Design of LED driver circuit based on IR53HD420

The following figure shows a novel circuit for IR53HD420 to drive light-emitting diodes. In the figure, VDC is 320V DC. It is obtained by rectifying and filtering 220V AC. The IR53HD420 output generates 320Vp- and is connected to a series combination of a compact fluorescent tube and a current-limiting inductor Ll, and is connected to a parallel capacitor to form an LC resonance, thereby heating. Lighting the fluorescent tube. And providing the tube current. The device works very well. The voltage of the compact fluorescent tube is typically 150Vp-p. Dozens of LEDs (LEDI-LED64) are connected in series. And they are connected to a bridge rectifier (D1-D4). At least in the on state, it can imitate the effect of a compact fluorescent lamp. The values ​​of Rt and Ct given in the figure make the bridge work at 70kHz. The current provided to 64 LEDs by this circuit is about 80mA. The LED current consists of a DC plus a small ripple current. Keeping the ripple current low is conducive to obtaining high efficiency and extending the service life of the LED. In one oscillation cycle, the voltage at the input of the LED rectifier remains unchanged. Therefore, the current on the inductor L1 is a triangular wave. This is good for EMC (electromagnetic compatibility). The LED average current equation is Iledaug = (1/2VDC-NxVFled), (4xfxLl), where VDC is the power supply, N is the number of LEDs in series. Vnm is the forward voltage drop of the LED. f is the oscillation frequency. L1 is the inductance of the current limiting inductor.