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IRPLHALO1E

P1-P3P4-P6P7-P9P10-P12P13-P15P16-P17
IRPLHALO1E
10 www.irf.com
Ignoring the output transformer we can assume for this calculation that the load is connected from
the half bridge to the mid point of the two output capacitors and that the voltage at this point will be
half the DC bus voltage. The RMS voltage of the DC bus is the same as that of the AC line so we
can see that the RMS voltage across the load shown in Figure 4, will be half the RMS voltage of the
line. The load is the maximum rated load of the convertor. The current in Rcs will be half the load
current given by :
Since the load is resistive the current waveform will have a sinusoidal envelope and so the peak
can be easily determined taking into account that the current is has a high frequency component
with an approximate 50% duty cycle :
Therefore:
For correct operation at maximum load the peak voltage should be 0.4V.
The calculation can be simplified by combining the formulae,
which can be simplified to:
Example
For a 100W convertor working from a 230VAC supply the current sense resistor would need to be
The nearest preferred value in the E12 range is 0.33 Ohms.
Ω=
×
324.0
100
230141.0
LOAD
AC
CS
P
V
R ⋅= 141.0
LOAD
CS
CS
P
V
R
⋅⋅
⋅
=
22
4.0
CSPKCSPKCS
RIV
×=
)()(
)()(
22
RMSCSPKCS
II ×=
AC
LOAD
RMSCS
V
P
I =
)(
IRPLHALO1E
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The power dissipation in Rcs should also be considered and is given by
In this case :
It is important to bear in mind that the resistor must be rated to handle this current in a high ambient
temperature as well as the high currents that occur for a short period under short circuit conditions.
IMPORTANT NOTE
The filter resistor RF should be 1K, which is needed to protect the CS input from negative going
transients. CF should be 1nF and is also necessary to filter out switching transients that can impair
the operation of the shutdown circuit.
Output Transformer Selection
The size of the core and design of the output transformer need to be seleted bearing in mind the
throughput power at maximum load. The core may be a toroid, a pair of E-cores, or any other
shape. The turns ratio can be easily determined by the ratio of half the AC line RMS input voltage
and the required output, which is usually (but not always) 12V. When the correct number of turns for
the primary has been determined, it is simple to calculate the number of secondary turns.
The number of turns required for the primary should be calculated so that at the minimum frequency
the transformer will not saturate at the peak voltage, i.e. half the peak DC bus voltage.
It is important to bear in mind that the peak voltage at the DC bus occurs during dimming with a
leading edge, triac type, dimmer where the firing angle is at the peak of the line. This is because a
transient is produced when the triac is fired and so an additional voltage is added to the DC bus
voltage. As a rule the MOSFET breakdown voltage may be used to calculate the primary turns such
that the core will not saturate at minimum frequency and maximum temperature.
If the core saturates a large current will flow in the half bridge and may trigger the short circuit
protection when the convertor is connected via a dimmer.
W062.033.0
230
100
2
=×
CS
AC
LOAD
CS
R
V
P
P ×
=
2
IRPLHALO1E
12 www.irf.com
The transformer selection procedure is as follows:
1. Select the correct core size for the maximum load at 30kHz. Use the manufacturer’s core
data to determine the maximum throughput power at this frequency.
2. Calculate the number of turns required at the primary
Where Bmax (maximum flux density in Teslas) can be obtained from the core manufactur-
ers data for the Ferrite material operating at a temperature of 100 degrees C and Ae is the
cross sectional area of the magnetic path in mm
2
.
Vpk may be considered to be 400V and Ton(max) may be considered to be 18uS.
3. Calculate the number of turns required at the secondary.
4. Determine the leakage inductance.
This can only be done by measurement. Simply short out the secondary and measure the
primary inductance with an LCR bridge. The amount of leakage inductance depends on the
physical construction of the transformer, the better the primary and secondary windings are
coupled, the lower the leakage inductance will be. However, in this application some leak-
age inductance is desirable to limit the primary current if the secondary is short circuited. It
is also important to consider that a high isolation breakdown voltage (4kV) is required in this
application between the primary and secondary windings, for safety certification of the con-
vertor. This means that winding the primary and secondary close together is not possible
and therefore, some leakage inductance will result.
In some designs the output transformer is purposely designed to have a high leakage
inductance in order to limit the primary current under short circuit conditions and therefore
allow smaller half bridge MOSFETs to be used. In this case the short circuit primary current
may be too small to trigger the short circuit protection, the result being that the system will
take longer to shut down because only the overload threshold is exceeded at the CS pin.
This is unlikely to cause any problems but should be considered.
)(
)(..2
rmsVac
rmsVoutNp
Ns =
AeB
TonVpk
Np
⋅⋅
⋅
=
(max)2
(max)
(min)
P1-P3P4-P6P7-P9P10-P12P13-P15P16-P17

IRPLHALO1E

Mfr. #:
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Manufacturer:
Infineon / IR
Description:
Power Management IC Development Tools Halogen Cnvrtr 220/230VAC
Lifecycle:
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