ANALOG CIRCUITS
(V18ECET08)
II-B.Tech II- Sem-ECE & ECT
(V18 Regulation)
Prepared
Dr.U Yedukondalu
UNIT-1
WAVE SHAPING CIRCUITS
Wave Shaping
Definition: It is the process of changing the shape of input signal with linear / non-linear circuits.
Types:
Linear Wave Shaping
Definition: The process where by the form of a non-sinusoidal signal is changed by transmission through a linear network is called Linear Wave Shaping.
Types:
Non-sinusoidal wave forms
Step Waveform
t
t=0
i
Vi=0 t<0
V =V t>0
A step voltage is one which maintains the value zero for all times t<0 and maintains the value V for all times t>0.
Vi
V
Pulse
The pulse amplitude is „V‟ and the pulse duration is tp.
0≤t≤tp
Otherwise
Vi=V
Vi=0
t=tp
t
Vi
V
t=0
0
Square Wave
T1
T2
Ramp
A waveform which is zero for t<0 and which increases linearly with time for t>0.
Vi
Vi =αt
Vi =αt , t>0
0
t
Exponential
0
t
where T is the time constant of the exponential input
Vi
V
High Pass RC Circuit
R
+
V
o
C
+
-
V
i
-
If f=low, Xc becomes high
C act as open circuit, so the Vo=0.
If f=high, Xc becomes low
C acts as short circuit, so we get the output.
The higher frequency components in the input signal appear at the output with less attenuation due to this behavior the circuit is called “High Pass Filter”.
XC =
1
2ΠfC
Sinusoidal input
+
V
O
i
+
Vin
_
_
C
R
V
in
R - j XC
V
i =
=
R -
in
j
2πf C
Vin
i=
R ⎡ 1-
j
⎤
⎢⎣
2πfRC⎥⎦
O
Vin ×R
Vin
V =i R=
=
1-
j
2πfRC
⎡ j ⎤
R ⎢⎣1- 2πfR C⎥⎦
Vo= iR
VO
1
=
Vin
1+j ⎛-f1⎞
⎜ f ⎟
⎝ ⎠
1
=
VO
V in
1 + ⎜
2
⎛ f1 ⎞
⎟
⎝ f ⎠
θ =-tan -1⎜⎛-f1⎞ = tan-1⎛⎜f1 ⎞
⎝ f ⎟⎠ ⎝f ⎟⎠
At the frequency f = f1
Vin
VO =
1 = 1 =0.707
1+1 2
A =0.707
At f = f1 the gain is 0.707 or this level corresponds to a signal reduction of 3
decibels(dB).
∴ f1 is referred to as Lower 3-dB frequency.
Square wave input
Tilt is defined as the decay in the amplitude of the output voltage wave due to the input voltage maintaining constant level
2
1
X 100
P =
1
V
V −V 1
1
- T 1
R C
= V 1. e
V
'
2
- T 2
RC
V
'
= V2 . e
1
2
V
'
- V = V
V - V ' = V
1 2
(1)
(2)
(3)
(4)
& because of
symmetry V1 = - V2
By substituting these in above equation (3)
•
-T2RC -
V=V1.e V2
-T2RC+
V=V1.e V1
-T2RC
V=V1(1+e )
I
Equation (1)
II
ForRC>>T theequation(I)&(II)becomesas
1
2
V1 ≅ V (1+ T
2 4RC 2 4RC
) & V1 ≅ V (1- T )
1
V1 -V1
Thepercentage tilt ‘P’ is definedby P=
V
2
× 100
High Pass RC Circuit acts as Differentiator:-
and the output signal across R is
V0 = iR
V0 = RC
Low Pass RC Circuit
C
X =
1
2Πf
C
If f=low, Xc becomes high
C act as open circuit, so we get the output.
If f=high, Xc becomes low
C acts as short circuit, so Vo=0.
As the lower frequency signals appear at the output, it is called as
“Low pass RC circuit”.
Sinusoidal input
in
V
× XC
j XC
O
V =
R +
j
1
2πfC
C
X =
in
V
×
1
j ω C
1
O
V =
R +
j ω C
wh
ere
O
Vin Vin
V=
=
jωRC+1 1+j2πfRC
CS
o
V = 1 i
O
Vin
V =
1+ j f
f
2
2
1
2πRC
where f =
A = VO
Vin
=
1
1 + j
f
f2
A
= 1
2
1 + ⎛ f ⎞
⎜ f ⎟
⎝ 2 ⎠
θ=- tan -1 ⎛ f
⎟
⎜f
⎝⎞ 2 ⎠
and
At the frequency f = f2
Vin
VO =
1 = 1 =0.707
1+1 2
A =0.707
At f = f2 the gain is 0.707 or this level corresponds to a signal reduction of 3 decibels(dB).
∴ f2 or fh is referred to as upper 3-dB frequency.
Square wave input
0 0to
The time required for the voltage to rise from 10 900 0of the final steady value is called “Rise Time”.
Vd.c.
V’
V01
V02
V’
V2
V2
V1
T1
V’’
T2
= V1 + (V1-V 1 ) . e -T 1 RC
The output voltage V01 & V02 is givenby
………………… (1)
………………… (2)
V01
V
02
= V11 + (V2-V 11 ) . e - T 2 RC
if we set
and
V01 = V2 at t=T1
V02 = V1 at t= T1+T2
1 1
V2= V +( V1-V ) e
11 11
1
V =V +( V2-V )
-
- T 1 RC
T
2
RC
e
Since the average across R is zero then the d.c voltage at the output is same as that of the
input. This average value is indicated as Vd.c.
Consider a symmetrical square wave with zero average value, so that
- T
V ⎡1 - e - T 2RC ⎤
V = ⎢ ⎥
2
⎢⎣1 + e
2RC ⎥⎦
T
V ⎡ e T 2RC - 1 ⎤
⎥
V2 = 2 ⎢
⎢⎣e
2RC + 1⎥⎦
T 4RC
2
2 e2x + 1
V = V . e2x - 1
where x =
2
V = V tan hx
2
Low pass RC circuit acts as an integrator
“Integrator”.
Vi =iR
For low pass RC circuit the output voltage Vo is given by
O
V
= 1
i dt
C ∫
O
V = 1 Vi
dt
C∫R
O
i
V = 1
V dt
RC∫
Advantages of Integrator over differentiator
computer applications for the following reasons.
differentiator may over load.
Non-Linear Wave Shaping
Definition:
The process where by the form of a sinusoidal signals are going to be altered by transmission through a non-linear network is called Non-linear Wave Shaping.
Non-linear elements (like Diode, transistor) in combination with resistors can function as clipper circuit.
Types:
Clipper Classifications
According to biasing, the clippers may be classified as
According to configuration used the clippers may be
and reference voltages
Contd…
According to level of clipping the clippers may be
Clipper
Voltage limiters
Current limiters Amplitude selectors Slicers
Unbiased Clippers( Parallel Positive Clippers)
be the negative portion of the input wave (assuming the
bottom node is grounded). When vi > 0, the diode is on (short-circuited), vi is dropped across R and vo=0. When vi
<0, the diode is off (open-circuited), the voltage across R is
zero and vo=vi.
Unbiased Clippers( Parallel Negative Clippers)
+ive cycle :- anode is at ground potential and cathode sees variable +ive voltage from 0 to +Vm So complete cycle, the diode is reverse biased and Vo =Vin.At positive peak Vo=+5V
-ive cycle :- anode is at ground potential and cathode sees variable -ive vols from 0 to –Vm. When magnitude of in put volatge i.e / Vin/ >Vd, the diode become forward biased and hence Vo =-Vd =0.7V
Series Positive Clipper
+ive cycle :- anode is at ground potential and cathode sees variable +ive voltage from 0 to +Vm.For comlpete, cycle, diode become reverse biased and hence Vo =0V
-ive cycle :- anode is at ground potential and cathode sees variable -ive voltage from 0 to –Vm. So in complete cycle, the diode is forward biased and Vo= Vin + Vd andAt negative peak, Vo= -Vm+ Vd = -5v
Series Negative clipper
+ive cycle :- anode is at positive potential from 0 to +Vm.For comlpete, cycle, diode become forward biased and hence vo= 5v
-ive cycle :- Cathode is at ground potential and cathode sees variable - ive voltage from 0 to –Vm. So in complete cycle, the diode is Reverse biased and negative peak, Vo= 0
Positive Shunt clipping with zero reference Rvoltage
D
Vo
Vi
Transfer characteristics equations:
VO=0for Vi>0 VO = Vi for Vi< 0
D– ON
VO=Vγfor Vi >Vγ VO=Vi for Vi < Vγ
D– OFF
[Ideal]
V
O
V
i
V
O
V
i
Slope =1
Vγ
Vγ
Input
Input
Output
Positive Shunt clipping with positive reference Voltage
D
Vi Vo
Transfer characteristics equations:
Vi < VR+Vγ
D – OFF
VO = Vi
D – ON
Vi > VR+Vγ
VO = VR+Vγ
Input
VR + Vγ VR + Vγ Output
VO
VR VO
Vi
Positive Shunt clipping with negative
reference voltage
R
D
VR
Vi
Vo
Transfer characteristics equation:
Vi > Vγ - VR
D – ON VO
D – OFF
= Vγ - VR Vi < Vγ - VR VO = Vi
VO
VO
Vi
V
i
Input
Output
Vi
Negative Shunt clipping with zero reference voltage
R
Vi
Vo
D
Transfer Characteristic
Equations:
Vi > -Vγ D – OFF
VO = Vγ
Vi < -Vγ D – ON
VO = -Vγ
-Vγ
-Vγ
VO
VO
Vi
Vi
Input
Output
Negative Shunt clipping with positive
reference voltage
R
D
VR
Vi
Vo
Transfer Characteristics
Equations:
D – ON
Vi < VR-Vγ
VO = VR-Vγ
Vi > VR-Vγ
D – OFF
VO = Vi
VR - Vγ
VO
Vi
Vi
VO
DOFF
DON
Negative Shunt clipping with negative reference voltage
R
D
V
R
Vi
Vo
Transfer characteristic
equations:
D – ON VO
Vi < -( Vγ + VR)
= -( Vγ + VR)
Vi < -( Vγ + VR) D – OFF
VO = Vi
V
O
VO
Vi
Vi
- (Vγ + VR
Input
Negative Series clipper with zero
reference
R
Vi
Vo
D
Transfer characteristicequations:
Vi<0 Vi>0 | D–OFF D–ON | VO=0 VO=Vi | IdealDiode |
Vi<Vγ | D–OFF | VO=0 | |
Vi>Vγ | D–ON | VO=Vi-Vγ | PracticalDiode |
VO
VO
Vi
Vi
Output
CLIPPING AT TWO
INDEPENDENT LEVELS
R
D
VR
Vi
Vo
D
VR
Transfercharacteristic equations:
Inp ut (Vi) | DiodeState | Outp ut (VO) |
Vi ≤ VR 1 | D1 –ON,D2 – OFF | VO =VR 1 |
VR < Vi < VR 1 2 | D1 –OFF,D2 – OFF | VO =Vi |
Vi ≥ VR2 | D1 –OFF,D2 – ON | VO=VR 2 |
V
O
VO
Vi
Vi
Input
Output
VR
1
Contd..
CLAMPING CIRCUIT
There are two types of clamping circuits.
Diode :- Clamper
Positive Clamper
The circuit for a positive clamper is shown in the figure. During the negative half cycle of the input signal, the diode conducts and acts like a short circuit. The output voltage Vo
⇒ 0 volts . The capacitor is charged to the peak value of input voltage Vm. and it behaves like a battery. During the positive half of the input signal, the diode does not conduct and acts as an open circuit. Hence the output voltage Vo ⇒Vm+ Vm This gives a positively clamped
voltage.
Vo ⇒Vm+ Vm = 2
V
m
Negative Clamper
During the positive half cycle the diode conducts and acts like a short circuit. The capacitor charges to peak value of input voltage Vm.
During this interval the output Vo which is taken across the short circuit will be zero During the negative half cycle, the diode is open. The output voltage can be found by applying KVL.
Biased Clamper
CLAMPING CIRCUIT THEOREM
∴ Af = Rf
Ar R
Consider a square wave input is applied to a clamping circuit under steady state condition
If Vf (t) is the output waveform in the forward direction, then from below figure
the capacitor charging current is
V
if = f
Rf
Therefore the charge acquired by the capacitor during the
forward interval
i dt =
V dt =
T1 T1
∫ f
1
Rf
A
f
Rf
∫ f
0 0
…………….. (1)
1 Ar
T2 T2
∫ir dt = R∫ Vr dt = R
…………….. (2)
T1 T2
In the steady-state the net charge acquired by the capacitor must be zero.
Therefore from equation (1) & (2) this equation says that
for any input waveform the ratio of the area under the output voltage curve in the forward direction to the reverse direction is equal to the ratio .
Clamping Circuit taking Source and Diode Resistances into account
Practical Clamping circuit
Effect of diode characteristics on clamping voltage
Synchronized Clamping