l7 clipper clamper logic gates

8/10/2019 L7 Clipper Clamper Logic Gates

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Clipper and Clamper Circuits


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Clippers

Clipper circuits, also called limiter circuits, are used to eliminate

portion of a signal that are above or below a specified level.

The level here is referred to as clip value, V. To find V, use KVL

at L1

The equation is : V VB- V= 0V = VB+ V

Then, set the conditions

If Vi > V, diode conducts, hence Vo = V

If Vi < V, diode off, open circuit, no current flow, Vo = Vi

L1

Vi

V = VB+ V


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Clippers

Other clipping circuits can be constructed by reversing the diode,

or the polarity of the voltage VB.

V = VB- V

conditions: Vi > V off, Vo = Vi

Vi < V conducts, Vo = V

V = - VB+ V

conditions: Vi > V conducts, Vo = V

Vi < V off, Vo = Vi

V = - VB- V

conditions: Vi > V off, Vo = Vi

Vi < V conducts, Vo = V


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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


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EXAMPLE 1

Consider the parallel clip circuit shown below.

Assume the VZ1= 6V and VZ2= 4V and V=

0.7V. Given vi = 10 sin t, sketch vO


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Clampers Clamping shifts the entire signal voltage

by a dc level.

Consider, the sinusoidal input voltage

signal, vI.

1st 900, the capacitor is charged up to

the peak value of Vi which is VM.

Then, as Vi moves towards thevecycle,

the diode is reverse biased.

Ideally, capacitor cannot discharge,

hence Vc = VM

By KVL, we get

NOTE: The input signal is shifted by a dc

level; note that the peak-to-peak value is

the same


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Clampers

STEP 1: Knowing what value that the capacitor is charged to. Using KVL,

VC+ VBVS= 0 VC= VMVB

STEP 2: When the diode is reversed biased and VCis already a constant

value

VOVS+ VC= 0 VO= VSVC.

A clamping circuit that includes an independent voltage sourceV

B.Peak value VM


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EXAMPLESclampers with ideal diode

C

+

Vo

-5V

+

Vi

-

Vi

t

-10

10

Vi

t-8

10C

+

Vo

-5V

+

Vi

-

Step 1: VC- VBVi= 0 VC= 10 + 5 = 15V

Step 2: VOVi+ VC= 0 VO= Vi15.

-5

-25


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What if the diode is non-ideal?

C

+

Vo

-5V

+

Vi

-

Vi

t

-10

10

The diode is a non-ideal

with V= 0.7V

Step 1: VC+ V

+ VBV

i= 0 V

C= 10 + 50.7 = 14.3V

Step 2: VOVi+ VC= 0 VO= Vi14.3.

-4.3

-24.3


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Multiple Diode Circuits


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DIODE ID VD

OFF 0 VD< V

ON ID> 0 VD= V


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OR GATE

V1 V2 VO

Vo = voltage

across R

D1 and D2 off; no current flow, 0 0 0

D1 off, D2 on, current flow,

VoV2 + V= 0

0 5V ( 1 ) 4.3V

D1 on, D2 off, current flow,

VoV1 + V= 0

5V ( 1 ) 0 4.3V

Both on, using both loops will

give the same equation

5V ( 1 ) 5V ( 1 ) 4.3V


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V1 V2 VO

Both on, using both loops will

give the same equation

0 0 0.7

D1 on, D2 off 0 5V ( 1 ) 0.7

D1 off, D2 on 5V ( 1 ) 0 0.7V

Both are off; open circuit no

current flowing through R since

no GND destination

5V ( 1 ) 5V ( 1 ) 5V

AND GATEVo = node

voltage

l7 clipper clamper logic gates

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