兩個(gè)簡(jiǎn)單的電路實(shí)現(xiàn)用電池供電的微處理器驅(qū)動(dòng)兩個(gè)LED。

　　本設(shè)計(jì)方案的基礎(chǔ)是使用三個(gè)電阻和一個(gè)微處理器I/O引腳作為輸入高阻抗或輸出，獨(dú)立地驅(qū)動(dòng)兩個(gè)LED工作的電路（參考文獻(xiàn)1）。本設(shè)計(jì)的想法聽(tīng)上去很好，主要出于微處理器缺少多余I/O引腳和簡(jiǎn)化設(shè)計(jì)的考慮。不幸的是，電路不能使于電池供電設(shè)計(jì)，因?yàn)槠湓谡９ぷ飨碌挠?mA泄漏電流，甚至在兩個(gè)LED都不工作的情況下也存在。本設(shè)計(jì)方案改進(jìn)了原電路，僅使用一個(gè)I/O引腳驅(qū)動(dòng)兩個(gè)LED，但存在低漏電流（圖1）。雖然電路使用了兩個(gè)二極管和一個(gè)電阻，但價(jià)格低且器件數(shù)少。

低靜態(tài)電流的LED" border="0" height="354" hspace="0" src="http://files.chinaaet.com/images/20100816/bdcdc76d-2bce-44ed-b514-92f3cd4130c1.jpg" width="419" />

　　兩個(gè)電路工作的基礎(chǔ)是二極管的非線性特征，電流隨通過(guò)其的電壓呈指數(shù)增長(zhǎng)。為描述其工作過(guò)程，假定微處理器管腳被設(shè)置為輸入，其余管腳為高阻狀態(tài)。在第一個(gè)電路中，假設(shè)LED需要約1.5V電壓才能工作，小信號(hào)二極管電壓降約為0.6V（圖1a）。所以，為使兩個(gè)LED都工作，理論上需要4.2V。實(shí)際上，LED在約4V電流80µA時(shí)開(kāi)始變暗，4.4V電流1mA時(shí)完全暗掉。對(duì)3.3V，漏電流僅為2.41µA。電路名義上的電壓稍小于3.3V，但如果那樣的話，應(yīng)該使用Schottky二極管。

　　第二個(gè)電路的電源電壓大于5V（圖1b）。使用圖中值，LED在7V 74µA電流下開(kāi)始變暗，8.5V 1mA電流下完全暗掉，5V電源1.53µA下仍關(guān)閉。為導(dǎo)通LED，必須配置微處理器的I/O引腳為輸出；輸出值為1導(dǎo)通下面的LED，輸出值為0導(dǎo)通上面的LED。如果兩個(gè)LED都必須表現(xiàn)為工作，可以編程使引腳在0和1之間以大于50Hz的頻率循環(huán)。為計(jì)算兩種情況的電阻值，使用下面的公式：R=(3.3V–V_D–V_LED)/I_LED（圖1a），和R=(V_CC–V_Z–V_LED)/I_LED（圖1b），在這里I_LED為L(zhǎng)ED所需的電流，V_D為I_LED電流流過(guò)二極管產(chǎn)生的電壓，V_Z為zener二極管電壓，V_LED為I_LED電流流過(guò)LED的前向電壓。應(yīng)該使用Schmitt觸發(fā)器或I/O引腳的模擬輸入來(lái)避免過(guò)電流。

　　英文原文：

　　One microcontroller pin drives two LEDs with low quiescent current

　　Two simple circuits allow a battery-powered microcontroller to drive two LEDs.

　　Antonio Muñoz, Laboratorios Avanzados de Investigación del I3A, Zaragoza, Spain, and Arturo Mediano, PhD, GEPM University of Zaragoza, Zaragoza, Spain; Edited by Charles H Small and Fran Granville -- EDN, 2/7/2008

　　The basis for this Design Idea is a circuit that uses three resistors and a microcontroller-I/O pin to work as input high impedance or output to independently drive two LEDs (Reference 1). The idea sounded good for this design, mainly because of the lack of spare I/O pins in the microcontroller and the simplicity of the implementation. Unfortunately, you cannot use the circuit in battery-powered designs because it exhibits a current leakage on the order of 2 mA even with both LEDs off. This Design Idea modifies that circuit, using only one I/O pin to drive the two LEDs but with a low current drain (Figure 1). Although the circuit uses a couple of diodes and a resistor, the price and the component count are low.

　　The basis for the operation of both circuits is the nonlinear characteristic of a diode, in which current grows exponentially with the voltage applied across it. To describe the operation, suppose that the microcontroller pin is configured as an input, leaving the pin in high impedance. In the first circuit, assume that LEDs need a voltage of approximately 1.5V to turn on and that the small-signal-diode voltage drop is approximately 0.6V (Figure 1a). So, to turn on both LEDs, you theoretically need 4.

2V. In practice, the LEDs start dimming at approximately 4V with a current of 80 µA and are fully on with 4.4V at a current of 1 mA. With 3.3V, leakage current is merely 2.41 µA. The nominal voltage for this circuit can be slightly lower than 3.3V, but, in that case, you should use Schottky diodes.

　　The second circuit is for supply voltages greater than 5V (Figure 1b). Using the values in the figure, the LEDs start dimming with 7V at 74-µA current and are fully on with 8.5V at 1 mA, remaining off for a 5V supply at 1.53 µA. To turn on the LEDs, you must configure the microcontroller’s I/O pin as an output; an output value of one turns on the lower LED, and a value of zero turns on the upper LED. If both LEDs must appear to be on, your program can cycle the port pin between one and zero with a frequency greater than 50 Hz. To calculate the value of the resistor in both cases, the following formulas apply: R=(3.3V–VD–VLED)/ILED (Figure 1a), and R=(VCC–VZ–VLED)/ILED (Figure 1b), where ILED is the desired LED-on current, VD is the voltage across the diode when an ILED current flows through it, VZ is the zener-diode voltage, and VLED is the forward voltage across the LED when an ILED current flows through it. You should use a Schmitt trigger or an analog input for the I/O pin to avoid excessive current draw.

　　Reference

　　Pefhany, Spehro, “Circuit Controls Two LEDs With One Microcontroller Port Pin,” Electronic Design, April 1, 2002.