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).
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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
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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
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The timing diagram in Figure 4 indicates the relationship of a dual-buck converter and lamp
current.
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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
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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
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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
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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.
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●Constant power control circuit
In this solution, the voltage on pin 3 of U2 (L6562A) is fixed, therefore, during the warmup time, pin 2 of U2 is clamped at its upper threshold, so the input peak current
detected by pin 4 is also fixed. Once the lamp power increases with the lamp voltage
increase, pin 2 decreases accordingly, the lamp works at a constant power state.
Constant power control assures the output power is constant and stabilizes lamp
luminosity without flicker. The lamp operates at the best rated lamp power. Here is an
indirect method to perform the constant power control for the lamp. As shown in
Figure 6, an Rs resistor is connected between the PFC stage and the full bridge stage.
If the output voltage of the PFC stage is constant, it means the current of Rs is
constant. With the proper controlling of the average current flow through Rs, the current
sources from the boost converter (PFC stage) become constant, and the output power
in full bridge stage is also constant,assume the power losses of the dual-buck
converter (full bridge stage) is constant. Therefore the lamp power has constant control
indirectly.
Figure 6.Indirect constant power control circuit
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The average current sense circuit is shown as Figure 7. R1 and C3 is the filter used to
obtain the average voltage on Rs. The Vf feedback signal is generated to control the on-time
of the dual-buck converter. In practice, the operating lamp voltage and current change
according to the age of the lamp, but the change in power loss to the dual-buck converter is
very small and therefore negligible. This indirect method achieves good performances in a
low power application.
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Figure 7.Average current sense circuit
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2.3 Ignition circuit
A high voltage is required to ignite the HID lamp. The ignition voltage depends on the type of
HID lamp. For a MHL (metal halide lamp) it is about 3-5 kV. For a hot lamp re-striking needs
about 20 kV. Immediate re-ignition of a hot lamp is not advised. Therefore, a cooling down
period for hot lamps is recommended. The ignition circuit is shown in Figure 8. The pulse
transformer (T1) is used to give the ignition pulse. The LIC01 is specially designed for high
voltage pulse generation purposes. At the beginning of turn-on S, with LIC01 in off-state,
bus voltage Vdc charges to C1 through R1 until it reaches the breakdown voltage (V
Figure 9) of LIC01. Once LIC01 breaksdown, C1 discharges and the crowbar current
occurs. LIC01 is latched to on-state. The LIC01 is turned off when discharging a current
lower than the holding current (IH). Then LIC01 returns to the off-state. In such a case, a
voltage pulse is generated on Lp. With the turn ratio 1:n of T1, the high voltage across C2 is
generated and remains constant for a very short time. Therefore the lamp obtains a high
voltage pulse to ignite. LIC01 returns to the off-state after C1 discharges and another
charging to C1 is restarted. After the lamp is ignited, S is turned off and there is no more
voltage pulse generation on Lp.
BO
in
Figure 8.Ignition circuitFigure 9.Electrical characteristics of LIC01
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Description of demonstration boardAN2835
3 Description of demonstration board
3.1 Specifications
The demonstration board is designed with open-lamp protection, specifications are shown in
Tab le 1 .
Table 1.Specifications
ParameterValueUnit
Line voltage range88 to 264Volt
Line frequency50 or 60Hz
Load with HID lamp70Watt
Lamp rate voltage85Vrms
Power factor0.98 minimum-
Efficiency0.88 minimum%
3.2 The PCB layout
Figure 10. Demonstration board top-side view
Figure 11. Demonstration board bottom-side view
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AN2835Description of demonstration board
3.3 Electrical schematic
Figure 12. Schematic diagram of demonstration board
1. Core: PC40EF25-Z or equivalent; bobbin: EF-25; winding: AWG30*4, 100 turns and AWG29*2,8 turns; air gap: about 1 mm
2. Core: PC40EF25-Z or equivalent; bobbin: EF-25; winding: AWG27*2, 150 turns; air gap: about 1.8 mm
3. Core: PC40EF25-Z or equivalent; bobbin: EF-25; winding: ~; air gap: about 1 mm
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AN2835Experimental results
4 Experimental results
Figure 13 shows the lamp current during start up. This warm-up current is higher than the
steady-state current. This current should be constant during the warm-up stage (the circled
area) before it enters the steady-state. During warm-up, the equivalent resistor for a 70 W
HID lamp can vary from 20 Ω to 70 Ω.
Figure 13. Lamp current at warm-up state
Since the HID lamp needs constant current control during the warm-up state and constant
power control during steady-state, designers, therefore, used a 30 Ω and 50 Ω dummy load
to evaluate the performance of the warm-up state. Figure 14 is the test with a 30 Ω dummy
load and Figure 15 is the test with a 50 Ω dummy load. Obviously the current values during
warm-up equal 1.1 A constantly, and therefore the constant current control is well achieved.
Figure 14. Load with 30 Ω during warm-upFigure 15. Load with 50 Ω during warm-up
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Upper: Output voltage 50 V/div
Lower: Output current 1.0 A/div
Upper: Output voltage 50 V/div
Lower: Output current 1.0 A/div
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For constant power control to the 70 W HID lamp, the rated voltage of the lamp is 85 V and
the rated current is 0.82 A. Therefore, the equivalent resistor for the lamp equals 103.6 Ω.
As the equivalent resistor for a new or old lamp varies, the typically varied range can have a
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Experimental resultsAN2835
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20% difference to the rated value. Therefore a 100 Ω and 140 Ω dummy load is chosen for
constant power evaluation on bench. Please refer to Figure 16 and 17 for test results of the
lamp voltage and lamp current.
Figure 16. Load with 100 Ω in steady-stateFigure 17. Load with 140 Ω in steady-state
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Upper: Output voltage 100 V/div
Lower: Output current 500 mA/div
Upper: Output voltage 200 V/div
Lower: Output current 500 mA/div
Figure 18 shows the input line voltage and current waveforms at 110 Vac and Figure 19 for
220 Vac, both bench measurements show the AC input simultaneous waveform, the input
current plot is very good and has extreme low distortion.
Figure 18. Steady-state at 110 Vac inputFigure 19. Steady-state at 220 Vac
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Upper: Output voltage 200 V/div
Lower: Output current 1.0 A/div
In steady-state, the input power (Pin), output power (Pout), operating efficiency and power
Upper: Input voltage 200 V/div
Lower: Input current 500 mA/div
factor under 110 Vac and 220 Vac is shown in Tab l e 3. Obviously the efficiency is over 88%
and the power factor is higher than 98%.
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AN2835Experimental results
Table 3.Test results of power factor and efficiency
PinPout
Conditions
Efficiency
(Pout/Pin)
PF
WattsWatts%∼
At 110 Vac77.868.988.50.99
At 220 Vac76.968.989.60.98
FromSection 2.3, we know that high voltage pulse generates continuously before it steps
into steady-state. Once the HID lamp is in open-circuit or absent from the system board,
there is no chance to step into steady-state. In such a condition, the ignition circuit is not
only continuously generating high pulse voltage, but also the full bridge circuit is working at
low frequency (about 200 Hz) as the output of U2 stays high before pin 4 of U2 detects a
current signal.
To avoid a hazard from 3~5 kV on the system board while the HID lamp is in open-circuit or
the HID lamp is absent from the system, the building of a timer to abort this high voltage
pulse generation may be important.
If it is necessary to abort the high voltage pulse generation, a NE555 timer (see Figure 20)
is used to set up a limited time (normally 5 minutes) to turn-off the circuit at the scheduled
time, or apply an MCU in digital solutions. The microcontroller gives more flexible and
precise time control compared to one with a simple hardware solution.
Figure 20. The timer circuit
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Experimental resultsAN2835
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4.1 Test with HID lamp
The test was done at 220 Vac line input with a HID lamp at room temperature. The test lamp
is a powerball HCI-T 70W/830 WDL from OSRAM. Tabl e 4 shows the test results. The input
power for the 70 W HID lamp is 76.4 W and the power factor achieved is 0.98.
Table 4.Test results of power factor
Results
Conditions
T
= 25 °C2200.34776.40.98
AMB
VinIinPinPF
VoltsA mperesWa t t s~
The lamp current during start up is shown in Figure 21. The detail of warm-up current is
shown in Figure 22 and the steady-state current is shown in Figure 23.
Figure 21. Lamp current during start up with HID lamp
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Figure 22. The lamp current in warm-up state Figure 23. The lamp current in steady-state
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Vertical: lamp current 1 A/div
Horizontal: time 2 ms/div
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Vertical: lamp current 1 A/div
Horizontal: time 2 ms/div
AN2835References
5 References
1.AN2747
2. L6562A datasheet
3. L6569 datasheet
4. L6385E datasheet
5. LIC01 datasheet
6. LM358 datasheet
7. STTH2L06 datasheet
8. STQ3NK50ZR datasheet
9. STP9NK50Z datasheet
10. STP15NM60ND datasheet
11. 2N7002 datasheet.
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Revision historyAN2835
6 Revision history
Table 5.Document revision history
DateRevisionChanges
13-May-20101Initial release
20/21 Doc ID 15073 Rev 1
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