How does a LED power driver buck-boost converter work?

by:MOSO     2019-12-14
For LED drivers in low voltage applications, a buck-boost converter is a good choice. The reason is that they can drive LED strings (boost and buck) with voltages higher and lower than the input voltage, high efficiency (easy to reach more than 85%), and discontinuous operation mode can suppress changes in input voltage (provided Excellent line voltage adjustment), peak current control mode allows the converter to adjust the LED current without complicated compensation (simplified design), it is easy to achieve linear and PWM LED brightness adjustment, switch transistor failure will not damage the LED, etc. Figure 2 shows the connection circuit of the buck, boost and buck-boost converter to the LED string. However, this method still has disadvantages: First, the peak current is controlled, because the buck-boost converter using discontinuous current mode is a converter with constant power. Therefore, any change in the LED string voltage will cause a corresponding change in the LED current; another problem is that the open-circuit state of the LED will cause high voltage in the circuit that damages the converter; in addition, an additional circuit is required to convert the constant power converter to constant Current converter and need to protect the converter under no load conditions. Figure 3 shows a specific buck-boost converter application circuit. The controller has an built-in oscillator for setting the switching frequency. At the beginning of the switching cycle, Q1 is turned on. As the input voltage VIN is added to the inductor, the inductor current (iL (t)) starts to rise from zero (initial steady state). When the induced current rises to a preset current value (ipk), Q1 turns off. The switch on time (ton) is determined by the following formula: ton = ipkL / VIN At this time, the total energy (J) stored in the inductor is: J = Li2pk / 2 In this way, although the switch is closed at this time, the current flowing through the inductor will not be interrupted. This will turn on the diode D1 and generate an output voltage (-Vo) across the inductor. This negative voltage will cause the inductor current to drop rapidly. After a certain time tOFF, the inductor current approaches zero. This time can be calculated by the following formula: tOFF = ipkL / VO In order for the converter to work in discontinuous conduction mode, the sum of the switch on time and the inductor current fall time must be less than or equal to the switching period TS, so as to ensure that the inductor current can start from zero in the next switching period. In fact, (tON + tOFF) can reach the maximum value with the minimum input voltage and the maximum output voltage. Therefore, ensuring that the converter operates in discontinuous conduction mode at these voltages can guarantee that the conditions listed in the following formula can be met under any conditions: tON + tOFF≤Ts The product of the power (Pin) inductance obtained by the converter from the input and the switching frequency f: Pin = fsLi2pk / 2 Assuming the voltage (VO) of the LED string is constant and the efficiency is 100%, then the LED current (iLED) is: iLED = PIN / VLED = Li2pkfs / 2V In peak current control mode, ipk is usually a fixed value. Therefore, the LED current is completely independent (in theory) from the input voltage. At a fixed ipk, the rise (fall) of the input voltage will cause the transistor's on-time to decrease (increas) in inverse proportion, which will provide good line voltage regulation. In practical applications, the delay between the detection of the current peak by the control IC and the actual shutdown of the GATE pin will cause the input power to vary. Shorter on-time results in more errors due to the delay time because the delay time will account for a significant portion of the on-time. In fact, the LED current is inversely proportional to the voltage of the LED string. A circuit with a nominal output of 20 V and 350 mA will generate 700 mA at a 10 V output voltage, which is obviously not the desired result. However, by making the switching frequency proportional to the output voltage, the above formula provides a way to convert a constant power converter into a constant voltage converter. Assuming fs = KVO, where K is constant, then: iLED = kLi2pk / 2 In this way, iLED will be independent of the input and output voltages. Another disadvantage of the flyback converter is that it is susceptible to the open-circuit state of the output. When the LED is open, the energy stored in the inductor will be transferred to the output capacitor at the end of each switch on time. In this way, a load lacking a capacitor discharge will cause the voltage across the capacitor to rise gradually, eventually exceeding the nominal value of the device and damaging the power stage. Therefore, additional voltage can be added to provide output voltage feedback and overvoltage protection.
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