In part 1, we explored the need for a high intensity flash in medium to low light conditions to ensure good pictures from 2-megapixel (and greater) camera phones. The battery presents the system limitation, as it cannot deliver the high current pulse required for adequate LED light output for high-resolution images. And if camera phone designers lengthen flash exposure times to compensate for the lack of light, it then results in blurry photos.
Part 1 outlined a dual-cell supercapacitor-based solution using high capacitance (0.4F to 1F), low ESR (less than 100 milliohms), thin (1 to 3 mm) prismatic supercapacitors that are compact enough to fit inside a camera phone to support the battery and deliver the pulse power to drive an LED to full light intensity.
Solution 1 outlined in article 1 placed a dual-cell supercapacitor at the output of a buck-boost converter, offering maximum power to overcome dark and blurry photos by supporting a flash photo up to 3 meters.
Solution 2, outlined here, uses a single-cell supercapacitor in series with the battery, and yields an effective flash up to 1.5 meters.
Solution 2, Supercapacitor in series with the battery
Fig 1 shows the solution 2 block diagram. The supercapacitor is in series with the positive terminal of the battery.
The advantages of this configuration are:
* Only a single-cell supercapacitor is required
* This is ~half the volume of the dual-cell supercapacitor required for solution 1 and is lower cost
* Since the supercapacitor +ve terminal voltage is always ≥ than the battery voltage, there is no supercapacitor inrush current
* Since there is only a single cell, no balancing circuit is needed
The disadvantage of this configuration is:
* The battery current = LED flash current, unlike solution 1 where battery current = supercapacitor charge current only (and can = 0 during LED Flash).
This configuration enables designers to achieve much higher LED current for a given battery current than would be possible using a "standard" topology of a current controlled boost converter or charge pump to directly drive the Flash LEDs. Consider a charge pump driving an LED at 70% efficiency with the LED current = 1A. Assume the LED maximum forward voltage1 = 4.8V and the battery voltage under load = 3.3V.