This application note describes the electronic lamp ballast for 70 W high intensity discharge
(HID) metal halide lamps (MHL) used for general indoor applications. The ballast is
composed of a boost converter and an inverter. The inverter is realized by a full bridge driver
with a power control circuit.
The booster converter for power factor correction (PFC) is controlled by the L6562A
controller (U1). The inverter is a full bridge topology driven by two pairs of half bridge buck
converters, L6385E (U3) and L6569 (U4), with the constant power control circuit L6562A
(U2).
In this note the dual-buck converter is introduced. One works in high frequency and the
other works in complementarity with necessary dead time at a lower frequency.
Warning:The demonstration board must be used in a suitable
laboratory by qualified personnel who are familiar with the
installation, use, and maintenance of electrical systems.
Intended use
The demonstration board is designed for demonstration purposes only, and must not be
used for domestic installations or for industrial installations. All technical data, including the
information concerning the power supply and working conditions, should only be taken from
the documentation included in the pack and must be strictly observed.
Installation
The installation instructions for the demonstration board must be taken from the present
document and strictly observed. The components must be protected against excessive
strain, and in particular, no components are to be bent, or isolating distances altered during
transportation, handling or use. The demonstration board contains electrostatically sensitive
components that are prone to damage through improper use. Electrical components must
not be mechanically damaged or destroyed (to avoid potential risk and personal injury).
Electrical connection
Applicable national accident prevention rules must be followed when working on the mains
power supply. The electrical installation must be carried out in accordance with the
appropriate requirements (e.g. cross-sectional areas of conductors, fusing, and PE
connections).
Board operation
A system architecture which supplies power to the demonstration board must be equipped
with additional control and protective devices in accordance with the applicable safety
requirements (e.g. compliance with technical equipment and accident prevention rules).
4/21 Doc ID 15073 Rev 1
AN2835The selected solution
2 The selected solution
2.1 The dual-buck converter topology
The fundamental application circuit in Figure 2 is composed of a PFC stage and a power
inversion stage. As the boost converter for power factor correction (PFC) is commonly used,
only the power inversion stage is introduced in this application note.
Figure 2.The fundamental diagram for the HID lamp ballast
!-V
The full bridge inverter consists of two half bridge buck converters. This is shown in
Figure 3. Both converters have the same L2 load, C2 and lamp. One of the buck converters
(S2 and S4) works in high frequency (several tens of kHz) and the second buck converter
(S3 and S5) works in complementarity with necessary dead time at a lower frequency (a few
hundred Hertz). This kind of full bridge stage is also called dual-buck converter.
Figure 3.The dual-buck converter
!-V
The timing diagram in Figure 4 indicates the relationship of a dual-buck converter and lamp
current.
Doc ID 15073 Rev 15/21
The selected solutionAN2835
Figure 4.The timing chart
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Description of the four operating states
●State 1 in Figure 3a (from t0 to t1, see Figure 4):
–S3 and S4 in off-state, S2 and S5 are turned on, C1 discharges through S2, L2,
Lamp and S5. Certain energy stores to L2 and C2.
●State 2 in Figure 3b (from t1 to t2, see Figure 4):
–S3 and S4 remain in off-state. S2 is working in high frequency in off-state, and S5
is working in low frequency and remains in on-state. The energy of L2 and C2
keeps on releasing through Lamp, S5 and the reversed body diode of S4. As S2 is
working at a higher frequency, state 1 and state 2 is repetitive until the S5 is
turned-off at t3.
●State 3 in Figure 3c(from t3 to t4, see Figure 4):
–S2 and S5 in off-state, S3 and S4 are turned on, C1 discharges through S3, Lamp,
L2 and S4. Certain energy stores to L2 and C2.
●State 4 inFigure 3d (from t4 to t5, seeFigure 4):
–S2 and S5 remain in off-state. S4 is working in high frequency in off-state, and S3
is working in low frequency and remains in on-state. The energy of L2 and C2
keeps on releasing through the reversed body diode of S2, then S3 and lamp. S4
is working at a high frequency from t3 to t6, therefore state 3 and state 4 is
repetitive until S3 is turned-off at t6. One full operating cycle is completed. Starting
from t6, the behavior of t0 is repeated again. From the above analysis, we realize
the lamp current flow to this dual-buck converter, and loads to the same L2
inductor, C2 output capacitor, and HID lamp. The lamp current at state 3 and state
4 is in the opposite direction.
2.2 The power control circuit
There are two main functions of the power control circuit. One is constant current control
during warm-up and the other is constant power control during steady-state operation.
●Constant current control
6/21 Doc ID 15073 Rev 1
Normally, the lamp current is higher during the warm-up stage than when working at
steady-state. But a warm-up current that is too high may cause the electrode to decay
and shorten the operating life of the lamp. If warm-up current is too low, the time to
steady-state is postponed. Therefore providing a value with 20% higher than the rate of
warm-up current during warm-up time is respected. The constant current control is
AN2835The selected solution
realized by controlling the peak inductor current of the dual-buck converter. Assume the
input voltage of the buck converter is V
input peak current is I
, as the buck converter is working at a critical discontinuous
in_pk
, the output voltage is Vbo, the duty cycle is D,
bi
mode, and the average input current is:
Equation 1
1
-- -
I
in
D⋅=
I
inpk
-
2
And the duty cycle is:
Equation 2
V
bo
D
---------=
V
bi
The input power becomes:
Equation 3
P
inVbiIin
⋅=
Thus the relationship between the input power (P
voltage (V
bo
) is:
), input peak current (I
in
) and output
in_pk
Equation 4
If the lamp is operating with a constant current source, once input peak current (I
selected, we observe the input power (P
) is proportional to the output voltage (Vbo).
in
in_pk
Despite the power losses, the output power is also proportional to the output voltage.
Figure 5.Input power, output voltage and input peak current
) is
!-V
In Figure 5, there are three different I
curves, this helps to choose the proper input peak
in_pk
current according to the different types of lamp. After warming-up, the lamp voltage
increases slowly to the minimum value of the rated power, the duty cycle increases
accordingly. And then the input peak current decreases. In order to power up the lamp in
steady-state, the circuit changes from constant current control function to constant power
control.
Doc ID 15073 Rev 17/21
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