Camera phones with over 2M (two million) pixels require a powerful flash to ensure high-quality photos in medium to low light conditions. However, the battery's capacity is limited because it cannot deliver high-current pulses to the LEDs to provide sufficient luminous output for bright, high-resolution images. Traditional LED flash drivers use current-controlled boost converters. As shown in Figure 1, consider an example driving two parallel LEDs, such as Lumileds Luxeon PWF1, each requiring 1A of current, producing approximately 20 lux (lumens) at a distance of 2 meters using highly consistent optics. The boost converter's output voltage = LED forward voltage + voltage drop across the current sensing resistor Rsense. The maximum forward voltage = 4.8V, and assuming 200mV across the current sensing resistor, then the boost converter output voltage = 5V. Assume the camera phone battery voltage = 3.3V (under load) and the boost converter efficiency is 85%. In this example, the battery current would be 5V / 3.3V / 85% x 2A = 3.6A, which exceeds the capacity of a typical telephone battery. Another solution is to use a xenon flash, but this requires: (1) a storage capacitor that is very bulky for a mobile phone and (2) a high voltage that causes telephone and security problems. Also, in the case of an LED flash, the same LED circuit with a lower current can be used for both image capture and flashlight functions. To overcome power limitations, some camera phone suppliers have used long flash exposure times to compensate for insufficient light, thus increasing the total light energy, but this results in blurry photos. Design Scheme Figure 1. Current-controlled boost converter as LED light driver Figure 2. Schematic diagram of a high-power LED supercapacitor solution Figure 3. Circuit diagram of the implementation of the CAP-XX supercapacitor solution 4. Photo taken with a typical camera phone in low light (left) and a photo taken with a phone using the CAP-XX supercapacitor solution (right). Two solutions provide sufficient LED flash power to eliminate dark, blurry photos caused by insufficient light. The CAP-XX supercapacitor, with its high capacitance (1F or more) and low ESR (<100mΩ internal resistance), can support the battery. It provides the pulse power required by the LED, and its thin cylindrical shape is suitable for space-constrained applications such as camera phones. Therefore, two solutions are proposed: Solution 1: placing the supercapacitor at the output of the boost converter; and Solution 2: connecting the supercapacitor in series with the battery. Figure 2 shows a block diagram of the CAP-XX solution, and Figure 3 shows the implemented circuit. A small, low-cost current-limiting charge pump pre-charges the supercapacitor to approximately 5.5V. Once charged, the supercapacitor activates a current switch to deliver a high-current flash pulse, the energy and power of which come from the supercapacitor, not the battery or the charge pump. During the flash pulse, the charge pump can be activated or deactivated, with the charge pump current limited to approximately 300mA. In Torch mode, the charge pump is activated, and the battery and charge pump now deliver a current lower than the charge pump's current limit. In the reference design, the CAP-XX selects the most powerful LED—the Lumileds LXCL-PWF1—which can handle a high pulse current of 1A for at least 200ms. Four PWF1s are driven here, each at 900mA. The total LED current of 3.6A is limited by a Micrel Mic 2545 current switch, chosen for its current capability and relatively small size. The circuit described below refers to Figure 3. When power is first applied, the Flash/Torch select line must be low or floating to activate U1 (SD6685 charge pump), thus turning off transistor M1 and pulling pin 5 (EN) of U1 high via R6. This depends on the size of the supercapacitor. It is necessary to wait 10-15 seconds for the supercapacitor to charge from 0V to full. When the supercapacitor is at 0V, a high inrush current will occur, followed by charging to a voltage close to Vin. This appears to short-circuit the charge pump output. Adding R11 to the circuit aims to limit the initial inrush current to at least 750mA. Note that the SP6685 is only used to charge the supercapacitor, so it is always in Torch mode (pin 4, Flash and Gnd connected). Once the supercapacitor is charged, Flash or Torch mode can be selected. When this signal is high (Flash mode), transistor M2 conducts, setting the constant current resistor for U2 (MIC2545) = R9//R10, which sets the LED flashing current. When the Enable input (pin 1 of U2) is high, U2 conducts, thus supplying current to the LED and causing it to light up. Because the supercapacitor has a very large capacitance, the flash pulse only releases a very small amount of energy from the supercapacitor, typically less than 1V. In other words, the recharge time between flash shots is short, typically about 2 seconds, which is shorter than the time required for the LED to cool down after the flash. Figure 5 illustrates the relationship between the supercapacitor voltage, battery current, and LED current during and after the flash pulse. D6 is added to the circuit to prevent the supercapacitor from discharging into the battery through U1 when U1 is disabled. The C value and ESR value of the supercapacitor are selected as follows: Total LED current (ILED) = 3.6A to maintain a 150ms flash pulse, represented as PWFLASH. • According to the Lumileds datasheet, the rated LED forward voltage at 0.9A is 3.75V, with a tolerance of 4.2V. • According to the Micrel datasheet, the Rdson resistance is <50mΩ, so the voltage drop across the MIC2545 current switch is <180mV. • Therefore, the minimum voltage across the supercapacitor at the end of the flash pulse is ≥4.2V + 0.18V = 4.38V ≥ 4.4V. • Vout (charge pump voltage) is set to 5.3V, so the allowable total voltage drop across the supercapacitor is Vd = 5.3V - 4.4V = 0.9V. • The voltage drop across the supercapacitor is Vd = ILED x (ESR + PWFLASH / C). • Rearranging the values: C ≥ ILED x PWFLASH / (Vd - ILED x ESR). In the above example, C ≥ 2A x 0.15s / (0.9V - ... Assuming the supercapacitor has an ESR of 100mΩ (2A x ESR), then C ≥ 2A x 0.15s / (0.9V - 2A x 0.1Ω) = 0.43F. The selected supercapacitor should have half of the assumed ESR value to prepare for aging over its lifespan. Therefore, the CAP-XX GS206 (0.55F, 50mΩ) meets the above requirements. Note that dual supercapacitors are used in the circuit to determine the required maximum voltage rating of 5.5V. 100mΩ is the optimal starting point for the ESR value; it may be necessary to repeatedly calculate between C and ESR to find a suitable supercapacitor. With sufficient headroom, set the charge pump output voltage to the lowest possible point. CAP-XX supercapacitors have very low leakage current, typically less than 1μA; however, when two capacitors are used in series, a balancing circuit is needed to ensure that any leakage difference between the two capacitors does not cause midpoint voltage drift and overvoltage in one of the capacitors. The simplest balancing circuit is a pair of balancing resistors, as shown in Figures 2 and 3. In terms of the implementation of the mobile phone camera flash circuit, the supercapacitor is charged to more than 5V before the flash pulse, and the appropriate value of the balancing resistor is 22kΩ. If the supercapacitor is to be maintained at the normal optimal voltage (~3.6V), the total leakage current + balancing circuit current will be about 80μA. CAP-XX has to admit that the good battery standby time is too high. Possible improvement methods include: (1) disabling the charge pump function when the phone is not in camera mode - in other words, when the phone is in standby mode, there is no supercapacitor + balancing circuit leakage. (2) using a high-impedance, low-current op-amp to create an active balancing circuit - the total current obtained from the CAP-XX reference design is less than 2μA. There are no restrictions on the charge pump part. SP6685 is chosen here because of its small size. It should be noted that most charge pump soft-start functions cannot properly handle the output of the supercapacitor because the supercapacitor discharge is like a short circuit for several seconds until the supercapacitor voltage is close to the charge pump output voltage. A simple solution is to add a current-limiting resistor (R11 in Figure 3) to the input of the charge pump. As a result , the CAP-XX scheme allows two supercapacitors, the circuit in Figure 3, and four replacement LEDs to be incorporated into existing brand-name camera phones without altering the original appearance. Figure 4 shows photos taken using the modified and CAP-XX modified phones. The unmodified circuit delivers 1W of flash power over 160ms, while the modified circuit delivers up to 15W of flash power in the same time. Figure 5 shows the battery, LED flash current, and supercapacitor voltage during and after the flash pulse. Note that the battery current never exceeds 300mA, even if the flash pulse is 4A, the supercapacitor provides a 3.7A difference. Option 2: Supercapacitor and Battery in Series Figure 6. Block Diagram of Option 2 Figure 6 is a block diagram of Option 2, where the supercapacitor is connected in series with the positive terminal of the battery. The advantages of this layout are: • Only a single supercapacitor is needed. • It requires approximately half the capacity of the supercapacitor needed in Option 1, resulting in lower cost. • Since the positive terminal voltage of the supercapacitor is often lower than the battery voltage, there is no supercapacitor inrush current. • Due to the single supercapacitor, a balancing circuit is not required. The disadvantages of this layout are: • Battery current = LED flashing current; this differs from Option 1, where battery current = only the supercapacitor charging current (and can be zero – during LED flashing). This method achieves a significantly higher LED current at a given battery current than using a current-controlled boost converter or charge pump directly driving the flashing LED in a "standard layout." Consider the case where a charge pump drives an LED with an LED current of 1A at 70% efficiency. Assume the maximum forward voltage of the LED is 4.8V and the battery voltage under load is 3.3V. Thus, without the supercapacitor, the battery current = 1A x 4.8V/3.3V/70% = 2A, which is far too much for the battery to comfortably deliver current and for the remaining circuitry to power the phone. With Scheme 2, the battery current is only 1A, representing a 100% improvement. See Figure 7 for the circuit implementation of Scheme 2. As with Scheme 1, when power is first applied, the Flash/Torch select line must be low or floating to activate U1. Depending on the size of the supercapacitor, it takes 3 to 6 seconds for the supercapacitor to fully charge from the battery voltage (0V across the supercapacitor) to the initial value of Vled (~5.1V or 1.2V across the supercapacitor). During charging, the peak battery current is limited to approximately 300mA, which decreases rapidly as the supercapacitor charges. The charge pump behaves differently in Scheme 2 than in Scheme 1 because Vout ≥ Vin (always), and the charge pump never treats the supercapacitor as a short circuit. Similar to Scheme 1, the charge pump (U1) is only used to charge the supercapacitor, so it is usually in Torch mode (pin 4 Flash connected to Gnd). Once the supercapacitor is charged, either Flash or Torch mode can be selected. When this signal is high (Flash mode), M2 is ON, which sets the current-carrying resistor of U2 (MIC2545) to = R9//R10 = 120Ω//1150Ω = 110Ω. This sets the LED current to approximately 2A. The high position in Flash/Torch also puts M1 ON, which disables the charge pump function (pulling pin 5 of U1 low). This is the approach in Scheme 2, because the battery has already supplied LED current through the supercapacitor. If the charge pump were allowed to start, it would result in the battery providing a large LED current and efficiency less than 100%. As in Scheme 1, the flash pulse discharges only a small amount from the supercapacitor, so the recharging time of the supercapacitor between flashes is very short, generally shorter than the time required for the LED to cool down (approximately 2 seconds). Figure 8 shows the supercapacitor voltage, battery current, and LED current during and after the flash pulse when the battery supplies charging current to the supercapacitor in preparation for the next flash. The process for selecting the supercapacitor's C and ESR values ​​is as follows: • To drive a Lumileds LXCL-PWF1 at 0.8A to generate a 250ms flash pulse. • Referring to the Lumileds datasheet, the forward voltage at 0.8A is 3.7V, with a tolerance of 4.1V. • Referring to the Micrel datasheet, the Rdson resistance is <50mΩ, so the voltage drop across the MIC2545 current switch is <40mV. • Therefore, the minimum voltage across the supercapacitor at the end of the flash pulse is 4.1V + 0.04V = 4.14V. • With Vout (charge pump voltage) set at 5.2V, the allowable total voltage drop across the supercapacitor is Vd = 5.2V - 4.14V = 1.06V. • Assuming the supercapacitor's ESR is 50mΩ, C ≥ 0.8A x 0.25s / (1.06V - 0.8A x 0.05Ω) = 0.196F. Selecting the supercapacitor GW109, with C = 250mF and ESR = 35mΩ, allows for a doubling of the allowable ESR value. Substituting ESR = 20mΩ and C = 250mF into the Vd expression, and checking if the voltage drop is less than 1.06V: Vd = 0.8A x (70mΩ + 0.25s / 0.25F) = 0.8 x 1.07 = 0.86, which is less than 1.06V. Therefore, the GW109 (250mF, 35) supercapacitor is selected. Figure 8. Battery, LED flash current, and supercapacitor voltage during and after the flash pulse in Scheme 2.