V2-C LATAM Session 2 Circuit serie theory.pdf

V2-C LATAM Session 2 Circuit serie theory

1. Session 2
The AGL and
series circuit
theory
Jonathan Teixeira

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What is Airfield Ground Lighting?
Definition
“All the lights, with their respective mounting systems, their power supply line
and control system used in order to “visually” support the landing, the
movement, taxiing and take-off of aircrafts.”
Hence, we divide the AGL system in 3 principal parts:
Airfield Lighting Control
System (ALCS & ILCMS)
Power supply
Lighting

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Airfield Lighting Control and Monitoring System (ALCMS)

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Airfield Lighting Control and Monitoring System (ALCMS)

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Which technologies & skills do we have to
consider when installing & maintaining an
ALCMS
?

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Electrical incoming Power Supply
1.Primary power source: public or
private network
2. Standby/Emergency power supply:
▪Diesel Generator Set
▪Uninterruptible
Power Supply ( UPS )
AGL Power Supply

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Diesel Generator UPS (Dynamic or Static)
AGL Power Supply

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Power Transformers & Low Voltage Distribution Boards
AGL Power Supply

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Constant Current Regulators
AGL Power Supply
Transistor ThyristorFerro resonance
FAA markets ICAO markets

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Field equipment related to power supply of AGL
AGL Power Supply

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Which technologies & skills do we have to
consider when installing & maintaining AGL
Power Supply Systems
?

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Power Distribution
AGL Series Circuits
Primary cable
Isolating Transformer
Secondary cable
Primary connectors
Secondary connectors

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Difference between Airfield Lighting Series Circuits and conventional
electrical circuits:
1)They use electrical series circuits
2)They are Constant Currentcircuits NOT constant voltage circuits
3)AGL Circuits are floating in reference to ground
4)All components must be watertight
5)In most of the world, a standard connector system is applied
AGL Series Circuits

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Which technologies & skills do we have to
consider when installing & maintaining AGL
Series circuits
?

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Which technologies & skills do we have to
consider when installing & maintaining an
overall AGL system
?

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Voltage depends on the load
The Series Circuit
Primarycircuit
Isolation
Tranformers
I1V nTOT
VVVV ...
21
 TOTAL
V 2
V

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Voltage depends on the load
The Series Circuit

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Thyristor CCR
Types of Constant Current RegulatorsBLOC K DIAGRAM OF A REGULATOR
CONBTROLLED BY SERIAL THYRISTORS
I prim.
Setpoint
V
prim
V
sec
Z
I
sec
7/8
6/8
4/8
2/8
TI
Tower (remote)
Local

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Thyristor CCR
Types of Constant Current Regulators
Gate ON long
after zero
crossing
Gate ON just
after zero
crossing
Gate
triggered ON
at zero
crossing
5.5A Output
2.8A Output

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Thyristor CCR
Types of Constant Current Regulators
20KW LC CCR, 6.6A, Step 5, Full Load

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Thyristor CCR
Types of Constant Current Regulators

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Thyristor CCR
Types of Constant Current RegulatorsPARAMETER UNITS TAP 8/8TAP 4/8
Thyristor conduction angleDeg 85 124
Mains Voltage Vrms 398 401
Mains Current Irms 14.6 7.42
Output Voltage Vrms 326 312
Output Current Irms 6.6 6.6
Mains power W 2605 2379
Mains power VA 5831 2979
Output Power W 2203 2096
Output Power VA 2450 2060
Power Factor (input) 0.45 0.79
Efficiency % 84.5 88.1

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Thyristor CCR
Types of Constant Current Regulators

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Thyristor CCR
Types of Constant Current Regulators
Disadvantages
•Input Power Factor/harmonic
content degrades with reduced
load or current step
•Efficiency decreases with reduced
load (use taps) or current step
Advantages
•Low cost
•Physically small

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Transistor CCR
Types of Constant Current Regulators
•A –Line input
•B –Diode Bridge andSensingPCB
•C –IGBT module andIGBT PCB
•D –PWM filter
•E –Maintransformer
•F –Output MeasurementPCB
•G –Lightningarrestors
•H –Series output connection
•I –CPU PCB
•J –User Interface
•K –Remote Control
•L –Power Supply
Red lines:current Green lines: signalwiresconnections
Black lines: low voltage connections

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Transistor CCR
Types of Constant Current Regulators

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Transistor CCR
1 phase IGBT Power Bridge –IGBT Bridge
2 IGBT’s form 1 H-bridge
IGBT work frequency = 12,5 kHz => T = 80 μs
Measuring cycle T = 4 μs
Types of Constant Current Regulators

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Transistor CCR
Types of Constant Current Regulators

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Transistor CCR
Types of Constant Current Regulators

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Transistor CCR
Types of Constant Current Regulators
Disadvantages
•Expensive
•Increased level of expertise needed
for troubleshooting
Advantages
•Excellent input power factor
under any circumstances
•Very low harmonics under
any circumstances
•Improved ILCMS
communication
•Increased lifespan of
downstream equipment

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Basic principle
Isolating Transformers
65W
65W
6.6A
6.6A
Secondary
Side
Primary
side
•I primary= I secondary
•Short Circuit = No load!
•Open Circuit = Full load!

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Benefits of using isolating transformers
Isolating Transformers
65W
65W
6.6A
6.6A
Secondary
Side
Primary
side
•Lowervoltage imposedon the fixture, improving
safetyand reducingcost.
•Withoutisolation transformersa lampopen
wouldmeanall lampsgo out.

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Limits of use
Isolating Transformers

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Limits of use
Isolating Transformers

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Limits of use
Isolating Transformers

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1. Lamp load
Correct sizing of CCR
Take for example a circuit using 45 Watt fittings. Then
look at each individual isolating transformer and
fitting.
•So from drawing below : value of lamp 1 = 45 W
Secondarycircuit
Primary
Circuit
1
3
L2
4
IsolatingTransformer

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2. Secondary cable load
Correct sizing of CCR
•Length ( l ) from transformer to fitting = 30 m
If the section of wire used = 4mm²
Resistance R in secondary circuit is
whereresistivity
•Therefore
•Lossesin sec. Cable = I²R= 6,6² x 0,27 = 11,76W
•Total power at secondary= sumof 1 and 2 = 56,76Ws
l
2 R
sec
 m
mm
2
3-
10 18

  27.0 R
sec

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Correct sizing of CCR
3. Transformer value
•Power at primaryside= power at secondaryx 1,25 = 56,76 x 1,25 = 70,79W
•( 1.25 represents the isolating transformer characteristics ie. efficiency coefficient )
4. Secondarycircuit in total
•If we take a series circuit containing 100 isolating transformers then:
Total power of transformers = 70,79 x 100 = 7079 W
RST Coef
15 1.51
25 1.43
45 1.23
65 1.22
100 1.19
150 1.17
200 1.15
300 1.13
500 1.13

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Correct sizing of CCR
5. Primarycableload
•If the total length of the primary circuit is 8000 m, and the section of cable is 6mm² then
the losses in this primary circuit will be :
•I²R = 6,6² x 24 = 1045W where
4. Total loadon regulator
•Total loadon regulator= sumof 5 + 6 = 8424Ws
l
R
prim  24

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Correct sizing of CCR
ThereforeREMEMBER:
•A benchmark when choosing a regulator is to select a KVA rating which is somewhat
greater than calculated so that 80% of this power rating is
>= calculated load
•Look at the table of power ratings. If a 12.5 KVA regulator is chosen, then 80% of this
value = 10000 VA and this is > the calculated load 8424 W, therefore O.K

41. Thanks
for yourattention
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