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ELG4139: Rectifiers and Controlled Rectifiers AC to DC Converters
Linear Rectifier Consist of:
• Transformer: steps ac voltage up or down.
• Rectifier Diodes: change ac to “bumpy” dc.
• Filter Network: includes capacitors and inductors, smooths out the bumps.
• Voltage Regulator: keeps the voltage constant.
• Protection: usually a zener diode circuit.
Example: Computer Power Supply
Example: Adjustable Motor Speed Drive
Power Supply Specifics: Half Wave Rectifier
Source: ARRL
Half-Wave Rectifier
High ripple factor.
Low rectification efficiency.
Low transformer utilization factor.
Power Supply Specifics Full Wave Center-Tapped Rectifier
Source: ARRL
Power Supply: Full Wave Bridge Rectifier
Source: ARRL
Filtering Capacitors are used in power supply filter networks. The
capacitors smooth out the rippled AC to DC.
Source: ARRL
Rectifier Performance Parameters
22dcrmsac VVV
acdc PP / Rectification Efficiency
dcrms VVFF /
11 2
2
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
Form Factor
Ripple factor
𝑃𝑎𝑐 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠
Example 1: A half-wave rectifier has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor.
mm
mdc
VVtdtVV ))0cos(cos(
2)sin(
2
1
0 R
V
R
VI mdc
dc
2)sin(
2
1
0
2 mmrms
VtVV
R
VI m
rms2
%53.40
2*
2
*
*
*
R
VV
R
VV
IV
IV
P
P
mm
mm
rmsrms
dcdc
ac
dc
57.12
2
m
m
dc
rms
V
V
V
VFF
211.1157.11 22 FFV
VRF
dc
ac
.
Three-Phase Diode Bridge Rectifier
Waveforms and Conduction Times of Three-Phase Bridge Rectifier
Three-Phase Full-Wave Rectifier
Example 2: A single-phase diode bridge rectifier has a purely resistive load of R=15 ohms and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) and, (d) Input power factor.
VV
tdtVV mmdc 956.190
2sin
1
0
AR
VI m
dc 7324.122
VV
tdtVV mmrms 132.212
2sin
12/1
0
2
%06.81rmsrms
dcdc
ac
dc
IV
IV
P
P
11.1dc
rms
V
VFF
482.011 2
2
222
FFV
V
V
VV
V
VRF
dc
rms
dc
dcrms
dc
ac
Input power factor = 1cosPowerReal
SS
SS
IV
IV
PowerApperant
Alternative! Controlled Switching Mode
• By using linear regulator, the AC to DC converter is not efficient and of large size and weight!
• Using Switching-Mode
• High efficiency
• Small size and light weight
• For high power (density) applications.
• Use Power Electronics!
Thyristors and Controlled Rectifiers
Controlled Rectifier Circuit
𝑉𝑑𝑐 =1
2𝜋 𝑉𝑝𝑠𝑖𝑛𝜔𝑡𝑑𝜔𝑡 =
𝑉𝑝2𝜋
𝜋
𝛼
1 + 𝑐𝑜𝑠𝛼
𝑉𝑟𝑚𝑠 =1
2𝜋 𝜋
𝛼
𝑉𝑝2𝑠𝑖𝑛2𝜔𝑡𝑑𝜔𝑡
1/2
=𝑉𝑝
2
1
𝜋𝜋 − 𝛼 +
𝑠𝑖𝑛2𝛼
2
1/2
Example: Consider the following SCR-based variable voltage supply. For RL=240 Ohm, derive the RMS value of the load voltage as a function of the firing angle, and then calculate the load power
when the firing angle is 0, /2, and .
Full-Wave Rectifiers Using SCR
𝑉𝑑𝑐 =2
2𝜋 𝑉𝑝𝑠𝑖𝑛𝜔𝑡𝑑𝜔𝑡 =
2𝑉𝑝𝜋
𝜋+𝛼
𝛼
𝑐𝑜𝑠𝛼
𝑉𝑟𝑚𝑠 =2
2𝜋 𝜋+𝛼
𝛼
𝑉𝑝2𝑠𝑖𝑛2𝜔𝑡𝑑𝜔𝑡
1/2
=𝑉𝑝
2 = 𝑉𝑠
With a purely resistive load, SCRs S1 and S2 can conduct from to , and SCRs S3 and S4 can conduct from + to 2.
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