ee380: electronic circuits labee380: electronic...
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EE380: Electronic Circuits LabEE380: Electronic Circuits Lab
Exp 6 : WaveshapingExp. 6 : Waveshaping
B. MazhariDept of EE IIT KanpurDept. of EE, IIT Kanpur
G-NumberB. Mazhari, IITK1
Design Methodology
B k l bl i t t f i l blBreak a complex problem into a set of simpler problems
Problem
Sub-problem-2 Sub-problem-nSub-problem-1
Hi hi l d iti
Sub-problem-1
Hierarchical decomposition
G-NumberB. Mazhari, IITK2
Objective
Design a circuit with the following characteristics:
VO
1.5
2.5
V1 5
-1.5
3 5
-3.5
VIN1.5
-1.5
3.5
-2.5
G-NumberB. Mazhari, IITK3
Triangular wave of variable frequency is easy to generate
Motivation
Triangular wave of variable frequency is easy to generate
4
VO
2
3
ge (V
)
V1 5
1.5
-1.5
3 5
2.5
-3.5
1
Volta
VIN1.5
-1.5-2.5
3.5
0
0.0 0.2 0.4 0.6 0.8 1.0
Time(ms)
G-NumberB. Mazhari, IITK4
If a 4V Tr. Wave is given as an input, we get a Sine likewaveform of 2.5V amplitude
4
5
2
3
0
1
2
ge (V
)
2
-1
0
Volta
g
-3
-2V
0 0 0 2 0 4 0 6 0 8 1 0-5
-4
0.0 0.2 0.4 0.6 0.8 1.0
Time(ms)
G-NumberB. Mazhari, IITK5
Total Harmonic distortion ~3%
6
Triangular wave input is increased to 5V
456
123
(V)
-101
tage
(
-3-2Vo
l
-6-5-4
0.0 0.2 0.4 0.6 0.8 1.0-6
Time(ms)
G-NumberB. Mazhari, IITK6Total Harmonic distortion ~7.2%
Design Methodology
B k l bl i t t f i l blBreak a complex problem into a set of simpler problemsVO
2.5
VIN1.5
1.5
-1.5
3.5
-3.5
Problem
-1.5-2.5
Sub-problem-2 Sub-problem-nSub-problem-1
G-NumberB. Mazhari, IITK7
Sub-problem-1
Design Methodology
VO
2.5
V1 5
1.5
-1.5
3 5
-3.5
VIN1.5
-1.5-2.5
3.5
VO VOVO
+ +VIN1.5 VIN
VIN
+ + 3.5slope = 1
G-NumberB. Mazhari, IITK8
slope = -0.5 slope = -0.5
Design MethodologyThus we need to design a circuit with the followingThus we need to design a circuit with the followingInput-output characteristics
VO
1.5
2.5
VIN1.5
-1.5
-1.5
3.5
-3.5
-2.5
VO
b
VIN
b
VIN VOUT
a,b
G-NumberB. Mazhari, IITK9
slope = a
Simpler and familiar Problem
VO
VO
VIN
slope = a
V
b
slope = a
VIN
How do we obtain the following?How do we obtain the following?
fG-NumberB. Mazhari, IITK
10
This looks like a rectifier !
RectifierVO
D
O
VIN R
G-NumberB. Mazhari, IITK11
RectifierVO
D1 D2
O
VIN R
G-NumberB. Mazhari, IITK12
Need Precision Rectifier VO
R D
VIN
G-NumberB. Mazhari, IITK13
Problem: Slew Rate
G-NumberB. Mazhari, IITK14
Opamp output
F = 0.1KHz
G-NumberB. Mazhari, IITK15
F = 0.1KHz
G-NumberB. Mazhari, IITK16
F = 1KHz
G-NumberB. Mazhari, IITK17
F = 10KHz
G-NumberB. Mazhari, IITK18
F = 10KHz
G-NumberB. Mazhari, IITK19
Provide feedback path during both halves of sinusoidalwaveform
RVO
R2
D1
D2
R
VIN
D1R1
G-NumberB. Mazhari, IITK20
F = 0.1KHz
G-NumberB. Mazhari, IITK21
F = 10KHz
G-NumberB. Mazhari, IITK22
F = 10KHz
G-NumberB. Mazhari, IITK23
VO
2KHigher Slope
VIN
D1
D2
1K
Slope = -2
G-NumberB. Mazhari, IITK24
Reverse diode polarityVO
R2
D1
D2
R1
VIN
1
G-NumberB. Mazhari, IITK25
How do we shift the curve along x axis?
VO
2K
VIN
D1
D2
1K
G-NumberB. Mazhari, IITK26
One possibility
2K
VO
2K
D
D2
VIN
D11K
IN
2 5V2.5V
G-NumberB. Mazhari, IITK27
Curve gets shifted up as well !
G-NumberB. Mazhari, IITK28
Another Alternative
V2K
1mA VO
1K
1mA
D
D2
1K
-1V
V
D11K
VIN
G-NumberB. Mazhari, IITK29
Simulation Result
G-NumberB. Mazhari, IITK30
Partial Characteristics
VO
2.5
V1 5
1.5
-1.5
3 5
-3.5
VIN1.5
-1.5-2.5
3.5
VO VOVO
+ +VIN1.5 VIN
VIN
+ + 3.5slope = 1
G-NumberB. Mazhari, IITK31
slope = -0.5 slope = -0.5
Summer
R5
R
VO
R1V1
VR2
V2
V V = a V +a V +a VV3R3 R4 a1,2 = -R5/R1,2
VO = a1V1+a2V2+a3V3
a3= (1+R5/R1ll R2)(1+R3/R4)-1
3 ( 5 1 2)( 3 4)
G-NumberB. Mazhari, IITK32