Full wave CenterTapped transformer

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Full wave CenterTapped transformer Rectifier
Circuit
Two diodes and a center-tapped transformer are
required.
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Operation of the CenterTapped Transformer
Rectifier Circuit
For the positive half of the AC
cycle For the negative half of the AC
cycle
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• No matter the value of the turns ratio the
  output is always for ideal
  diode
• or for
  practical model
• Where Vsec is the end to end voltage of the
  secondary transformer winding
• PIV gt2Vm for ideal diode and PIV gt2Vm 0.7 for
  practical model
• Note that Vm here is the transformer secondary
  voltage to the tap.
• VDC 0.636(Vpout) (since the area above the axis
  is double the area obtained for the half wave
  rectifier)
• Ir.m.sIpout/v2

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Example

• Show the voltage waveforms across each half of
  the secondary winding and across RL when a 100 V
  peak sine wave is applied to the primary winding.
  Also, what PIV rating must the diodes have?

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• Each diode must have a minimum PIV rating of 50
  V (neglecting diode drop). The waveforms are
  shown in Figure

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Full-Wave RectificationBridge Network

• The dc level obtained from a sinusoidal input can
  be improved 100 using a process called full-wave
  rectification.
• The most familiar network is bridge configuration
  with 4 diodes.

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Operation of the Bridge Rectifier Circuit
For the positive half of the AC cycle
For the negative half of the AC cycle
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FULL WAVE RECTIFICATION

• The rectification process can be improved by
  using more diodes in a Full Wave Rectifier
  circuit.
• Full Wave rectification produces a greater DC
  output.

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Example

• Determine the output voltage for the bridge
  rectifier in Figure . What PIV rating is required
  for the silicon diodes? The transformer is
  specified to have a 12 V rms secondary voltage
  for the standard 110 V across the primary.

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Solution

• The peak output voltage is (taking into account
  the two diode drops)
• VPsec 1.414Vrms 17 V
• VPout VPsec -1.4 15.6V
• The PIV for each diode is
• PIVVPout 0.7V15.6V0.7V16.3V

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POWER SUPPLY FILTERS(PEAK DETECTORS)

• The pulsating dc output of a half-wave rectifier
  or the output of a full-wave rectifier must be
  filtered to reduce the large voltage variations.
  The figure below illustrates the filtering
  concept showing a nearly smooth dc output voltage
  from the filter. The small amount of fluctuation
  in the filter output voltage is called ripple.

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Capacitor Filter

• A half-wave rectifier with a capacitor filter is
  shown in Figure . RL represents the equivalent
  resistance of a load. We will use the half-wave
  rectifier to illustrate the principle, and then
  expand the concept to full-wave rectification.
• During the positive first quarter-cycle of the
  input, the diode is forward-biased , allowing the
  capacitor to charge to within 0.7 V of the input
  peak.

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• When the input begins to decrease below its peak,
  as shown in part (b), the capacitor retains its
  charge and the diode becomes reversebiased.
  During the remaining part of the cycle, the
  capacitor can discharge only through the load
  resistance at a rate determined by the RLC time
  constant, which is normally long compared to the
  period of the input.

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• The larger the time constant, the less the
  capacitor will discharge. During the first
  quarter of the next cycle, the diode will again
  become forward-biased when the input voltage
  exceeds the capacitor voltage by approximately
  0.7 V.
• The variation in the output voltage due to the
  charging and discharging is called the ripple
  voltage.
• The smaller the ripple, the better the filtering
  action

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• For a given input frequency, the output frequency
  of a full-wave rectifier is twice that of a
  half-wave rectifier. This makes a full-wave
  rectifier easier to filter.
• The full-wave rectified voltage has a smaller
  ripple than does a half-wave voltage for the same
  load resistance and capacitor values.

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Ripple Factor

• The ripple factor is an indication of the
  effectiveness of the filter and is defined as
• where Vr is the peak-to-peak ripple voltage and
  VDC is the dc value of the filter's output
  voltage.
• The lower the ripple factor, the better the
  filter. The ripple factor can be lowered by
  increasing the value of the filter capacitor or
  increasing the load resistance.

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• For a full-wave rectifier with a sufficiently
  high capacitance filter, if VDC is within 10 of
  the peak rectified input voltage, then the
  expressions for the peak-to-peak ripple voltage,
  Vr and VDC are as follows
• where Vp(in) is the peak rectified full-wave
  voltage applied to the filter
• and f is 60 (50) Hz for a half-wave rectifier or
  120(100) Hz for a full-wave rectifier

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Example

• Determine the ripple factor for the filtered
  bridge rectifier .

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