DLPS052 October 2015 DLPA3000
PRODUCTION DATA.
The illumination driver of the DLPA3000 is a buck converter with two internal low-ohmic N-channel FETs (see Figure 7). The theory of operation of a buck converter is explained in Understanding Buck Power Stages in Switchmode Power Supplies (SLVA057). For proper operation, selection of the external components is very important, especially the inductor L_{OUT} and the output capacitor C_{OUT}. For best efficiency and ripple performance, an inductor and capacitor should be chosen with low equivalent series resistance (ESR).
Several factors determine the component selection of the buck converter, such as input voltage (SYSPWR), desired output voltage (V_{LED}) and the allowed output current ripple. Configuration starts with selecting the inductor L_{OUT}.
The value of the inductance of a buck power stage is selected such that the peak-to-peak ripple current flowing in the inductor stays within a certain range. Here, the target is set to have an inductor current ripple, k_{I_RIPPLE}, less than 0.3 (30%). The minimum inductor value can be calculated given the input and output voltage, output current, switching frequency of the buck converter (ƒ_{SWITCH}= 600 kHz) and inductor ripple of 0.3 (30%):
Example: V_{IN}= 12 V, V_{OUT}= 4.3 V, I_{OUT}= 6 A results in an inductor value of L_{OUT}= 2.7 µH
Once the inductor is selected, the output capacitor C_{OUT} can be determined. The value is calculated using the fact that the frequency compensation of the illumination loop has been designed for an LC-tank resonance frequency of 15 kHz:
Example: C_{OUT}= 41.7 µF given that L_{OUT}= 2.7 µH. A practical value is 2 × 22 µF. Here a parallel connection of two capacitors is chosen to lower the ESR even further.
The selected inductor and capacitor determine the output voltage ripple. The resulting output voltage ripple V_{LED_RIPPLE} is a function of the inductor ripple k_{I_RIPPLE}, output current I_{OUT}, switching frequency ƒ_{SWITCH} and the capacitor value C_{OUT}:
Example: k_{I_RIPPLE}= 0.3, I_{OUT}= 6 A, ƒ_{SWITCH}= 600 kHz and C_{OUT}= 44 µF results in an output voltage ripple of V_{LED_RIPPLE}= 8.5 mVpp
As can be seen, this is a relative small ripple.
It is strongly advised to keep the capacitance value low. The larger the capacitor value the more energy is stored. In case of a V_{LED} going down, stored energy needs to be dissipated. This might result in a large discharge current. For a V_{LED} step down from V_{1} to V_{2}, while the LED current was I_{1}. The theoretical peak reverse current is:
For the single-LED case, it is advised to keep C_{OUT} at maximum 44µF.
Two other components need to be selected in the buck converter. The value of the input-capacitor (pin ILLUM_A_VIN) should be equal to or greater than the selected output capacitance C_{OUT}, in this case >44 µF. The capacitor between ILLUM_A_SWITCH and ILLUM_A_BOOST is a charge pump capacitor to drive the high side FET. The recommended value is 100 nF.