remains in the negative saturation atate until v1 goes
negative to the point that it equals –Vth = Vsat-. Then OP Amp
goes to the positive saturation (vo =Vsat+)because the difference
between the V+ and V- terminals becomes positive.
vo
vo
+Vcc
Vth
+Vcc
-Vth
v1
-Vcc
v1
-Vcc
vo
+Vcc
-Vth
-Vcc
Vth
v1
Linear electronic
96
A Square-wave Oscillator by using Astable
Multivibrator
A square waveform can be generated by arranging for a bistable
multivibrator to switch states periodically. This can be done by
connecting bistable with an RC circuit in a feedback loop.
R
vc

C
vo
vf
R1
+
R2
Let vo = Vsat+, the capacitor will charge toward this level through
resistor R. Thus the voltage across C, which is applied to the
negative input terminal of Op-Amp, will rise exponentially
toward Vsat+ with time constant  = C R. In this case V+ = Vth.
When the capacitor voltage (Vc) reaches the positive threshold
(Vth) the Op-Amp will switch to other stable state on which vo =
Vsat-. The capacitor will then start discharging and its voltage
will decrease exponential toward Vsat-. When the capacitor
voltage equal to the negative threshold (-Vth), the Op-Amp will
switch to other stable state on which vo = Vsat+
- During the charging interval T1: the voltage of negative
terminal (capacitor voltage) at any time is:
V  Vc  Vsat  ( Vsat  Vsat ) exp(  t /  )
At
Vc=Vsat+

V  
 1    sat  

 Vsat  
T1   ln

1 






Linear electronic
97
During the discharging interval T2: the voltage of negative
terminal (capacitor voltage) at any time is:
V  Vc  Vsat  ( Vsat  Vsat )exp( t /  )
Put,
Vc=Vsat-

V  
 1    sat  

 Vsat  
T2   ln

1 






If Vsat+ = -Vsat-, Then:
 1 
T  2 ln
 1 



L+ = Vsat+
and
L- =Vsat-
Linear electronic
98
Lecture No. 24
Triangle Waveform Generator
The exponential waveforms generated in the astable circuit can
be changed to triangular by replacing the RC circuit with an
integrator. The integrator causes linear charging and discharging
of capacitor, thus providing a triangular waveform.
v
Let, L+ = Vsat+
and
L- =Vsat-
Let the output of the bistable circuit be at L+, a current equal
to L+/R will flow into the resistor R and through capacitor C
causing the output of the integrator to linearly decrease with
slop of –L+/CR. This will continue until the integrator output
reaches the lower threshold VTL of the bistable circuit, at which
point the bistable circuit will switch states, its output becoming
negative and equal to L-.
Linear electronic
99
At this the current through R and C will reverse direction. The
integrator will strart to increase linearly with a positive slop
equal to |L-| / RC. This will continue until the integrator output
voltage reaches the positive threshold of the bistable circuit Vth.
At this point, the bistable circuit switches, its output becomes
(L+).
- During T1 interval :
VTH  VTL
L
 
T1
CR
T1  CR
VTH
 VTL 
L
- During T2 interval:
VTH  VTL  L

T2
CR
T2  CR
VTH
 VTL 
 L
Linear electronic
100
Example: Consider the triangular waveform generator circuit
with Op Amp have saturation voltage  10V . If a capacitor
C=0.01F and resistor R1 = 10k are used. Find the values of R
and R2 such that the frequency of oscillation is 1kHz and the
triangular waveform has a 10V peak to peak amplitude
Sol
Because The waveform with 10V peak to peak , then one peak
is equal to 5V, i.e
VTH = 5V and VTL= -5V
VTH =  Vsat+
 = VTH / Vsat+
 = 5/10 = 0.5
But

R1
R1  R2
0.5 
10
10  R2
T1 =T2 because Vsat+ = VsatT = 1/f = 1/1kHz = 1msec
and
T1= 0.5msec
T1  CR
VTH
 VTL 
L
0.5  10  3  0.01  10  6  R
5  ( 5 )
10
0.5  10 3
4
R

5

10
0.01  10  6
R= 50k
R2= 10k