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[Ideas For Design]
White LED Voltage Booster Uses 555 Timer IC

Anthony H. Smith  |   ED Online ID #8968  |   October 28, 2004


This voltage-booster circuit for driving one or more white LEDs uses a 555 timer as its main element (see the figure). The timer, IC1, functions as a resettable astable multivibrator where R1, R2, and C2 are the timing components.

When the supply voltage, VS, is first applied, D1 conducts and reservoir capacitor C1 charges to a voltage just less than VS. Initially, transistor Q2 is off, IC1's RESET input is high, and the OUTPUT pin goes high, allowing C2 to charge up via R1. During this time, the DISCHARGE pin is pulled up by R4, which turns on Q1, and the current (IL) in inductor L1 starts to ramp up. Because Q1 is saturated, both D3 and the LED are reverse-biased.

When the voltage on C2 crosses the threshold voltage at pin 6 of IC1, both the OUTPUT and DISCHARGE pins go low, and Q1 turns off. The resulting "back EMF" generated across L1 instantaneously raises the LED's anode voltage (VA) above VS, thereby illuminating the LED. Diode D3 is now forward-biased and pulls up IC1's supply voltage (V+) to a level some 2 to 4 V higher than VS.

C2 is now quickly discharged via D2 and R2, ready for the next cycle. Provided the values of R5 and R6 are chosen correctly, Q2 turns on while the LED is RESET input. When the energy stored in L1 is exhausted, the LED and D3 again become reverse-biased, and VA falls to a low level. Q2 now turns off, allowing IC1 to commence another cycle, and C2 again begins to charge via R1. The process repeats thousands of times a second, so the LED appears to be continually illuminated.

The circuit uses three "tricks" to optimize performance. First, the bootstrapping provided by D3 boosts the timer's supply voltage, enabling the circuit to continue working even when VS drops below 1 V. In addition, it provides enhanced base drive for Q1 via R4. Second, the "feedback" provided via Q2 ensures that a new cycle begins as soon as L1's energy is depleted, thereby maximizing the average LED current. Third, Q1 is driven from the timer's open-drain DISCHARGE terminal, rather than from the OUTPUT pin. Therefore, the base drive doesn't depend on the current-source capabilities of the 555's output terminal.

Transistor Q1, which should be a low-saturation type, is driven on for a time tON, given by:

tON = K×R1×C2    (s)

where K is a constant that depends on the particular type of 555 timer used. The LED's peak current is roughly equal to the maximum inductor current, IL(max), where:

IL(max)= [(VS — VCE(sat))/L1] × tON    (A)

If Q1's saturation voltage is low, say less than 50 mV, we can ignore VCE(sat) and simplify the expression to:

IL(max)= (VS/L1) × tON    (A)

Thus, for a particular value of VS, the values of R1, C2, and L1 may be selected to obtain the largest value of IL(max) that produces maximum LED brightness without exceeding its peak current rating.

Resistors R5 and R6 should be chosen to ensure that Q2 is off when VA = VS (the case when power is first applied), and on when the LED is forward-biased (VA > VS). Q2 itself should be a small-signal device with good current gain.

For efficient, low-voltage operation, a CMOS timer such as Intersil's ICM7555 or Texas Instruments' TLC555 should be used. These types are specified to function with a supply voltage as low as 2.0 V. Plus, their internal discharge transistors are able to pull pin 7 down to around 100 mV or less, ensuring that Q1 can be turned fully off.

A test circuit built using a TLC555 for IC1, and Q1 = ZTX649, Q2 = BC546, L1 = 100 µH, R5 = 56 kΩ, and R6 = 10 kΩ was found to start up with VS as low as 1.0 V. The circuit produced excellent brightness in a Lumileds (www.lumileds.com) LXHL-PW01 white LED. Transistor Q1's "on" time (tON) was around 20 µs, resulting in a peak inductor current of around 300 mA at VS = 1.5 V. However, this could easily be altered by changing the values of C2 and R1, and/or L1. Performance was equally impressive using an ICM7555, although the minimum startup voltage was slightly higher at 1.2 V.

The circuit is ideal for single-cell applications, because it continues to generate adequate LED intensity even when the supply voltage falls below 1.0 V. Also, two or more LEDs may be connected in series, although there will be a corresponding reduction in brightness.


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    Reader Comments

    ideas are much satisfied

    ganesh kumar -February 02, 2008

    nice ideas.. i am very interested while seeing these things..

    sabir khan -February 02, 2008

    well

    Saif -July 31, 2007

    ok

    USMAN -June 26, 2007

    dig S,H,I,T

    Anonymous -June 05, 2006

    dig ****

    gorge -June 05, 2006

    Could have been a great resource!

    TSIDOCK -March 05, 2006

    Just follow positive and negative! Current flows only in one direction.

    NOYB -February 06, 2006

    What they said

    Anonymous -December 22, 2005

    http://www.edn.com/contents/images/90758f1.pdf

    Take a look here.

    (from here: http://www.edn.com/article/CA90758.html?1=1)

    alz -November 14, 2005

    "Why do many of the schematics under design ideas have black components on black background? "

    use paint to open the schematics and use "invert colors"

    alz -November 14, 2005

    Clever idea with some clever tricks, but you guys really need to fix your problems with the schematics.

    Otherwise I'd have given it at least a 4, maybe 5.

    Steve Greenfield -November 10, 2005   (Article Rating: )

    Seems that nobody on the publisher's side reads the comments. Simply wasted room with black components on black background. H E L L O !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Is there anybody out there to fix this ??????????????????????????????????????????????????????????????????????????????????????????????????????????????

    Anonymous -September 07, 2005

    Seems that nobody on the publisher's side reads the comments. Simply wasted room with black components on black background. H E L L O !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Is there anybody out there to fix this ??????????????????????????????????????????????????????????????????????????????????????????????????????????????

    Anonymous -September 07, 2005

    Why do many of the schematics under design ideas have black components on black background?

    Stephen McDonald -August 17, 2005

    Circuits is not visuable?

    What is this

    Ajit Gadkari -June 09, 2005

    Great circuit idea but can't see the components in the figure. Not sure which way the diodes are oriented or exactly where they connect to.

    Anonymous -June 02, 2005   (Article Rating: )

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