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LAB#2b: HALF-WAVE RECTIFIER CIRCUIT WITHOUT AND ..., Summaries of Electronic Circuits Analysis

Objectives: 1. To construct a half-wave rectifier circuit and analyze its output. 2. To analyze the rectifier output using a capacitor in shunt as a filter.

Typology: Summaries

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Uploaded on 09/27/2022

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LAB#2b: HALF-WAVE RECTIFIER CIRCUIT WITHOUT AND WITH FILTER
Objectives:
1. To construct a half-wave rectifier circuit and analyze its output.
2. To analyze the rectifier output using a capacitor in shunt as a filter.
Overview:
The process of converting an alternating current into direct current is known as rectification.
The unidirectional conduction property of semiconductor diodes (junction diodes) is used for
rectification. Rectifiers are of two types: (a) Half wave rectifier and (b) Full wave rectifier. In a half-
wave rectifier circuit (Fig. 1), during the positive half-cycle of the input, the diode is forward biased
and conducts. Current flows through the load and a voltage is developed across it. During the negative
half-cycle, it is reverse bias and does not conduct. Therefore, in the negative half cycle of the supply,
no current flows in the load resistor as no voltage appears across it. Thus the dc voltage across the load
is sinusoidal for the first half cycle only and a pure a.c. input signal is converted into a unidirectional
pulsating output signal.
Fig.1: Half-wave rectifier circuit
Since the diode conducts only in one half-cycle (0-π), it can be verified that the d.c. component in the
output is Vmax/π, where Vmax is the peak value of the voltage. Thus,
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LAB#2b: HALF-WAVE RECTIFIER CIRCUIT WITHOUT AND WITH FILTER

Objectives:

  1. To construct a half-wave rectifier circuit and analyze its output.
  2. To analyze the rectifier output using a capacitor in shunt as a filter. Overview: The process of converting an alternating current into direct current is known as rectification. The unidirectional conduction property of semiconductor diodes (junction diodes) is used for rectification. Rectifiers are of two types: (a) Half wave rectifier and (b) Full wave rectifier. In a half- wave rectifier circuit (Fig. 1), during the positive half-cycle of the input, the diode is forward biased and conducts. Current flows through the load and a voltage is developed across it. During the negative half-cycle, it is reverse bias and does not conduct. Therefore, in the negative half cycle of the supply, no current flows in the load resistor as no voltage appears across it. Thus the dc voltage across the load is sinusoidal for the first half cycle only and a pure a.c. input signal is converted into a unidirectional pulsating output signal. Fig.1: Half-wave rectifier circuit Since the diode conducts only in one half-cycle (0-π), it can be verified that the d.c. component in the output is Vmax/π, where Vmax is the peak value of the voltage. Thus,

max V^ V^ max^0. 318 V dc    The current flowing through the resistor, R

V

I (^) dc  dc and power consumed by the load, P IdcR 2 . Ripple factor: As the voltage across the load resistor is only present during the positive half of the cycle, the resultant voltage is "ON" and "OFF" during every cycle resulting in a low average dc value. This variation on the rectified waveform is called "Ripple" and is an undesirable feature. The ripple factor is a measure of purity of the d.c. output of a rectifier and is defined as 1 1. 21

  1. 318

2 2 2 2 2 2    

dc rms dc rms dc dc output ac V

V

V

V V

V

V

r In case of a half-wave rectifier Vrms = Vmax/2 = 0.5Vmax. (How?) Rectification Efficiency: Rectification efficiency, η, is a measure of the percentage of total a.c. power input converted to useful d.c. power output.

R

r R r V

V

I r R

I R

V I V I

dcpowerdeliveredtoload acpowerat input ac d d^ d dc dc dc ac ac 1

2 max 2 max 2 2

Here rd is the forward resistance of diode. Under the assumption of no diode loss (rd<<), the rectification efficiency in case of a half-wave rectifier is approximately 40.5%. Filters: The output of a rectifier gives a pulsating d.c. signal (Fig.1) because of presence of some a.c. components whose frequency is equal to that of the a.c. supply frequency. Very often when rectifying an alternating voltage we wish to produce a "steady" direct voltage free from any voltage variations or ripple. Filter circuits are used to smoothen the output. Various filter circuits are available such as shunt capacitor, series inductor, choke input LC filter and π-filter etc. Here we will use a simple shunt capacitor filter circuit (Fig. 2). Since a capacitor is open to d.c. and offers low impedance path to a.c. current, putting a capacitor across the output will make the d.c. component to pass through the load resulting in small ripple voltage.

Procedure: i) Configure the half-wave rectifier circuit as shown in the circuit diagram. Note down all the values of the components being used. ii) Connect the primary side of the transformer to the a.c. Mains and secondary to the input of the circuit. iii) Measure the input a.c. voltage (Vac) and current (Iac) and the output a.c. (Vac), d.c. (Vdc) voltages using multimeter for at least 3 values of load resistor (Be careful to choose proper settings of multimeter for ac and dc measurement). iv) Multiply the Vac at the input by √2 to get the peak value and calculate Vdc using the formula Vdc = Vmax/ π. Compare this value with the measured Vdc at the output. v) Feed the input and output (in DC coupling mode) to the two channels of oscilloscope. We will use oscilloscope here only to trace the output waveform. Save the data for each measurement using SAVE/LOAD or STORAGE button of the oscilloscope. vi) Calculate the ripple factor and efficiency. vii) Connect an electrolytic capacitor (with –ve terminal connected to ground) across the output for each load resistor and measure the output a.c. and d.c. voltages once again and calculate the ripple factor. Trace the input and output waveforms in oscilloscope and notice the change. viii) Repeat the above measurement foe all values of capacitors and study the output. Observations:

  1. Code number of diode = ________
  2. Input Voltage: Vac = _________ Volt Table(I): Half wave rectifier w/o filter Sl. No Load R (kΩ) Input Current Iac (mA) Output Voltage Ripple Factor r Efficiency η (V^2 dc/R)/VacIac (%) Vac (Volt) Vdc (Volt) Vmax/ π (Volt) 1 2 3

Table(II): Half wave rectifier with filter (C = ____ μF) (Make separate tables for each capacitor) Sl. No Load R (kΩ) Output Voltage Ripple Factor Vac (Volt) Vdc (Volt) r 1 2 3 (III) Input and output waveforms: Waveforms without Filter: R = ______ Input Output (Paste data here) Waveforms with Capacitor Filter: C = ______ μF R = ______ Input Output (Paste data here) Discussions: Precautions: