Electronic Clipper Circuits and Rectifier Comparisons

Posted on Feb 19, 2026 in Communications Electronic Engineering

1. Define Clipper, and explain clipping at two independent levels?

A.A clipper (also known as a limiter) is an electronic circuit used to remove or “clip” portions

Of a signal that exceed a certain threshold voltage level without distorting the remaining

Part of the waveform. Clippers are used in various applications, such as signal conditioning,

Waveform shaping, and protecting circuits from voltage spikes

Clipping at Two Independent Levels:

Clipping at two independent levels involves limiting the signal so that it does not exceed a specified

Upper voltage level and does not go below a specified lower voltage level. This type of clipper

Is also known as a dual-diode clipper or a window clipper

Circuit Description:

A typical dual-diode clipper circuit consists of two diodes, two reference voltage sources (one for

The upper threshold and one for the lower threshold), and resistors for current limiting. The circuit

Is designed to clip both the positive and negative extremes of the input signal independently

1. Positive Clipping:

O When the input voltage exceeds the upper threshold voltage VU, the diode

Connected to VU becomes forward-biased and starts conducting

O Any input voltage above VU is clipped off

2. Negative Clipping:

O When the input voltage goes below the lower threshold voltage VL, the diode connected to

VL becomes forward-biased and starts conducting

O Any input voltage below VL is clipped off

Applications:

Signal Conditioning: Protecting sensitive components from voltage spikes

Waveform Shaping: Modifying the shape of signals in communication systems

Audio Processing: Limiting audio signals to prevent distortion

Voltage Regulation: Ensuring voltage levels stay within safe limits

2.Compare half wave rectifier, full wave rectifier and bridge rectifier?

Comparison of Rectifiers

3.Prove the efficiency of full wave rectifier is 81.2%?

To prove the efficiency of a full-wave rectifier, we need to calculate the ratio of the

DC power output to the AC power input. Efficiency (n) is given by:

N=PDC/PAC X 100%

Step-by-Step Calculation:

1. Calculate DC Power Output (PDC):

. The DC component of the output voltage VDc for a full-wave rectifier is:

VDC=2Vm/

Where Vm is the peak voltage of the AC input

· The DC power output is given by:

PDC=V2DC/RL=92VM/   )2/RL=4V2m/  2RL

2. Calculate AC Power Input (PAC):

. The total RMS voltage of the input is:

VRMS = Vm/underroot symbol 2

· The AC power input is given by:

PAC = V2RMS/RL=9VM/UNDERROOT2)2/RL=V2m/2RL

3. Calculate Efficiency (n):

. Now we can find the efficiency by taking the ratio of PDc to PAC:

N = PDC/PAC × 100%

N = 4V2m/underroot2RL/V2m x 100%

N=4V2m.2RL/N2RL.V2m X 100%

N=8/pie3 x 100%

4. Simplify:

. Simplifying the above equation:

N= 8/pie2 × 100%

N =8/9.8696 × 100%

N~~ 0.81×100%

N~81.2%

4. A HWR has a load of 3.5kΩ.If the diode resistance and secondary coil

Resistance together have a resistance of 800Ω and the input voltage has

A single voltage of peak value 240V calculate?

A) Peak, average, RMS value of current flowing

B) DC power output

C) AC power input

D) Efficiency of rectifier

A.To solve the given problem for a Half Wave Rectifier (HWR), we’ll follow a

Systematic approach using the given values:

Load resistance RL = 3500 o

. Diode and secondary coil resistance Rs= 800 o

. Total resistance Rtotal = RL + Rs = 3500 o + 800 o = 4300 o

. Peak input voltage Vm = 240 V

(a) Peak, Average, and RMS Value of Current:

1. Peak Current (Imax):

Imax=Vm/Rtotal=240V/4300o ~ ~   0.0558 A

2. Average Current (Iavg):For a half-wave rectifier:

Iavg = Imax/pie ~ ~  0.0558 A/3.1416  ~ ~ 0.0178 A

3. RMS Current (Irms):For a half-wave rectifier:Irms = Imax/2 ~ ~ 0.0558 A/2 ~ ~  0.0279 A

(b) DC Power Output (PDc)

PDC = I2avg x RL = (0.0178 A)2 X 3500 o ~ ~  3.35w

(c) AC Power Input (PAC)

PAC = I2avg X Rtotal = (0.0279 A)2 x 4300o ~ 3.35 W

(d) Efficiency of Rectifier (n)

M=PDC/PAC X 100% = 1.11 W/3.35W X 100% ~ ~  33.13%

Summary of Results

· Peak Current (Imax): 0.0558 A

· Average Current (Iavg): 0.0178 A

. RMS Current (Irms): 0.0279 A

. DC Power Output (PDC): 1.11 W

. AC Power Input (PAC): 3.35 W

. Efficiency (n): 33.13%