Electronic Design

Dual-Gate Inverter Oscillator Saves Power, Boosts LED Brightness

An LED's brightness is directly proportional to the current through it, which creates a challenge for low-voltage and battery-powered applications. But it's possible to increase an LED's brightness without increasing the system's power requirements.

The solution described here uses a high peak current to obtain a bright LED, and a low average current to minimize the power consumption. The LED oscillator circuit achieves these requirements by providing a low-duty cycle waveform with a short-duration "ON" time, and a long "OFF" time (Fig. 1).

Pulsed LEDs can be brighter than direct-drive LEDs for two reasons. First, the human eye functions as both a peak detector and an integrator. Therefore, the eye perceives a pulsed LED's brightness somewhere between the peak and the average brightness. Also, reviewing the LED's relative-efficiency versus peak-current curves shows another reason.

For example, at a peak pulsed current of 30 mA, the emerald-green Agilent Technologies HLMP LEDs are approximately 30% brighter than the equivalent dc drive circuit. Note that the pulsed circuit doesn't always produce a brighter LED, and the dc circuit will produce a brighter LED for peak currents under 10 mA.

The oscillator circuit is derived from a conventional two-inverter oscillator that has a duty cycle of approximately 50%. But the LED oscillator circuit has two RC time constants, so both the duty cycle and frequency can be ad-justed. R2 and C2 control the "ON" time of the LED pulse, while R1 and C1 control the "OFF" time. To en-sure oscillation at power-up, R4 was added in parallel with C2 to provide a dc path through the capacitor. If NAND gates are used instead of inverters, the circuit can be modified to include ON/OFF control for applications like status indicator lamps (Fig. 2). Figure 3 shows V1, the oscillator's LED drive voltage.

The equations describing the oscillation frequency and duty cycle are obtained by analyzing the discharge times of the RC networks formed at each inverter. To simplify the calculation, R3, R4, and the LED aren't included in the analysis.

These equations are developed to predict the time it takes the RC circuits to discharge to the inverter's threshold-switching voltage, assumed to be one-half the supply voltage. A function of the output current, the actual VOH decreases as the output current increases. The general equation for an RC circuit discharging to the logic switching threshold voltage (VTH) with an initial voltage (Vi) is:

The assumptions listed below result in an equation that can be solved for time (t). Assume VTH = 0.5 × VCC and Vi = VOH ≈ VCC. Then:

While the LED's "ON" time is controlled by the discharge time (t1) at inverter U1A, the discharge time (t2) at inverter U1B controls the LED's "OFF" time:

t1 ≈ 0.693 × R2C2
   ≈0.693 × (12,000 Ω)(0.01 µF)
   ≈83.2 µs

t2 ≈ 0.693 × R1C1
   ≈0.693 × (12,000 Ω)(0.1 µF)
   ≈832 µs

The time period (T) of the oscillator is equal to the sum of the charge times in the first and second RC stages:

T = t1 + t2 = 83.2 µs + 832 µs = 915 µs

f = (1/T) = (1/915 µs) = 1.09 kHz

The duty cycle (DS) for the oscillator at V1 is proportional to the ratio of the two time constants that are set by capacitors C1 and C2:

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