Transcript For negative half cycle

3RD SEM EC
ENROLLMENT
NO:130600111021
HAIDERY HUSAINA G.
CLIPPER
 The
clipping circuits using diodes have
the ability to “clip” off or remove a
portion of the input signal without
distorting the remaining part of the
waveform.
1) Series clippers
2) Parallel clippers
●The purpose of the diode is that when it is
turn on, it provides the clip value
●Clip value = V’. To find V’, use KVL at L1
●The equation is : V’ – VB - V = 0 
●V’ = VB + V
Then, set the conditions
If Vi > V’, what happens? 
diode conducts, hence
Vo = V’
If Vi < V’, what happens? 
diode off, open circuit,
no current flow, Vo = Vi
Series clipper
The basic cofiguration
of a series clipper is to “clip off” the
half cycle (negative or positive) of
the applied voltage.
BLOCK DIAGRAM
Series clippers
Positive clipper
Negative clipper




In the positive half cycle of the
sinusoidal input,the diode is forwad
biased.Being an ideal diode,it acts as a
closed switch and connects the load
across the input.The load voltage is
therefore equal to the input voltage in the
positive half cycle.
In the negative half cycle of the input the
diode is reverse biased and acts as an
open circuited switch.The load voltage is
therefore zero during the negative half
cycle.
The negative half cycle is thus “clipped
off “or “shaved off” by the series
negative clipper.
Vice versa for “positive seires clipper”
Parallel clipper
circuit:
In
this circuit,the clipping
diode is connected in
branch which is parallel to
the load .
The diode is assumed to be
an ideal one.Resistor R
controls the current flowing
through diode D.
BLOCK DIAGRAM
Parallel clipper
Positive clipper
Negative clipper






For positive half cycle:
Diode is forwad biased.
The output voltage will therefore be
zero in the positive half cycle of the
input voltage . Thus the positive
half cycle “clipped off”.
For negative half cycle:
The diode does not conduct and
therefore acts as an open switch.
The load voltage is equal to the
instantaneous input voltage.Thus
negative half cycle appears as it is
across the load.


By reversing the
direction of the diode in
the parallel positive
clipper we can get the
parallel negative clipper
The diode will conduct
only in the negative half
cycle of the input to clip it
off.The positive half cycle
will appear as it is across
the load.
What if the diode isVi non-ideal?
C
+
10
+
-4.3
Vi
t
Vo
-10
-
5V
-
-14.3
The diode is a nonideal with V = 0.7V
-24.3
Step 1: VC + V - VB – Vi = 0  VC = 10 + 5 – 0.7 = 14.3V
Step 2: VO – Vi + VC = 0  VO = Vi – 14.3.
THANKING
YOU