本設計方案的基礎是使用三個電阻和一個微處理器I/O引腳作為輸入高阻抗或輸出，獨立地驅動兩個LED工作的電路（參考文獻1）。本設計的想法聽上去很好，主要出于微處理器缺少多余I/O引腳和簡化設計的考慮。不幸的是，電路不能使于電池供電設計，因為其在正常工作下的有2mA泄漏電流，甚至在兩個LED都不工作的情況下也存在。本設計方案改進了原電路，僅使用一個I/O引腳驅動兩個LED，但存在低漏電流（圖1）。雖然電路使用了兩個二極管和一個電阻，但價格低且器件數少。
低靜態電流的LED" border="0" height="354" hspace="0" src="http://files.chinaaet.com/images/20100816/bdcdc76d-2bce-44ed-b514-92f3cd4130c1.jpg" width="419" />
兩個電路工作的基礎是二極管的非線性特征，電流隨通過其的電壓呈指數增長。為描述其工作過程，假定微處理器管腳被設置為輸入，其余管腳為高阻狀態。在第一個電路中，假設LED需要約1.5V電壓才能工作，小信號二極管電壓降約為0.6V（圖1a）。所以，為使兩個LED都工作，理論上需要4.2V。實際上，LED在約4V電流80µA時開始變暗，4.4V電流1mA時完全暗掉。對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，應該使用Schottky二極管。
第二個電路的電源電壓大于5V（圖1b）。使用圖中值，LED在7V 74µA電流下開始變暗，8.5V 1mA電流下完全暗掉，5V電源1.53µA下仍關閉。為導通LED，必須配置微處理器的I/O引腳為輸出；輸出值為1導通下面的LED，輸出值為0導通上面的LED。如果兩個LED都必須表現為工作，可以編程使引腳在0和1之間以大于50Hz的頻率循環。為計算兩種情況的電阻值，使用下面的公式：R=(3.3V–VD–VLED)/ILED（圖1a），和R=(VCC–VZ–VLED)/ILED（圖1b），在這里ILED為LED所需的電流，VD為ILED電流流過二極管產生的電壓，VZ為zener二極管電壓，VLED為ILED電流流過LED的前向電壓。應該使用Schmitt觸發器或I/O引腳的模擬輸入來避免過電流。
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.