#the4thLichuangcompetition#Automotive high-intensity gas discharge lamp HID ballast

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[Content to be filled in at the registration stage ↓, required] Briefly introduce the work: 45W output automotive HID high-intensity gas discharge lamp xenon lamp ballast (ballast). Parameter requirements: Input: 9-18V. Output: 40 - 120V (100HZ) AC square wave. Stable bulb tube voltage 85V±17V. (Square wave AC can extend the service life of the lamp and no light flickering can be seen) Input power: 50W. Output power: 45W. Maximum startup current: < 6A Efficiency: > 90% at 13.2V. Input anti-reverse protection: Yes. Output open circuit protection: Yes. Under-overvoltage protection: Yes. Secondary short circuit: Yes. No-load bus clamp voltage: < 450V. Ignition high voltage pulse: <= 25KV. Fast strobe: It requires 0.5 seconds to turn on and off quickly, so that the lights can light up normally and the ballast will not be damaged. Constant power: When the input voltage changes and the secondary voltage changes, the output power deviation is < 1W. Refill light function: The light bulb needs to be relit after it suddenly goes out. Chip selection: UC2843 + STM8S003 +LM321. Solution selection: Uncommon tapped inductor BOOST + low frequency full bridge + 25KV pulse high voltage ignition (current limiting and constant current in the front stage, constant power in the rear stage).

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1. Details of the work; (I study these purely as a personal hobby. I learned all the software and hardware by myself online. Due to my limited knowledge and the fact that I have not touched electronic products for more than 5 years, I still need to learn and design many things.) In the lighting industry, semiconductor LED Previously, HID gas discharge lamps were regarded as good light sources, with high luminous efficiency, low light attenuation, and high color rendering. However, the shortcomings are also obvious. The bulb has a negative resistance characteristic when started, and high-frequency lighting causes acoustic resonance (the phenomenon of serious light jitter). It also requires high-voltage lighting, constant power control, etc., which places very high requirements on the ballast (ballast). Although most lighting currently uses LED light sources, HID is still used in some special occasions. For example: car high-speed rail subway headlights, large drone lighting, plant growth lights, dysprosium lamps for photography, etc. This design is developed for a xenon lamp ballast. This solution can be slightly modified and transplanted to other gas discharge lamps for use. The solution uses analog + digital control. The front stage uses UC2843 (UC2843 has a wider operating temperature) to control current limiting, constant current, voltage and current acquisition, and constant power calculation. Each protection is controlled by STM8S003 microcontroller, and the secondary current acquisition and amplification uses LM321 operation. Amplifier, after amplification, it is sent to the inside of the microcontroller for constant power operation. Brief block diagram of the scheme: The general working process: After the input is filtered by the capacitor, the voltage is divided and entered into the microcontroller to detect whether there is over-voltage or under-voltage. After the voltage is normal, the full bridge on the secondary side is bootstrap charged, and a pair of diagonal bridges are turned on. UC2843 starts to work, the secondary clamping ring works to limit the no-load voltage to about 400V (to prevent the no-load voltage from being too high and causing damage to the subsequent stage), the high-voltage ignition circuit starts to work, and the bulb is broken down (if the bulb is not broken down after 5 tests) , identify short circuit or open circuit and protect the lock). The current limiting function of the front stage works, limiting the maximum starting current to 13.2V (<6A), and detecting whether it is a hot lamp or a cold lamp. If the cold lamp is used, the maximum power will be started quickly, and the output will gradually decrease to 45W. The subsequent stage calculates the voltage_current and Start constant power and work normally, detect (over and under voltage, open circuit and short circuit protection). If the heat lamp is used, it will quickly drop from the maximum power to the output of 45W, and start constant power and so on.

2. Describe the challenges faced by the work and the problems it solves; > (1) Efficiency: The traditional solution is low-voltage flyback + full bridge, 50W power, and it is very difficult to achieve 90% efficiency. The advantages and disadvantages of several options are introduced below, and you can choose the option that meets your requirements. 1: Conventional solution: flyback + full bridge, the full bridge is not drawn anymore, and the efficiency of the front stage is analyzed. Advantages: 1: The voltage can be stepped up and down, the secondary is in a short-circuit state, and the current can also be limited. 2: After the front stage is closed, the rear stage is completely closed. 3: Cold start current is not affected by different tube pressures of different secondary bulbs. 4: The output power error of different bulbs and bulb tube pressures is small, and the power can be simply compensated. Disadvantages: 1: Low efficiency. 2: The front-stage MOS tube and secondary diode have large peaks. 3: The front-stage MOS tube, secondary diode, and secondary full-bridge MOS tube require higher withstand voltage.

2: Special solution: non-isolated tapped inductor BOOST + full bridge, the full bridge is not drawn, and the efficiency of the front stage is analyzed. Advantages: 1: High efficiency, MOS tube peak voltage + BOOST boost voltage = output voltage. Equivalent to BOOST, which only needs to be raised to about 50V! 2: The MOS tube and secondary diode are less affected by the leakage inductance of the transformer, and the peak value is relatively low, so the MOS tube withstand voltage needs to be relatively low. 3: The front-stage MOS tube, secondary diode, and secondary full-bridge MOS tube require lower withstand voltage. Disadvantages: 1: It can only boost the voltage. When the secondary is short-circuited, it is equivalent to the input being short-circuited through the inductor, and the current limit is limited! 2: After the front stage is closed, the rear stage is not closed. The entire bridge will also need to be closed. 3: Cold start current is affected by different tube pressures of different secondary bulbs. 4: The output power error of different bulbs and bulb tube pressures is relatively large, so constant power must be output.

3: Super special solution: flyback + half-bridge commutation solution, conventional flyback, I won’t draw it anymore, let’s briefly analyze the features: 1: There are 2 less MOS tubes and drivers than the traditional full-bridge commutation. There is an extra diode and filter capacitor. 2: It is twice as high as the bus voltage of the full bridge (voltage at point AB), and the two MOS tubes need to double the withstand voltage. 3: Secondary working current and voltage sampling is a problem. If GND is point C, then the lower MOS tube can be driven directly, and the upper MOS tube can be driven in a bootstrap mode. It is relatively simple, but the lamp voltage DB is not easy to collect, and DB It is an AC square wave. When working in the lower cycle, the series resistor at point C can collect the operating current. However, when working in the upper cycle, the current cannot be collected unless a transformer is used, but it is too complicated. If point B is GND, then the lower MOS tube is difficult to drive. 4: The above three points are not the most difficult. If the secondary current cannot be collected, it will not be collected. The front current * front voltage is used to perform constant power. Although there is an error in the efficiency of each transformer, it is not large, but If the AB voltage is set too high, the internal resistance of the two MOS tubes is not much different from the four in the full bridge. If the AB point voltage is clamped to 400V, then there will be problems with current connection. (The role of current connection will be explained below) 4: Super special solution: double flyback high-frequency chain difficulties: 1: Q2 Q3 is not easy to drive 2: The timing requirements are high, and there will be slight timing errors. Q2 Q3 must be broken 3: Secondary operating current and voltage Sampling is a problem. Summary: Considering efficiency comprehensively, the most likely efficiency to reach 90% is (non-isolated tapped inductor BOOST + full bridge)

> (2) Current connection. Ordinary current connection circuits have losses. In order to achieve 90% efficiency, this part of the loss needs to be solved. Why do we need current to connect circuits? Reason: When the lamp is broken down by high voltage, because the circuit response speed cannot keep up, there is no large current to keep up after the breakdown, causing the lighting to fail. Therefore, current is needed to connect the circuit. At the moment of startup, R1 charges the capacitor, and the capacitor stores energy. Mainly proportional to the square of voltage and capacitance. The increase in capacitor capacity will cause the charging speed to slow down and R1 loss to increase, so a better way is to increase the charging voltage. After the lamp breaks down (equivalent to a short circuit), the capacitive energy is provided by M7 and 100R to provide large current to the bulb. R2 100R acts to reduce the discharge rate. If the first lighting fails, the capacitor energy is still enough for the second lighting. If the discharge is too fast without adding 100R, it can only be lit once! (The biggest drawback of this current connection is that after normal operation, C2 Will continue to work and cause losses)

Many methods have been thought of to solve the loss of ordinary current connecting circuits. The more feasible methods are shown in the figure below. Charging goes through R2, one-way TVS D2 charges the capacitor, and discharge flows out through the internal diode of TVS D2 to R2. When the voltage is lower than 130V during normal operation, C2 does not work and there is no loss. However, this circuit also has flaws. Because the C2 capacitor is not a real conductor, it is equivalent to a short circuit only at the moment of charging, so the voltage can only reach 250V. The voltage energy of 250V is still a little worse. It is better to replace C2 with 4.7UF. The only solution is the test. The method is to increase the bus voltage to about 430V, and the charging voltage can reach 290V, which is enough.

> (3) Pre-stage current sampling (functions to limit the maximum current and can maintain constant current. This resistor has relatively large losses, so it is necessary to find ways to reduce the sampling resistor to improve efficiency) Resistor sampling is more commonly used, and the temperature change of this resistor has a slightly smaller impact, but R1 has a relatively large loss when working, especially when the current increases, the loss will become more serious, so finding a way to remove the loss of the sampling resistor is a more effective way to improve the overall efficiency. (1> R1 is reduced, and the current is amplified by the op amp, but the circuit is more complicated, and the response speed will be slower after adding the op amp. 2> Transformer sampling, the transformer is slightly larger and more expensive, so it is not suitable .3> Relatively novel tube voltage drop current sampling)

When the MOS tube is turned on, because the MOS tube has internal resistance, there will be a voltage drop at points AB. In fact, it is similar to an ordinary sampling resistor, except that the internal resistance of the MOS tube is used as a sampling resistor. It can be used as a lossless current sampling, but this detection method is flawed. The MOS tube itself is a heat source, and the internal resistance will change with the temperature. If the front stage uses this method to maintain constant current, the temperature change will inevitably cause power failure. Changes and solutions include adding temperature compensation. The specific temperature compensation will be noted in the schematic diagram, so I won’t go into it again.

(4) Difficulty: (constant power) Conventional solutions generally use simulation methods. When inputting 9-18V, some power is compensated at each stage to approximately form constant power. However, it is generally impossible to achieve very precise results, and when outputting different powers , the software needs more changes. The voltage of the downstream tube will also affect the power error. Since there are two car lights, if the power deviation is large, the brightness of the two lights will be different. In order to solve this problem, constant power control is needed because the output power needs to be achieved. . Conventional constant power method: 1. PID current value comparison, use given value/secondary voltage = to obtain the secondary reference current value, and then compare it with the secondary current collection value, and PID operation difference control PWM control UC2843 pin 1. This method is more accurate, and both positive and negative differences can be controlled, but PID always fails due to limited capabilities. 2. PID power value comparison, use secondary voltage secondary current = to get the current actual power, then compare it with the given power value, PID operation difference output PWM control UC2843 pin 1, this method is similar to the above method, the positive and negative extreme differences The value can be controlled. 3. Constant power of software and hardware, use secondary voltage secondary current = to get the current actual power, and then output PWM and send it to UC2843 pin 2 for comparison with the internal reference 2.5V. If it exceeds 2.5V, the power will be reduced, and then To achieve stability, this method is easier to implement and debugging is not complicated, so this method is used in this design. However, this method is also flawed. Constant power only works when the power value is > 2.5V, and it does not work when the power value is < 2.5V.

> (5) The most difficult point: design of pulsed high-voltage package. Schematic diagram

The principle is very simple. Start the full bridge and conduct it diagonally. The midpoint of the two bridges is a DC voltage of about 430V. The negative electrode charges C1 through D1 R1. When the voltage reaches the breakdown voltage of the discharge tube, the electric energy stored in the capacitor is instantly discharged. The high-voltage package low-voltage coil generates 25KV pulse high voltage from the secondary, which then breaks down the bulb. Because the high voltage generated is too high, the only option is to use a series boost, which has a relatively small impact on the entire bridge. After the lamp is lit, the bus voltage drops, which is not enough to breakdown the discharge tube, and the high-voltage package does not work. It seems to be a very simple principle, but it is not that easy to do it well. Difficulty 1. The generated voltage is relatively high (25KV). Difficulty 2. If too many turns are added to the secondary, it means that the internal resistance of the high-voltage package will be large. Because the current needs to pass through the secondary of the high-voltage package during normal operation, the internal resistance cannot be too large. , otherwise the loss will be relatively large. Difficulty 3. Lighting requires not only the height of the high voltage, but also a more important parameter, the high voltage maintenance time (pulse width). Difficulty 4. Because the high voltage is relatively high, ordinary manganese-zinc materials with relatively high magnetic permeability will not work because of the manganese-zinc resistance. The rate is small and will be broken down by high voltage, so you can only choose nickel-zinc material with low magnetic permeability. Difficulty 5: It is extremely difficult to achieve high voltage, sufficient pulse, and small enough internal resistance so that the high voltage does not discharge the magnetic core. Currently, the internal resistance of 25KV 500NS on the market is above 3Ω. This design needs to be controlled at 1.2Ω. Around, it’s very difficult. Based on the above issues, I have also studied magnetic materials for a long time and finally came to a conclusion;

1. Nickel-zinc magnet rods with the same diameter have higher magnetic permeability, higher voltage, and the same pulse width.
2. For nickel-zinc materials with the same magnetic permeability, the larger the diameter, the higher the high voltage and the larger the pulse width.

Since I am not capable of doing magnetic research, I finally determined the parameters based on the magnetic rods available on the market. The diameter of the magnetic rod is 10*40, and the wire diameter is 0.45. The 5-slot skeleton (each slot has a voltage resistance of 5000V, and the enameled wire must withstand a voltage of 3000V). Magnetic Conductivity 250, turns ratio, 3:250. See the figure below for the winding method;

The center slot should be as close to the magnetic rod as possible. It should not be too thick and the gap should not be too large. It should be wound as flat as possible. The final product needs to be filled with epoxy resin. The final product has an internal resistance of 1.2Ω 25KV 520NS pulse width. Measurement of internal resistance of high-voltage package: schematic diagram: PCB board: first attempt at arcuate wiring. 3. Describe the key points involved in the hardware and software parts of the work; Software:

Hardware: 4. Material list of the work; Attached competition BOM 5. Upload the picture of the work; (the PCB must have the competition logo and take a photo and upload it, if not, it will be deemed as giving up the competition) Efficiency calculation picture: Waveform test: PCB picture: Physical picture: Heat dissipation structure , board + thermal grease + thermal silicone sheet + thermal aluminum plate + screw fixation Dimensions: 67 X 58 X 15 Unit: mm Loading photos:

6. Demonstrate your work and record it as a video for upload; (The content of the video must include: introduction to the work; functional demonstration; performance test; close-up of the competition logo on the PCB. Failure to do so will be deemed as giving up the competition) Demonstration video: https://v. youku.com/v_show/id_XNDM1NTQzOTUxNg==.html?spm=a2hzp.8244740.0.0 Loading video: https://v.youku.com/v_show/id_XNDM1NTM0OTU0OA==.html?spm=a2hzp.8244740.0.0 7. Open source documentation. See the attachment, which contains the source code of the schematic PCB file. The PCB and schematic are both drawn by Protel  99SE.