Application Note Electrical properties of GaN LEDs & Parallel connections I. Outline For reference on current and forward voltage, please consult Nichia’s LED Product Guide 2004. In this example, Figure 1 provides the Forward Voltage of NSCW215 as 3.6V (Typ), 4.0V (Max) at IF=20mA. This indicates for a group of LEDs, the forward voltage - forward current characteristic is not a single fixed value. In a group of LEDs operating at 20mA the voltage will be an average of 3.6 volts with a maximum of 4.0 volts. But for any given LED in the group the VF may be any value between 2.8 and 4.0 volts. Typ. (V) 3.6 Max. (V) 4.0 Light Emitting Diode Figure 1 Forward Voltage of NSCW215 (IF=20mA) Therefore, when constructing a system where GaN-basis LEDs are connected in parallel, please note the following conditions. ･Even if you energize the same voltage to the same product, the current can be different. ･If the current value is different, the luminous intensity is also different. ･If the current value is different, the color can also be different. II．Parallel connection of GaN-basis LED i) Example 1 (A circuit with 2 resistors) Designing following circuit with LEDs with different forward voltage characteristics. 5V I2 I1 90 Ω V R1 90 Ω V R2 V d1 V d2 LED “A” LED “B” @2 0 m A @2 0 m A VF = 3 . 2 V VF = 3.8 V Figure 1: Circuit Diagram Information in this document is tentative and subject to change without notice for improvement. 1/4 (STS-KSE3694) April 15, 2004 Application Note Supposition Suppose as follows: Power-supply voltage: 5V LED ”A” side: 3.2V at 20mA with 90Ω LED ”B” side: 3.8V at 20mA with 90Ω The current value ofＩ１ side is calculated as follows. Postulation#1 I1 = 10mA VR1 = 10mA × 90 Ω = 0.9 V Vd1 = 5 V - 0.9 V = 4.1 V Refer to Fig.2 which shows: VF = 3.06V at 10mA If (mA) 20 Max Mini 25 Middle Light Emitting Diode Therefore,Vd1 should be 3.06V. As the result, 10mA of Postulation#1 is not correct. 15 10 5 0 0 1 2 3 4 5 6 Vf (V) Postulation#2 I1 = 15 mA Figure 2：Forward Voltage - Forward Current VR1 = 15 mA × 90 Ω = 1.35 V Vd1 = 5 V – 1.35 V = 3.65 V Refer to Fig.2 which shows: VF = 3.14V at 15mA Therefore,Vd1 should be 3.14 V As the result, Postulation#2 is not correct but the “15mA “ is more accurate than “10mA”. Postulation#3 I1 = 20 mA VR1 = 15 mA × 90 Ω = 1.8 V Vd1 = 5 V – 1.8 V = 3.2 V Postulation#3 equal to the “Supposition” The same way of calculation will result as I2=15mA. Information in this document is tentative and subject to change without notice for improvement. 2/4 (STS-KSE3694) April 15, 2004 Application Note ii) Example 2 (A circuit with 1 resistor) The circuit with one resistance that is cost-effective. 5V I1 I2 Light Emitting Diode V d1 Supposition Suppose as follows: V d2 LED”A” LED “B” @2 0 m A @2 0 m A Vf = 3.2 Vf = 3.8 V Figure 3：Circuit diagram Power-supply voltage: 5V LED“A” side: 3.2V at 20mA LED”B” side: 3.8V at 20mA The current value ofＩ１ side is calculated as follows. Postulation#1 I1 = 20 mA Vd1 = 3.2 V Vd1 = Vd2 Therefore, Vd2 = 3.2 V The current value ofＩ2 is 2mA which is derived from Figure 2 Therefore, Ｉ１:Ｉ2 =10：1 In Fig.3, Postulation#2 I1 = 20 mA Vd1 = Vd2= 3.8 V The current, I1, exceeds the scope of the graph of current vs. voltage shown in Fig.2. An LED used in this manner may also exceed Nichia's defined absolute maximum current. The luminous intensity of LED"A" compared to LED"B" will be noticeably different. * Please refer to Nichia's product specifications for reference on the maximum allowable forward current. * Please do not exceed the absolute maximum current when using Nichia LEDs. Information in this document is tentative and subject to change without notice for improvement. 3/4 (STS-KSE3694) April 15, 2004 Application Note 3. Summary When two LEDs of similar luminous intensity are driven at the same current, they will appear identical. This can be achieved using a parallel connection where both LEDs are at the same forward voltage. However, if the voltage changes the luminous intensity will change according to the forward current change of each LED. As shown in Figure 1, adjusting the voltage with separate resistors will reduce variation in forward current resulting giving similar luminous intensity values. For the example of Figure 1, both resistors were the same thus the luminous intensity would not be the same. If the resistors were matched to the VF of the LED, the same luminous intensity could be achieved. Theoretically one could consider very narrow VF binning as a solution. Practically this is not an answer. When the range of available product becomes small, the ability to ship decreases and the cost drastically increases. The best Light Emitting Diode answer in this case is one which considers VF variation as a function of the product and strives to limit its effect on the system. Information in this document is tentative and subject to change without notice for improvement. 4/4 (STS-KSE3694) April 15, 2004