Diode Circuits or Uncontrolled Rectifier

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Diode Circuits or Uncontrolled Rectifier

• Rectification The process of converting the
  alternating voltages and currents to direct
  currents

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The main disadvantages of half wave rectifier
are

• High ripple factor,
• Low rectification efficiency,
• Low transformer utilization factor, and,
• DC saturation of transformer secondary winding.

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Performance Parameters
rectification effeciency
form factor
ripple factor
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Single-phase half-wave diode rectifier with
resistive load.
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the load and diode currents
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Example 1 The rectifier shown in Fig.2.1 has a
pure resistive load of R Determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d)
Peak inverse voltage (PIV) of diode D1.
.

(d) It is clear from Fig2.2 that the PIV is
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Half Wave Diode Rectifier With R-L Load
Fig.2.3 Half Wave Diode Rectifier With R-L Load
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Single-Phase Full-Wave Diode Rectifier
Center-Tap Diode Rectifier
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PIV of each diode
Example 3. The rectifier in Fig.2.8 has a purely
resistive load of R Determine (a) The efficiency,
(b) Form factor (c) Ripple factor (d) TUF (e)
Peak inverse voltage (PIV) of diode D1 and(f)
Crest factor of transformer secondary current.
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The PIV is
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Single-Phase Full Bridge Diode Rectifier With
Resistive Load
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Example 4 single-phase diode bridge rectfier has
a purely resistive load of R15 ohms and, VS300
sin 314 t and unity transformer ratio. Determine
(a) The efficiency, (b) Form factor, (c) Ripple
factor, (d) The peak inverse voltage, (PIV) of
each diode, , and, (e) Input power factor.
The PIV300V
Input power factor
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Full Bridge Single-phase Diode Rectifier with DC
Load Current
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Example 5 solve Example 4 if the load is 30 A
pure DC
Input Power factor
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Effect Of LS On Current Commutation Of
Single-Phase Diode Bridge Rectifier.
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Example 6 Single phase diode bridge rectifier
connected to 11 kV, 50 Hz, source inductance
Ls5mH supply to feed 200 A pure DC load,
find (i) Average DC output voltage, (ii) Power
factor. And (iii) Determine the THD of the
utility line current.
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Three-Phase Half Wave Rectifier
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ThePIV of the diodes is
Example 7 The rectifier in Fig.2.21 is operated
from 460 V 50 Hz supply at secondary side and the
load ? resistance is R20. If the source
inductance is negligible, determine (a)
Rectification efficiency, (b) Form factor (c)
Ripple factor (d) Peak inverse voltage (PIV) of
each diode.
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The PIV
Vm650.54V
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Three-Phase Half Wave Rectifier With DC Load
Current and zero source induct
New axis
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Example 8 Solve example 7 if the load current is
100 A pure DC
The PIV
Vm650.54V
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Three-Phase Full Wave Rectifier With Resistive
Load
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Example 10 The rectifier shown in Fig.2.30 is
operated from 460 V 50 Hz supply and the load
resistance is R20ohms. If the source inductance
is negligible, determine (a) The efficiency, (b)
Form factor (c) Ripple factor (d) Peak inverse
voltage (PIV) of each diode .
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The PIV
Vm650.54V
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Three-Phase Full Wave Rectifier With DC Load
Current
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Example 11 Three phase diode bridge rectifier
connected to tree phase 33kV, 50 Hz supply has 8
mH source inductance to feed 300A pure DC load
current Find Commutation time and commutation
angle. DC output voltage. Power factor. Total
harmonic distortion of line current.
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