Transcript Clipper

Recall Lecture 6
• Rectification – transforming AC signal into a
signal with one polarity
– Half wave rectifier
• Full Wave Rectifier
– Center tapped
– Bridge
• Rectifier parameters
– Duty Cycles
– Peak Inverse Voltage (PIV)
Clipper Circuits
Standard Clipper Circuits
●
Clipper circuits, also called limiter circuits, are used to eliminate
portion of a signal that are above or below a specified level – clip value.
●
The purpose of the diode is that when it is turn on, it provides the clip
value
i.
Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on
ii.
The equation is : V’ – VB - V = 0  V’ = VB + V
VI
V’ = VB + V
L1
iii.
Then, set the conditions
If
v > V’, what happens? 
diode conducts, clips and hence o = V’
If
v < V’, what happens? 
diode off, open circuit, no current flow, o =
I
I
v
v v
I
EXAMPLE
For the circuit shown below sketch the waveform of the output
voltage, vo. The input voltage is a sine wave where vI = 5 sin t.
VB is given as 1.3 V. Assume V = 0.6 V
1.3 V
Solution
i.
Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on
ii.
The equation is : V’ – 1.3 - 0.6= 0  V’ = 1.9 V
1.3 V
iii.
Then, set the conditions
If
If
v > V’, what happens? 
I
v < V’, what happens? 
v
I
diode conducts, clips and hence
v
o
= V’ = 1.9 V
diode off, open circuit, no current flow,
I
V’ = 1.9V
v =v
o
I
EXAMPLE
For the circuit shown below sketch the waveform of the output
voltage, vou. The input voltage is a sine wave where vI = 10 sin t.
Assume V = 0.7 V
+
+
vI
vo
-
-
Solution
i.
Find the clip value = V’. To find V’, use KVL at L1 assuming the diode is on
ii.
The equation is : V’ – 4 + 0.7= 0  V’ = 3.3 V
iii.
Then, set the conditions
If
v > V’, what happens? diode off, open circuit, no current flow, v = v
If
v < V’, what happens?  diode conducts, clips and hence v
o
I
o=
I
VI
L1
V’ = 3.3V
I
V’ = 3.3 V
Parallel Based Clippers

Positive and negative clipping can be performed simultaneously
by using a double limiter or a parallel-based clipper.

The parallel-based clipper is designed with two diodes and two
voltage sources oriented in opposite directions.

This circuit is to allow clipping to occur during both cycles;
negative and positive
Clipper in Series
ECE 1201
The input signal is an 8 V p-p square wave. Assume, V = 0.7 V. Sketch the output
waveform, vo.
c
b
a
vo
Initial stage: node a = 0V
Hence, node b = 1.5 V
So, node c of the diode must be at least 2.2 V
• Hence, the positive cycle of the square wave
has met the condition of  2.2 V
• Perform KVL as usual:
– 4 + 0.7 + 1.5 + vo = 0
vo = 1.8 V
4
1.8
-4
• What if now the input is change to 4 sin t?
• It has to wait for the input signal to reach 2.2 V.
Before that, the output, vo is zero as diode is off.
• Perform KVL as usual:
– 4 + 0.7 + 1.5 + vo = 0
vo = 1.8 sin t V
Peak value is 1.8 V
• So, basically, clipper in series clips at zero. It is
similar to half wave where the diode only
turns on during one of the cycle.
Clipper in Series
Problem 3.11
Figure P3.11(a) shows the input voltage of the circuit as shown in Figure
P3.11(b). Plot the output voltage vo of these circuits if V = 0.7 V
c
P3.11(a)
Initial stage: node a = 0V
Hence, node b = - 0.7 V
So, node c of the diode must be at least -1.9 V
b
P3.11(b)
a
• Hence, the negative cycle of the square wave
has met the condition of  |1.9| V
• Perform KVL as usual:
+ 3 - 0.7 - 1.2 + vo = 0
vo = - 1.1 V
vo
-
3V
+
-1.1
-1.1
• What if now the input is change to 5 sin t ?
• It has to wait for the input signal (negative cycle)
to reach  |1.9| V. Before that, the output, vo is
zero as diode is off.
• Perform KVL as usual:
+ 5 - 0.7 - 1.2 + vo = 0
vo = -3.1 sin t V
Peak value is approximately -3.1 V