Correctly select the peripheral components of LED constant current source

In order to maintain strict hysteresis current control when designing an LED constant current source, the inductor must be large enough to ensure that energy can be supplied to the load during the HO and ON periods to avoid a significant drop in load current, causing the average current to fall below the expected value.

　　First, let's look at the effect of inductance. Assuming there is no output capacitor (COUT), the load current and the inductor current are exactly the same, which can more clearly illustrate the effect of inductance. The figure below shows the effect of inductance value on frequency within the range of input voltage. It can be seen that the input voltage has a great influence on the frequency, and the inductance value has a great influence on reducing the frequency when the input voltage is low. (Note: It may not be exactly the same as the reference figure, it is just for illustration)

　　The figure above shows the frequency response under different inductance values. The figure below shows that when the inductance decreases, the change of load current increases significantly within the range of input voltage change.

　　The following figure shows the frequency change curve according to different output voltages and different inductance values.

　　The figure below shows that when the inductance decreases, the fluctuation of the load current increases significantly within the range of output voltage.

　　The LED driving circuit generates audible noise (or microphonic noise) that is audible to the human ear. Usually, white light LED drivers are switching power supply devices (buck, boost, charge pump, etc.), and their switching frequencies are all around 1MHz, so in typical applications of the driver, no audible noise is generated. However, when the driver is switching, if the frequency of the PWM signal happens to fall between 200Hz and 20kHz, the inductance and output capacitor around the white light LED driver will generate audible noise to the human ear. Therefore, when designing, avoid using low frequencies below 20kHz.

　　We all know that a low-frequency switching signal acting on an ordinary wire-winding coil will cause mechanical vibrations between the coils in the inductor. The frequency of the mechanical vibrations falls exactly at the above frequency, and the noise emitted by the inductor can be heard by the human ear. The inductor generates part of the noise, and the other part comes from the output capacitor.

　　The value of the inductor is mostly based on your experience. The main conditions for choosing the appropriate inductor value are that the circuit works in the appropriate frequency range, the appropriate switching frequency reduces the number of MOS switches, reduces the heat generated by MOS, and avoids interference with the same frequency of the PCB circuit; choose the appropriate inductor internal resistance, which is the main factor of inductor heating, thereby improving the circuit efficiency; choose the appropriate current value, sometimes the volume and cost are the main restrictive factors, but it should still be greater than 2 times the peak current (usually 65%), even if the board-level space is very precious, it is necessary to ensure 30% of the reserved space margin, which can effectively reduce the internal resistance and heat generation; poor quality and loosely wound inductors will also have noise; when the unshielded inductor is installed in a metal casing, the line oscillation frequency will change, thereby generating noise, and the inductor needs to be shielded at this time; in addition, when the wavelength of the shielded interference signal is close to a certain size of the metal casing, the metal casing can easily become a large resonant cavity, that is, the electromagnetic wave will reflect back and forth in the metal casing and will overlap each other.

　　In order to obtain the best efficiency, a ferrite core inductor should be selected. An inductor that can handle the required peak current without causing saturation should be selected, and the DCR (copper wire resistance) of the inductor copper wire should be low. This is to reduce I2R power consumption. Remember that the insulation layer of the inductor copper wire cannot withstand 160 degrees or long-term high temperature environments. SMT sometimes has an impact, which will cause serious changes in the inductance value. Carefully understand the temperature tolerance requirements of the supplier's products.

　　EMC inductor selection:

　　EMC inductors are used in input and output filters to reduce conducted interference and are designed to be below the limits of EMC standards. All inductors require iron powder cores rather than ferrite. They can handle larger currents before saturation, and the appropriate current value needs to be selected based on the load.

　　When making filter inductors, when choosing the core material, in addition to preventing the core saturation problem, the core's constant permeability characteristics must also be considered. It should be pointed out that some designers tend to only pay attention to the inductance index and choose materials with high permeability to reduce the number of turns of the coil, but when the rated current of the inductor is large, whether the inductance is reduced, to what extent, and whether it will reach saturation, they pay little attention. This should be avoided.

　　Iron powder cores are widely used because of their high saturation flux density, good constant magnetic permeability characteristics and low price.

　　Output capacitor device selection:

　　The output can use output capacitors to achieve precise control of the target frequency and current. The capacitor can reduce the frequency over the entire input voltage range, and a small 4.7μF capacitor can significantly reduce the frequency. The current regulation can also be improved by increasing the capacitance value. It can be easily seen from the picture below that there is an inflection point on the graph, and increasing the capacitance value has little effect on the adjustment of the operating frequency and output current.

　　Increasing the output capacitor (COUT) essentially increases the amount of energy that the output stage can store, which means that the current can be supplied for a longer period of time. Therefore, by slowing down the di/dt transients of the load, the frequency is significantly reduced. With the output capacitor (COUT), the current in the inductor will no longer be consistent with the current seen by the load. The inductor current will still be a perfect triangle shape, and the load current will have the same trend, but all the sharp corners have become smoother and all the peaks have been significantly reduced, as shown in the figure below.

　　The application design uses low ESR (equivalent series resistance) ceramic capacitors on the output to minimize output ripple. Use X5R or X7R type material dielectrics, which can maintain their capacity unchanged over a wider voltage and temperature range compared to other dielectrics. For most high current designs, a 4.7 to 10uF output capacitor is sufficient. Converters with lower output currents only need to use a 1 to 2.2uF output capacitor.

　　Input capacitor selection:

　　Generally, a capacitor is set at the input of the driver IC to solve the EMI problem of the line switching frequency on the power supply part. Sometimes people mistakenly think that it is set for power supply filtering, but this is not the case. Because the rectifier diode is widely used, the price has become very low and stable, and there is no cost advantage in integrating it into the IC, so most people do not consider the rectifier and filter part as a whole.

　　If an electrolytic capacitor provides additional bypassing or the input power supply impedance is very low, a smaller, inexpensive Y5V capacitor will also work well. General constant current devices have very fast rise and fall time pulses to absorb current from the input power supply. The input capacitor reduces the synthetic voltage ripple at the input and forces the switching current into a tight local loop, thereby minimizing EMI. The input capacitor must have low impedance at the switching frequency to do this job efficiently, and it must have a sufficient rated ripple current. Usually the ripple current will not be greater than 1/2 times the load current.

　　Ceramic capacitors are preferred due to their small size and low impedance (low equivalent series resistance or ESR). Low ESR produces very low voltage ripple, and ceramic capacitors can handle larger ripple currents than other capacitor types of the same value. X5R or X7R dielectric ceramic capacitors should be used. Electrolytic capacitors with a capacitance value greater than 1/3 of the reference value can be used instead, but factors such as size and life are not very suitable for matching with LEDs . Tantalum capacitors are prone to failure due to excessive surge current and are not recommended for use here.

　　Schottky diode selection:

　　Usually, the switch conversion type LED constant current driver IC transmits current during the MOS tube off period. The reverse withstand voltage of the selected diode should be determined according to the maximum output voltage pulse value of the line, which should be greater than this value. The forward current of the diode does not have to be equal to the switch current limit. The average current flowing through the diode is If, which is a function of the switch duty cycle, so a diode with a forward current IF=I*(1-D) should be selected. Usually, the duty cycle of the current transmitted by the diode when the power switch is turned off is usually less than 50%, so the current value can be selected to be equal to the drive current. If PWM is required to adjust the grayscale, it may also be important to consider the diode leakage from the output during the PWM low level period (there is gas on the hot spot).

　　The output diode in the boost converter flows current during the off period of the switch tube, and the diode has to withstand a reverse voltage equal to the regulator output voltage. The normal operating current is equal to the load current, and the peak current is equal to the inductor peak current.

　　Id (diode current) = Il (inductor current) = (1 + X/2) * Iout (maximum current) / 1-Dmax

　　The power dissipated by the diode is:

　　Pd=Iout(max)*Vd

　　Keep the diode lead length short and follow the correct switch node layout to avoid excessive ringing and increased power consumption. The higher the withstand voltage, the better. It should be appropriate. The Vf value of a high-voltage Schottky diode will be higher, the power consumption will be greater, and the price will be higher. The Vf value of a relatively high-voltage and high-current model will be lower, and the cost will be slightly higher. If there is no cost pressure, it can be considered.