Transient analysis of clamping circuits

ANIL NEERUKONDA INSTITUTE OF TECHNOLOGY & SCIENCES(A)
Department of Electronics and Communication Engineering
ECE 216 Electronic circuits and Analysis-I
➢Academic year : 2020-21
➢Class & Section : 2/4 ECE-C
➢Name of the Faculty : Mr.D.Anil Prasad
ANIL PRASAD DADI/DEPT OF ECE/ANITS

ANIL NEERUKONDA INSTITUTE OF TECHNOLOGY & SCIENCES(A)
Department of Electronics and Communication Engineering
Non Linear Wave shaping circuits
Transient Analysis

Transient Analysis
V0=Vi +Vm

Transient Analysis
V0=Vi -Vm

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t0
10
RL
C
+
-
VS V0
RS
t=0+ VS=10V
VC(0
-)=0=VC(0+) Capacitor is initially uncharged
Diode is FB
+
-
RL
+
-
VS V0
RS
-
C
Rf
i
(+)0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
t=0+ VS=10V
VC(0
Diode is FB
+
-
+
-
VS V0
RS
-
C
Rf
i
0
(+)0

+
-
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
t=0+ VS=10V
VC(0
Diode is FB
+
-
VS
V0
-
C Rs
Rf
+
V0=VS
𝑅𝑓
𝑅𝑓+𝑅𝑠
V0=10
100
100+100
V0=5V
5
i
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
0<t<T/2 VS=10V Diode is FB
V0(t)=Vf - (Vf-Vi)𝑒−
(𝑡−𝑡𝑖)
𝜏
Vf=0V ti=0 Vi=5V τ=(Rs+Rf)C
5
+
-
VS
V0
-
Rs
Rf
i
+
C
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
+
-
VS
V0
-
Rs
Rf
i
+
C
V0(t)=5𝑒−
𝑡
𝜏 V0(t=
𝑇
2
)=5𝑒−
𝑇
2𝜏 V0(t=
𝑇
2
)≈ 3V
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
+
-
VS
V0
-
Rs
Rf
i
+
V0(t=
𝑇
2
)=5𝑒−
𝑇
2𝜏 V0(t=
𝑇
2
)≈ 3V VC(t=
𝑇
2
)=?
+ VC -
VC(t=
𝑇
2
)= Vs − Vi VC(t=
𝑇
2
)= 10 − (3 + 3) VC(t=
𝑇
2
)= 4
VC(t=
𝑇
−
2
)=VC(t=
𝑇
+
2
)=4
Vi
-
+
0

+ VC -
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
V0(t=
𝑇
2
)≈ 3V VC(t=
𝑇
−
2
)=VC(t=
𝑇
+
2
)=4 V0(t=
𝑇
+
2
)=?
t=
𝑇
+
2
VS=0V Diode is RB
RL
+
-
VS V0
RS
-
Rr
0
(+)0

+ VC -
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
V0(t=
𝑇
2
)≈ 3V VC(t=
𝑇
−
2
)=VC(t=
𝑇
+
2
)=4 V0(t=
𝑇
+
2
)=?
V0(t=
𝑇
+
2
)= Vs − Vc
= 0 − 4 = −4 V0(t=
𝑇
+
2
)= -4V
t>
𝑇
+
2
VS=0V Diode is RB
RL
+
-
VS V0
RS
-
since drop across Rs≈0
0
+

-
-
+ +
RS
+
-
+
-
+ VC -
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
CVS V0
RS
-5
3
RLVS
V0
+
-
Rs
Rfi
+
VS
V0
C
0<t<T/2 VS=10 Diode FB
V0(t=
𝑇
2
)=5𝑒−
𝑇
2𝜏 V0(t=
𝑇
2
)≈ 3VV0(t)=5𝑒−
𝑡
𝜏
VC(t=
𝑇
−
2
)=VC(t=
𝑇
+
2
)=4
V0(t=
𝑇
+
2
)= Vs − Vc
t=
𝑇
+
2
VS=0V Diode is RB
Rs≈0
V0(t=
𝑇
+
2
)=?
V0(t=
𝑇
+
2
)= -4V
-4
0

+
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
T/2<t<T VS=0V Diode is RB
V0(t)=Vf - (Vf-Vi)𝑒−
(𝑡−𝑡𝑖)
𝜏
Vf=0V ti=
𝑇
2
Vi= -4V τ=(Rs+RL)C
3
RS
-
+
-
+ VC - RLVS V0
0

+
Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
T/2<t<T VS=0V Diode is RB
3
RS
-
+
-
+ VC - RLVS V0V0(t)= -4𝑒−
(𝑡−
𝑇
2
)
𝜏 V0(T)= -4V
VC(t= T)= Vs − V0 VC(t= T)= 0 − −4 = 4V
VC(t= T -)= VC(t= T +) = 4V
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
t=T+ VS=10V Diode is FB
V0(T)= -4V VC(t= T -)= VC(t= T +) = 4V
V0(T+)= ? Vi= Vs − Vc = 10 − 4 = 6V
V0(T+)= 3V
3
V0
+
T/2<t<T
i
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
T<t<3T/2 VS=10V Diode is FB
V0(t)= 3𝑒−
(𝑡−𝑇)
𝜏
V0
V0(t=
3𝑇
2
)= 1.8V
Vc(t=
3𝑇
2
)= Vs-Vi Vc(t=
3𝑇
2
)= 10-(1.8+1.8)
Vc(t=
3𝑇
−
2
)= Vc(t=
3𝑇
+
2
)= 6.4
+
i
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
V0(t=
3𝑇
+
2
)=?
t=
3𝑇
+
2
VS=0V Diode is RB
+
-
VS
-
Rs
RL
+
6.4V
V0
V0(t=
3𝑇
+
2
)=Vs-Vc =-6.4
V0(t)= 3𝑒−
(𝑡−𝑇)
𝜏 V0(t=
3𝑇
2
)= 1.8V
Vc(t=
3𝑇
−
2
)= Vc(t=
3𝑇
+
2
)= 6.4
T<t<3T/2 + -
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
+
-
VS
-
Rs
RL
+
6.4V
V0V0(t=
3𝑇
+
2
)= -6.4
V0(t)= 3𝑒−
(𝑡−𝑇)
𝜏 V0(t=
3𝑇
2
)= 1.8V
Vc(t=
3𝑇
−
2
)= Vc(t=
3𝑇
+
2
)= 6.4
T<t<3T/2 + -
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
+
-
VS
-
Rs
RL
+
C
V0
+ -3T/2<t<2T VS=0V Diode is RB
Vc(t=2T-)=Vs-V0=0-(-6.4)=6.4
Vf=0V ti=
3𝑇
2
Vi= -6.4V τ=(Rs+RL)C
V0(t)= -6.4𝑒−
(𝑡−
3𝑇
2 )
𝜏 V0(t=2T)= -6.4V
Vc(t=2T-)= Vc(t=2T+)= 6.4
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
+
-
VS
-
Rs
Rf
+
+ 6.4 -
V0
3T/2<t<2T V0(t=2T-)= -6.4V
Vc(t=2T-)= Vc(t=2T+)= 6.4 V0(t=2T+)= ?
t=2T+ VS=10V Diode is FB
Vi=Vs-Vc =10-6.4=3.6
Vi
-
+
V0(t=2T+)= 1.8V
i
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
+
-
VS
-
Rs
Rf
+
Vc
V0
-
2T<t<5T/2 VS=10 Diode is FB
Vi
-
+
Vf=0V ti= 2T Vi= 1.8V τ=(Rs+Rf)C
V0(t)= 1.8𝑒−
(𝑡−2𝑇)
𝜏 V0(
5𝑇
2
)= 1.1V
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
+
-
VS
-
Rs
RL
+
7.8V
V0
-
2T<t<5T/2
Vi
-
+
V0(
5𝑇
2
)= 1.1V Vc(
5𝑇
−
2
)= Vc(
5𝑇
+
2
)= 7.8V V0(
5𝑇
+
2
)= ?
V0(t=
5𝑇
+
2
)= Vs − VC = 0 − 7.8
V0(t=
5𝑇
+
2
)= -7.8V
0
Since Rs≈0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
5T/2<t<3T VS=0 Diode is RB
Vf=0V ti=
5𝑇
2
Vi= -7.8V τ=(Rs+RL)C
V0(t)= -7.8𝑒−
(𝑡−
5𝑇
2
)
𝜏 V0(3T)= -7.8V
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
5T/2<t<3T
V0(3T)= -7.8V
Vc(3T)=VS-V0=0-(-7.8)=7.8V
Vc(3T-)= Vc(3T+)= 7.8V
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
V0(3T)= -7.8V Vc(3T-)= Vc(3T+)= 7.8V
V0(3T+)= ?
t=3T+ VS=10V Diode is FB
Vi=Vs-Vc =10-7.8=2.2
V0(t=3T+)= 1.1V
5T/2<t<3T
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T
7𝑇
2
t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
V0(3T+)= 1.1V
V0(
7𝑇
2
)= 0.66 Vc(
7𝑇
2
)= 8.68V
t=
7𝑇
2 VS=0V Diode is RB
V0=Vs-Vc =0-8.68=-8.68
V0(t=
7𝑇
+
2
)= -8.68V
3T<t<7T/2
0

Transient Analysis
VS
𝑇
2
T
3𝑇
2
2T
5𝑇
2
3T
7𝑇
2
t
10
RL
C
+
-
VS V0
RS
+
-
5
-4
3
0

The clamping circuit Theorem
• The clamping circuit theorem states that under steady-state
conditions for any input waveform, the ratio of the area under the
output voltage curve in the forward direction to that in the reverse
direction is equal to the ratio Rf /RL.
• This theorem enables us to find the voltage level to which the output is
clamped.
𝐴𝑓
𝐴𝑟
=
𝑅𝑓
𝑅 𝐿

• To derive the clamping circuit theorem, consider a typical steady-state
output for the clamping circuit

• In the interval t1 to t2 D is ON. Hence during this period, the charge builds
up on the capacitor C.
• Let if is the diode current , the charge gained by the capacitor during the
interval t1 to t2 is:
q1= ‫׬‬𝑡1
𝑡2
𝑖 𝑓 𝑑𝑡 =
1
𝑅 𝑓
‫׬‬𝑡1
𝑡2
𝑉𝑓 𝑑𝑡

• In the interval t2 to t3 D is OFF. Hence the capacitor discharges and charges
lost by C is:
q2= ‫׬‬𝑡2
𝑡3
𝑖 𝑟 𝑑𝑡 =
1
𝑅 𝐿
‫׬‬𝑡2
𝑡3
𝑉𝑟 𝑑𝑡

• At steady-state, the charge gained is equal to the charge lost. In other
words, q1 = q2
1
𝑅 𝑓
‫׬‬𝑡1
𝑡2
𝑉𝑓 𝑑𝑡=
1
𝑅 𝐿
‫׬‬𝑡2
𝑡3
𝑉𝑟 𝑑𝑡
𝐴𝑓 = න
𝑡1
𝑡2
𝑉𝑓 𝑑𝑡
𝐴𝑟 = න
𝑡2
𝑡3
𝑉𝑟 𝑑𝑡
𝐴𝑓
𝑅𝑓
=
𝐴𝑟
𝑅 𝐿
q1=
1
𝑅 𝑓
‫׬‬𝑡1
𝑡2
𝑉𝑓 𝑑𝑡
q2=
1
𝑅 𝐿
‫׬‬𝑡2
𝑡3
𝑉𝑟 𝑑𝑡
𝐴𝑓
𝐴𝑟
=
𝑅𝑓
𝑅 𝐿

Problem #1
• A unsymmetrical square wave with T1=1ms and T2=1us has an amplitude of 10V. This
signal is applied to the circuit shown in figure Rf=50 RL=50K. Find the level to which
output is clamped to
T1
T2
0 V1
RL
C
-
VS V0
-
+ +
Af
Ar
10V

T1
Problem #1
signal is applied to the circuit shown in figure Rf=50 RL=50K
T2
0 V1
RL
C
-
VS V0
-
+ +
Af
Ar
𝐴𝑓
𝐴𝑟
=
𝑉1 𝑥𝑇1
10 − 𝑉1 𝑥𝑇2
=
𝑉1 𝑥1000μ
10 − 𝑉1 𝑥1μ
𝑅𝑓
𝑅 𝐿
=
50
50𝐾
=10-3 𝐴𝑓
𝐴𝑟
=
𝑅𝑓
𝑅 𝐿
𝑉1 𝑥1000
10−𝑉1 𝑥1
=10-3 V1=10-5
10V
(10-V1)

Problem #2
• A unsymmetrical square wave with T1=1us and T2=1ms has an amplitude of 10V. This
T2
T1
0
V1
RL
C
-
VS V0
-
+ +
Af
Ar
𝐴𝑓
𝐴𝑟
=
𝑉1 𝑥𝑇1
10 − 𝑉1 𝑥𝑇2
=
𝑉1 𝑥1μ
10 − 𝑉1 𝑥1000μ
𝑅𝑓
𝑅 𝐿
=
50
50𝐾
=10-3 𝐴𝑓
𝐴𝑟
=
𝑅𝑓
𝑅 𝐿
𝑉1 𝑥1
10−𝑉1 𝑥1000
=10-3 V1=5V
10V
(10-V1)

Problem #3
T2
T1
0
V1
RL
C
-
VS V0
-
+ +
Af
Ar
𝐴𝑓
𝐴𝑟
=
𝑉1 𝑥𝑇1
10 − 𝑉1 𝑥𝑇2
=
𝑉1 𝑥1000𝑢
10 − 𝑉1 𝑥1μ
𝑅𝑓
𝑅 𝐿
=
50
50𝐾
=10-3 𝐴𝑓
𝐴𝑟
=
𝑅𝑓
𝑅 𝐿
𝑉1 𝑥1000
10−𝑉1 𝑥1
=10-3
V1=10-5V
10V(10-V1)

5V
Design of diode clamper
• Design a diode clamper to restore the positive peaks of the input signal to a voltage level
of 5V. Assume Vγ=0.5V f=1KHz, Rf=1K Rr=200K and RLC=10T
• When diode conducts
❑RL>>Rf
❑Let RL=kRf
• When diode doesn’t conduct
❑Rr>>RL
❑Let Rr=kRL
RL
C
-
VS V0
-
+ +
RL =
R𝑟
RL
𝑅 𝑓

5V
Design of diode clamper
• Design a diode clamper to restore the positive peaks of the input signal to a voltage level
of 5V. Assume Vγ=0.5V f=1KHz, Rf=1K Rr=200K and RC=10T
RL
C
-
VS V0
-
+ +
RL =
R𝑟
RL
𝑅 𝑓
R2
L = 𝑅 𝑟 𝑅 𝑓
RL = 𝑅 𝑟 𝑅 𝑓 RL = 14.14KΩ
RC=10T C=
10x1
f xRL
C=0.707μF
Let VR=VR1+ Vγ Where VR1=5V is given VR=5+0.5=5.5V

Transient analysis of clamping circuits

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