For white LEDs to maintain constant luminosity, they're typically driven with a constant dc current source. In portable applications with a single-cell lithium-ion (Li-ion) source, the sum of the voltage drop across the white LED and the current source can be lower or higher than the battery voltage. This means that whereas a single-cell Li-ion battery can directly power a red LED, a white LED sometimes requires a boosted battery voltage. The easiest way to boost the voltage is via a step-up dc-dc converter.

The advantage of this method is very high efficiency over all load and input voltage conditions, because the input voltage can be boosted to the sum of the LED forward voltage and current-source headroom voltage. This significantly optimizes efficiency at the expense of cost and pc-board area. As a matter of fact, the inductor footprint can be almost twice as big as the driver IC's footprint. Moreover, the inductor is a source of electromagnetic-interference (EMI) that can affect, in a mobile phone, performance of the display and of the radio.

Basic Principles Of Charge Pumps
An alternative method of boosting the battery voltage is to use a charge pump, also called switched-capacitor converter. Capacitors store electrical charge or energy for release at some predetermined rate and at some predetermined time. If an ideal capacitor is charged with an ideal voltage source, V G (Fig. 1a), the charge storage occurs instantaneously, corresponding to a Dirac impulse function for the current (Fig. 1b). The total stored charge is :

Q = CV_G

Real capacitors have equivalent series resistance (ESR) and equivalent series inductance (ESL), neither of which affects the capacitor's ability to store energy. They do, however, seriously impact the overall efficiency of the switched-capacitor voltage converter. Figure 1c shows an equivalent circuit for the charge of a real capacitor, where R SW is the switch resistance. The charging current path also will have a series inductance that can be reduced with proper component layout.

As soon as the circuit is energized, transient conditions of an exponential nature occur until a steady-state condition is reached. The capacitor parasitics limit the peak charge current and increase the charge transfer time (Fig. 1d). In other words, the capacitor charge buildup can't occur instantaneously, which means that the initial voltage variation across the capacitor equals zero.

Charge pumps use this property of capacitors (Fig. 2). The voltage conversion is achieved in two phases. During the first phase, switches S1, S2, and S3 are closed, whereas switches S4 to S8 are open. Therefore, C1 and C2 are stacked and, assuming C1 = C2, charged to half the input voltage:

V _C1+ - V _C1- = V _C2+ - V _C2- = V_IN /2

The output load current is supplied by the output capacitor C_HOLD . As this capacitor discharges and the output voltage falls below the desired output voltage, the second phase is activated to boost the output voltage above this value. During the second phase, C1 and C2 are in parallel, tied between V_IN and V_OUT . Switches S4 to S7 are closed, whereas switches S1 to S3 and S8 are open. Because the voltage drop across the capacitor can't change instantaneously, the output voltage jumps to one-and-a-half times the value of the input voltage:

V_C1+- V_C1- = V_C2+- V_C2- = V_OUT- V_IN = V_IN /2 => V_OUT = 3V_IN /2

This accomplishes voltage-boost operation. The duty cycle of the switching signal is usually 50%, since this value generally yields the optimal charge-transfer efficiency. Closing switch S8 and leaving switches S1 to S7 open achieves a voltage conversion with a gain of 1X.

Constant-Current LED Driver
The device in Figure 3 is based on an adaptive charge pump with gains of 1X and 1.5X. The input to the charge pump is connected to the V_IN pin, and the output is connected to the V_OUT pin. The charge pump has both open- and closed-loop modes of operation. In open-loop mode, the voltage at V_OUT is equal to the gain times the voltage at the input. When the charge pump operates in closed-loop mode, the voltage at V_OUT is regulated to a constant voltage (V _REG) . Internal current sources control current for each LED, arranged in a common-anode configuration. Peak drive current is programmed through the external resistor (R_SET ) .

Low-dropout regulators achieve current regulation (Fig. 4). The error amplifier takes the voltage across R₂ (V₂) and compares it against the reference voltage (V_REF ) . It then adjusts the LED current (I_DX ) via the series-pass element, an nMOS transistor, to the value required to drive the error signal (V_ERR = V_REF- V₂ ) as close as possible to zero. V_REF is:

V_REF = I_SET x R1

where:

I_SET = 1.25 V/R_SET

Setting V_REF = V2, the current through each LED is:

I_DX = (R1/R2) x I_SET

This holds true only if V_OUT- V_LED is sufficiently high to keep the pass element from saturating. In fact, the current sources require a minimum voltage, called headroom voltage (V_HR ) , across them to supply the desired regulated current through the LEDs. The headroom voltage is modeled with a resistor, R_HR (Fig. 5):

V_HR = R_HR x I_DX

The gain transitions are actively selected to maintain current regulation based on the LED forward voltage drop, the voltage across the current sources, and the input voltage (Fig. 3, again). As a result, the devices can stay in the most efficient gain mode of 1X over the widest range of the input voltage, reducing the power consumed from the battery.

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Reader Comments I have read through your article and was wondering if you know of a good way to step up the output voltage to 28 volts with a minimum input of 12 volts. I am looking to make a step up switching current regulator that can perform this step up and handle a max current of 700 mA. I will require 20 Watts of power for the system. If you have the time or interest please advise. Thank you Glenn Tucker gct_florida@hotmail.com Glenn Tucker -February 20, 2006 I have read through your article and was wondering if you know of a good way to step up the output voltage to 28 volts with a minimum input of 12 volts. I am looking to make a step up switching current regulator that can perform this step up and handle a max current of 700 mA. I will require 20 Watts of power for the system. If you have the time or interest please advise. Thank you Anonymous -February 20, 2006 vwery interesting Anonymous -February 16, 2006   (Article Rating: ) good article on led charg pumps Anonymous -February 01, 2006   (Article Rating: )
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