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ZXLD1321 датащи(PDF) 16 Page - Diodes Incorporated |
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ZXLD1321 датащи(HTML) 16 Page - Diodes Incorporated |
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16 / 20 page ![]() ZXLD1321 Issue 1 - January 2008 16 www.zetex.com © Zetex Semiconductors plc 2008 Operation in bootstrap mode Diagram B : Bootstrap mode Operation of bootstrap LED driver This is used when the input voltage is less than 2 volts. Note that the chip VIN now goes to the cathode of the Schottky diode D1. The control loop can not operate at this low voltage, so the chip goes into a start-up mode, where the output transistor is switched on and off at nominally 50kHz with a 50:50 duty cycle with about 10mA of base current into the power transistor (20% of nominal). The emitter current is still sensed by R1 and the "ON" part of the duty cycle will be terminated either when the emitter sense voltage reaches 10mV (corresponding to 20% of the set peak current) or the ON part of the duty cycle finishes after 10µs. There is no control of the LED current yet, the circuit just operates in Boost mode. Eventually, the reservoir capacitor C3 charges up to 2V and the chip goes into "Normal" mode, where it delivers 50mA to the base of the power transistor and the control loop works normally. It will continue to charge C3 until the LED current is correctly established, with the chip now running from a voltage equal to the LED forward drop (around 3.6V for one LED) even though the supply is still below 2 volts. Once the circuit has reached this condition, the rest of the description of the operation is the same as for the Boost operation. Like the Boost circuit, this circuit has an ON phase, where the coil is connected from the supply to ground and an OFF phase, where the coil current flows through the LED via the Schottky diode D1. The current therefore only flows into the LED circuit during the OFF phase, although the reservoir capacitor C2 should keep current flowing in the LED(s) continuously. ADJ is set between 50mV and 500mV to give between 10% and 100% power respectively. Making R2 = zero gives a base current to the output transistor of 50mA nominal and making R2 = 1.68k Ω gives 10mA nominal. The reduced base current will lower supply current and hence improve efficiency in lower power applications. Making R1 = 50m Ω gives a peak coil current of 1 Amps. The internal power transistor turns on until the coil current builds up to the peak. At this point the transistor switches off and the coil current continues to flow in the LED(s) via Schottky diode D1. |
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